Answer:
C) 21 m/s
Explanation:
The general formula of the Doppler effect is:
[tex]f'=(\frac{v+v_r}{v+v_s})f[/tex]
where
f' is the apparent frequency
f is the original frequency
v is the speed of the wave in the medium
[tex]v_r[/tex] is the velocity of the receiver, positive if the receiver is moving towards the source
[tex]v_s[/tex] is the velocity of the source, positive if the source is moving away from the receiver
Here we have
f = 440 Hz
f' = 415 Hz
v = 343 m/s
[tex]v_r = 0[/tex] (the observer is stationary)
[tex]v_s[/tex] is positive since we are considering when the train has passed the observer, so it is moving away from him
So we can rewrite the formula as
[tex]f'=(\frac{v}{v+v_s})f[/tex]
And solving for [tex]v_s[/tex], we find the speed of the train
[tex]v_s = v(\frac{f}{f'}-1)=(343 m/s)(\frac{440 Hz}{415 Hz}-1)=20.7 m/s \sim 21 m/s[/tex]
By applying Doppler's effect of a wave, the speed of this train is equal to 21 m/s.
Given the following data:
Maximum frequency = 440 Hz.
Apparent frequency = 415 Hz.
Speed of sound in air = 343 m/s.
Observer speed = 0 m/s (since his stationary).
What is Doppler effect?Doppler effect can be defined as the change in frequency of a wave with respect to an observer that is in motion and moving relative to the source of the wave.
How to calculate the speed of the train.Mathematically, Doppler's effect of a wave is given by this formula:
[tex]F_o = \frac{V+V_r}{V+V_s}F[/tex]
Substituting the given parameters into the formula, we have;
[tex]415 = \frac{343+0}{343+V_s} \times 440\\\\142345+415V_s=150920\\\\415V_s=150920-142345\\\\415V_s=8575\\\\V_s=\frac{8575}{415} \\\\V_s=20.66[/tex]
Speed = 20.66 ≈ 21 m/s.
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A thick, spherical shell made of solid metal has an inner radius a = 0.18 m and an outer radius b = 0.46 m, and is initially uncharged. A point charge q = 5.00 C is placed at the center of the shell. What is the electric field strength in the region r < a? Express your answer in terms of 1/r2. Tries 0/8 What is the electric field strength in the region a < r < b? Express your answer in terms of 1/r2. Tries 0/8 What is the electric field strength in the region b < r? Express your answer in terms of 1/r2. Tries 0/8 What is the induced charge density at r = a? (in C/m^2) Tries 0/8 What is the induced charge density (in C/m2) at r = b?
(a) [tex]E(r) = \frac{q}{4\pi \epsilon_0 r^2}[/tex]
We can solve the different part of the problem by using Gauss theorem.
Considering a Gaussian spherical surface with radius r<a (inside the shell), we can write:
[tex]E(r) \cdot 4\pi r^2 = \frac{q}{\epsilon_0}[/tex]
where q is the charge contained in the spherical surface, so
[tex]q=5.00 C[/tex]
Solving for E(r), we find the expression of the field for r<a:
[tex]E(r) = \frac{q}{4\pi \epsilon_0 r^2}[/tex]
(b) 0
The electric field strength in the region a < r < b is zero. This is due to the fact that the charge +q placed at the center of the shell induces an opposite charge -q on the inner surface of the shell (r=a), while the outer surface of the shell (r=b) will acquire a net charge of +q.
So, if we use Gauss theorem for the region a < r < b, we get
[tex]E(r) \cdot 4\pi r^2 = \frac{q'}{\epsilon_0}[/tex]
however, the charge q' contained in the Gaussian sphere of radius r is now the sum of the charge at the centre (+q) and the charge induced on the inner surface of the shell (-q), so
q' = + q - q = 0
And so we find
E(r) = 0
(c) [tex]E(r) = \frac{q}{4\pi \epsilon_0 r^2}[/tex]
We can use again Gauss theorem:
[tex]E(r) \cdot 4\pi r^2 = \frac{q'}{\epsilon_0}[/tex] (1)
where this time r > b (outside the shell), so the gaussian surface this time contained:
- the charge +q at the centre
- the inner surface, with a charge of -q
- the outer surface, with a charge of +q
So the net charge is
q' = +q -q +q = +q
And so solving (1) we find
[tex]E(r) = \frac{q}{4\pi \epsilon_0 r^2}[/tex]
which is identical to the expression of the field inside the shell.
(d) [tex]-12.3 C/m^2[/tex]
We said that at r = a, a charge of -q is induced. The induced charge density will be
[tex]\sigma_a = \frac{-q}{4\pi a^2}[/tex]
where [tex]4 \pi a^2[/tex] is the area of the inner surface of the shell. Substituting
q = 5.00 C
a = 0.18 m
We find the induced charge density:
[tex]\sigma_a = \frac{-5.00 C}{4\pi (0.18 m)^2}=-12.3 C/m^2[/tex]
(e) [tex]-1.9 C/m^2[/tex]
We said that at r = b, a charge of +q is induced. The induced charge density will be
[tex]\sigma_b = \frac{+q}{4\pi b^2}[/tex]
where [tex]4 \pi b^2[/tex] is the area of the outer surface of the shell. Substituting
q = 5.00 C
b = 0.46 m
We find the induced charge density:
[tex]\sigma_b = \frac{+5.00 C}{4\pi (0.46 m)^2}=-1.9 C/m^2[/tex]
A projectile is fired up into the air at a speed of 165 m/s at an angle of 75° relative to the horizontal. Ignore air drag. A.) Determine the MAXIMUM HEIGHT (above where it was thrown) the projectile will reach. B.) Determine how LONG the projectile will be in the air. Assume it lands at the same height it was launched. C.)Assuming it lands at the same height it was launched at, determine how FAR the projectile will land from where it was launched. D.) Assuming it lands at the same height it was launched at, determine how FAST the projectile will be going upon impact. E.) Assuming it lands at the same height it was launched at, determine the ANGLE (relative to the horizontal) of impact. °
A) 1296.3 m
The initial velocity of the projectile is
[tex]u = 165 m/s[/tex]
and the angle is
[tex]\theta=75^{\circ}[/tex]
So, we can find the initial vertical velocity of the projectile, which is given by
[tex]u_y = u sin \theta = (165 m/s)sin 75^{\circ}=159.4 m/s[/tex]
The motion of the projectile along the vertical axis is a uniformly accelerated motion, with constant acceleration
[tex]g=-9.8 m/s^2[/tex]
where the negative sign means the direction is towards the ground. The maximum height is reached when the vertical velocity becomes zero: therefore, we can use the following SUVAT equation
[tex]v_y ^2 - u_y^2 = 2gh[/tex]
where
[tex]v_y = 0[/tex] is the final vertical velocity
h is the maximum height
Solving for h, we find
[tex]h=-\frac{-u_y^2}{2g}=\frac{-(159.4 m/s)^2}{2(-9.8 m/s^2)}=1296.3 m[/tex]
B) 32.5 s
In order to determine how long the projectile will be in the air, we need to find the time t at which the projectile reaches the ground.
We can find it by analyzing the vertical motion only. The vertical position at time t is given by
[tex]y(t) = u_y t + \frac{1}{2}gt^2[/tex]
By substituting y(t) = 0, we find the time at which the projectile reaches the ground. We have:
[tex]0= u_y t + \frac{1}{2}gt^2\\0 = t(u_y + \frac{1}{2}gt)[/tex]
which has two solutions:
t = 0 --> beginning of the motion
[tex]u_y + \frac{1}{2}gt=0\\t=-\frac{2u_y}{g}=-\frac{2(159.4 m/s)}{-9.8 m/s^2}=32.5 s[/tex]
C) 1387.8 m
The range of the projectile can be found by analyzing the horizontal motion only.
In fact, the projectile travels along the horizontal direction by uniform motion, with constant horizontal velocity, given by:
[tex]u_x = u cos \theta = (165 m/s) cos 75^{\circ}=42.7 m/s[/tex]
So, the horizontal position at time t is given by
[tex]x(t) = u_x t[/tex]
If we substitute
t = 32.5 s
which is the time at which the projectile reaches the ground, we can find the total horizontal distance covered by the projectile.
So we have:
[tex]x= u_x t = (42.7 m/s)(32.5 s)=1387.8 m[/tex]
D) 165 m/s
The speed of the projectile consists of two independent components:
- The horizontal velocity, which is constant during the motion, and which is equal to
[tex]v_x = u_x = 42.7 m/s[/tex]
- The vertical velocity, which changes during the motion, given by
[tex]v_y = u_y + gt[/tex]
Substituting
[tex]u_y = 159.4 m/s[/tex]
and
[tex]t=32.5 s[/tex]
We find the vertical velocity when the projectile reaches the ground
[tex]v_y = 159.4 m/s + (-9.8 m/s^2)(32.5 s)=-159.4 m/s[/tex]
which is the same as the initial vertical velocity, but with opposite direction.
Now that we have the two components, we can calculate the actual speed of the projectile:
[tex]v=\sqrt{v_x^2+v_y^2}=\sqrt{(42.7 m/s)^2+(-159.4 m/s)^2}=165 m/s[/tex]
and the final speed is exactly equal to the initial speed, since according to the conservation of energy, the projectile has lost no energy during the motion.
E) [tex]-75^{\circ}[/tex]
The angle of impact is given by
[tex]\theta = tan^{-1} (\frac{|v_y|}{v_x})[/tex]
where
[tex]|v_y| = 159.4 m/s[/tex] is the final vertical velocity
[tex]v_x = 42.7 m/s[/tex] is the final horizontal velocity
We have taken the absolute value of [tex]v_y[/tex] since [tex]v_y[/tex] is negative; this means that the resulting angle will be BELOW the horizontal. So we have:
[tex]\theta = tan^{-1} (\frac{159.4 m/s}{42.7 m/s})=75^{\circ}[/tex]
which means [tex]-75.0^{\circ}[/tex], below the horizontal.
The Schwarzschild radius is the distance from an object at which the escape velocity is equal to the speed of light. A black hole is an object that is smaller than its Schwarzschild radius, so not even light itself can escape a black hole. The Schwarzschild radius ???? depends on the mass ???? of the black hole according to the equation ????=2????????????2 where ???? is the gravitational constant and ???? is the speed of light. Consider a black hole with a mass of 5.7×107M⊙. Use the given equation to find the Schwarzschild radius for this black hole. Schwarzschild radius: m What is this radius in units of the solar radius
1. [tex]1.69\cdot 10^{11} m[/tex]
The Schwarzschild radius of an object of mass M is given by:
[tex]r_s = \frac{2GM}{c^2}[/tex] (1)
where
G is the gravitational constant
M is the mass of the object
c is the speed of light
The black hole in the problem has a mass of
[tex]M=5.7\cdot 10^7 M_s[/tex]
where
[tex]M_s = 2.0\cdot 10^{30} kg[/tex] is the solar mass. Substituting,
[tex]M=(5.7\cdot 10^7)(2\cdot 10^{30}kg)=1.14\cdot 10^{38} kg[/tex]
and substituting into eq.(1), we find the Schwarzschild radius of this black hole:
[tex]r_s = \frac{2(6.67\cdot 10^{-11})(1.14\cdot 10^{38} kg)}{(3\cdot 10^8 m/s)^2}=1.69\cdot 10^{11} m[/tex]
2) 242.8 solar radii
We are asked to find the radius of the black hole in units of the solar radius.
The solar radius is
[tex]r_S = 6.96\cdot 10^5 km = 6.96\cdot 10^8 m[/tex]
Therefore, the Schwarzschild radius of the black hole in solar radius units is
[tex]r=\frac{1.69\cdot 10^{11} m}{6.96\cdot 10^8 m}=242.8[/tex]
An electron moving at right angles to a 0.14 T magnetic field experiences an acceleration of 6.5×1015 m/ s 2 . What is the electron's speed? Express your answer using two significant figures. By how much does its speed change in 1 ns( 10 −9 s) ?
Answer:
[tex]2.64\cdot 10^5 m/s[/tex], the speed does not change
Explanation:
The magnetic force on the electron is equal to the product between its mass and its acceleration:
[tex]qvB = ma[/tex]
where
q is the electron charge
v is the electron speed
B = 0.14 T is the magnetic field
m is the electron's mass
[tex]a=6.5\cdot 10^{15}m/s^2[/tex] is the acceleration (centripetal acceleration)
Solving for v, we find
[tex]v=\frac{ma}{qB}=\frac{(9.11\cdot 10^{-31} kg)(6.5\cdot 10^{15} m/s^2)}{(1.6\cdot 10^{-19} C)(0.14 T)}=2.64\cdot 10^5 m/s[/tex]
The speed of the electron does not change, because the acceleration is a centripetal acceleration, so it acts perpendicular to the direction of motion of the electron; therefore, no work is done on the electron by the magnetic force, and therefore the electron does not gain kinetic energy, which means that its speed does not change.
If a wheel falls from an airplane that is flying horizontally at an altitude of 500 m, how long will it take for the wheel to strike the ground?
10 s
50 s
80 s
100 s
A spring is stretched from x=0 to x=d, where x=0 is the equilibrium position of the spring. It is then compressed from x=0 to x=−d. What can be said about the energy required to stretch or compress the spring? View Available Hint(s) A spring is stretched from to , where is the equilibrium position of the spring. It is then compressed from to . What can be said about the energy required to stretch or compress the spring? More energy is required to stretch the spring than to compress it. The same amount of energy is required to either stretch or compress the spring. Less energy is required to stretch the spring than to compress it.
Answer:
The same amount of energy is required to either stretch or compress the spring.
Explanation:
The amount of energy required to stretch or compress a spring is equal to the elastic potential energy stored by the spring:
[tex]U=\frac{1}{2}k (\Delta x)^2[/tex]
where
k is the spring constant
[tex]\Delta x[/tex] is the stretch/compression of the spring
In the first case, the spring is stretched from x=0 to x=d, so
[tex]\Delta x = d-0=d[/tex]
and the amount of energy required is
[tex]U=\frac{1}{2}k d^2[/tex]
In the second case, the spring is compressed from x=0 to x=-d, so
[tex]\Delta x = -d -0 = -d[/tex]
and the amount of energy required is
[tex]U=\frac{1}{2}k (-d)^2= \frac{1}{2}kd^2[/tex]
so we see that the amount of energy required is the same.
Which of the following is a reason why microbes were found to be problematic?
They were found everywhere.
They could cause infection.
They killed specific bacteria.
They did not respond to antibiotics.
Answer:
I believe the answer is B.
Explanation:
Microbes could cause infectious diseases like for Ex: Flu and measles or heart diseases.
Hope my answer has helped you! :)
Microbes are found to be problematic for two reasons that they cause infection and sometime these organisms does not respond to antibiotics.
Explanation:
The microbes can cause infections in humans is a problem. Fungi, helminths and protozoans are the major cause of infectious diseases. Resistance to an antibiotic occurs when an ability to defeat drugs created to kill them are developed in a microbe. When a microbe becomes resistant, the antibiotics cannot fight against them and hence they multiply.
Antibiotic resistance is an urgent threat to the health of the public. They become a threat, when a microbe become resistant to antibiotic and it will be very difficult to destroy them. The resistance to antibiotics can cause serious issues like disability or even cause death.
The temperature of an object is changed when heat is added to or extracted from it. Determine the final temperature (in °C) of a 65.0 g mass of aluminum initially at 60.0°C if 1,050 J of heat energy is extracted from it. The specific heat of aluminum is 900 J/(kg · °C).
Answer:
[tex]42.1^{\circ}C[/tex]
Explanation:
The heat extracted from the aluminium is given by:
[tex]Q=mC_s (T_f-T_i)[/tex]
where we have
Q = -1050 J is the heat extracted
m = 65.0 g = 0.065 kg is the mass
Cs = 900 J/kgC is the specific heat of aluminum
[tex]T_i = 60.0^{\circ}[/tex] is the initial temperature
By solving for [tex]T_f[/tex], we find the final temperature:
[tex]T_f = \frac{Q}{m C_s}+T_i=\frac{-1050 J}{(0.065 kg)(900 J/kg C)}+60.0^{\circ}C=42.1^{\circ}C[/tex]
To find the final temperature, you use the formula for heat transfer, Q = mcΔT, and specific heat value. Given that 1050J of heat is extracted, which is negative heat transfer (-1050 J), plug in the values to get ΔT and then add this to the initial temperature to get the final temperature.
Explanation:The subject of the question is the concept of heat transfer and specific heat in Physics. The specific heat of a substance is the amount of heat required to raise one gram of the substance by one degree Celsius, which is a property intrinsic to the substance. Different substances require different amounts of heat to change their temperature.
To find the final temperature, you use the formula for heat transfer, Q = mcΔT, where Q is heat transferred, m is mass, c is the specific heat, and ΔT is the change in temperature (final temperature - initial temperature). Here, we're asked to find the final temperature of a 65 g (or 0.065 kg) mass of aluminum initially at 60.0°C when 1050 J of heat energy is extracted.
The sign of Q needs to represent that heat energy is extracted from the aluminum, so Q is -1050 J. Now, you can plug in the values and solve the equation for ΔT first, then add the initial temperature to find the final temperature.
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A rifle bullet with mass 8.00 g strikes and embeds itself in a block with mass 0.992 kg that rests on a frictionless, horizontal surface and is attached to a coil spring. The impact compresses the spring 15.0 cm. Calibration of the spring shows that a force of 0.750 N is required to compress the spring 0.250 cma) Find the magnitude of the block's velocity just after impact. b) What was the initial speed of the bullet?
Final Answer:
The magnitude of the block's velocity just after impact is 0.008 times the initial velocity of the bullet. The initial speed of the bullet is 1.25 m/s.
Explanation:
To solve this problem, we can use the principle of conservation of momentum. The mass of the bullet is 8.00 g, or 0.008 kg, and the mass of the block is 0.992 kg. Let's denote the velocity of the block just after impact as V and the initial velocity of the bullet as u. The equation for conservation of momentum is:
m_bullet * u = (m_bullet + m_block) * V
Substituting the given values, we have:
0.008 kg * u = (0.008 kg + 0.992 kg) * V
0.008 u = 1 * V
Now, let's find the magnitude of the block's velocity just after impact:
V = 0.008 u
Given that the impact compresses the spring 15.0 cm, which requires a force of 0.750 N, we can use Hooke's Law to find the spring constant:
Force = spring constant * displacement
0.750 N = k * 0.150 m
k = 5 N/m
Now, to find the initial speed of the bullet, we can use the principle of conservation of mechanical energy. The energy stored in the spring when compressed 0.250 cm is given as 5.0 J. Let's denote the initial speed of the bullet as v_i. The equation for conservation of mechanical energy is:
0.5 * k * (0.250 m)^2 = 0.5 * m_bullet * (v_i)^2
Substituting the given values, we have:
0.5 * 5 N/m * (0.0025 m^2) = 0.5 * 0.008 kg * (v_i)^2
0.00625 J = 0.004 kg * (v_i)^2
(v_i)^2 = 0.00625 J / 0.004 kg
(v_i)^2 = 1.5625 m^2/s^2
v_i = 1.25 m/s
a) The final velocity of the block is approximately 0.0000808 * v_bullet.
b) The initial speed of the bullet is approximately 3.68 m/s.
a) Final Velocity of the Block (v_final):
Given data:
Mass of the bullet (m_bullet): 8.00 g (convert to kg: 0.008 kg)
Mass of the block (m_block): 0.992 kg
Compression of the spring (x): 15.0 cm (convert to meters: 0.15 m)
Initial velocity of the bullet (v_bullet): to be determined
Using the conservation of linear momentum:
m_bullet * v_bullet = (m_bullet + m_block) * v_final
0.008 kg * v_bullet = (0.008 kg + 0.992 kg) * v_final
Solving for v_final:
v_final = (0.008 kg * v_bullet) / (1 kg + 0.992 kg)
v_final ≈ 0.0000808 * v_bullet
b) Initial Speed of the Bullet (v_bullet):
Given data:
Force required to compress the spring (F): 0.750 N
Compression of the spring (x): 0.250 cm (convert to meters: 0.0025 m)
Spring constant (k): to be determined
The work done in compressing the spring is given by W = (1/2) * k * x^2, and this work is equal to the initial kinetic energy (KE_initial) of the bullet:
KE_initial = W = (1/2) * k * x^2
The kinetic energy is also related to the initial speed of the bullet:
KE_initial = (1/2) * m_bullet * v_bullet^2
Setting the two expressions for kinetic energy equal to each other:
(1/2) * k * x^2 = (1/2) * m_bullet * v_bullet^2
Solving for v_bullet:
v_bullet = sqrt((k * x^2) / m_bullet)
Now, using the given information that F = kx, where F is the force required to compress the spring:
k = F / x
Substitute this value of k back into the equation for v_bullet:
v_bullet = sqrt((F * x) / m_bullet)
Substitute the known values:
v_bullet = sqrt((0.750 N * 0.0025 m) / 0.008 kg)
v_bullet ≈ 3.68 m/s
8. A package was determined to have a mass of 5.7 kilograms. What's the force of gravity acting on the package on earth?
The force of gravity on earth is 9.807, therefore you multiply earth’s gravitational force by the 5.7 kilograms which will give you 5.7 x 9.807=55.86 Newtons
can someone help me?!!!!!
Answer:
magnitude: 21.6; direction: 33.7 degrees
Explanation:
When we multiply a vector by a scalar, we have to multiply each component of the vector by the scalar number. In this case, we have
vector: (-3,-2)
Scalar: -6
so the vector multiplied by the scalar will have components
[tex](-3\cdot (-6), -2 \cdot (-6))=(18,12)[/tex]
The magnitude is given by Pythagorean's theorem:
[tex]m=\sqrt{18^2+12^2}=21.6[/tex]
and the direction is given by the arctan of the ratio between the y-component and the x-component:
[tex]\theta = tan^{-1} (\frac{12}{18})=33.7^{\circ}[/tex]
can someone help me?!!!!!
Answer:
56 m/s
Explanation:
The time we are considering is
t = 15 s
The vertical velocity of the projectile is given by
[tex]v_y(t) = v_{0y}-gt[/tex]
where
[tex]v_{0y}=100 m/s[/tex] is the initial vertical velocity
[tex]g=9.8 m/s^2[/tex] is the acceleration due to gravity
Substituting t=15 s, we find the vertical velocity of the projectile at that time:
[tex]v_y = 100 m/s - (9.8 m/s^2)(15 s)=-47 m/s[/tex]
where the negative sign means the direction is now downward.
The horizontal velocity does not change since there are no forces acting along that direction, so it remains constant:
[tex]v_x = 30 m/s[/tex]
So, the magnitude of the velocity at the moment of impact is
[tex]v=\sqrt{v_x^2 +v_y^2}=\sqrt{(30 m/s)^2+(-47 m/s)^2}=55.8 m/s \sim 56 m/s[/tex]
An object oscillates back and forth on the end of a spring. Which of the following statements are true at some time during the course of the motion?a. The object can have zero acceleration and, simultaneously, nonzero velocity.b. The object can have zero velocity and, simultaneously, zero acceleration.c. The object can have nonzero velocity and nonzero acceleration simultaneously.d. The object can have zero velocity and, simultaneously, nonzero acceleration.
Answer:
a. The object can have zero acceleration and, simultaneously, nonzero velocity.
c. The object can have nonzero velocity and nonzero acceleration simultaneously.
d. The object can have zero velocity and, simultaneously, nonzero acceleration.
Explanation:
For an object in simple harmonic motion, the total mechanical energy (sum of elastic potential energy and kinetic energy) is constant:
[tex]E=U+K=\frac{1}{2}kx^2 + \frac{1}{2}mv^2[/tex] (1)
where
k is the spring constant
x is the displacement
m is the mass
v is the speed
We can also notice that the force on the spring is given by Hook's law:
[tex]F=-kx[/tex]
And since according to Newton's law we have F = ma, this can be rewritten as
[tex]ma=-kx\\a=-\frac{k}{m}x[/tex]
which means that the acceleration is proportional to the displacement.
So by looking again at eq.(1), we can now states that:
- when the displacement is zero, x=0, the acceleration is zero, a=0, and the velocity is maximum
- when the velocity is zero, v=0, the acceleration is maximum, which occurs when the displacement is maximum
- in all the other intermediate situations, both velocity and acceleration are non-zero
So the correct answers are
a. The object can have zero acceleration and, simultaneously, nonzero velocity.
c. The object can have nonzero velocity and nonzero acceleration simultaneously.
d. The object can have zero velocity and, simultaneously, nonzero acceleration.
The object oscillating on a spring can have zero acceleration and nonzero velocity, zero velocity and nonzero acceleration, or nonzero velocity and nonzero acceleration at some point during the oscillation. However, it's not possible for it to have both zero velocity and zero acceleration at the same time. These properties are due to the physics of simple harmonic motion.
Explanation:Let's look at the motion of an object oscillating on a spring, a scenario that involves simple harmonic motion:
The object can have zero acceleration and, simultaneously, nonzero velocity. This happens at the equilibrium point (x = 0), where the object's speed is max and acceleration is zero. The object can have zero velocity and, simultaneously, nonzero acceleration. At the maximum displacement (amplitude points x = A or x = -A), the velocity of the object is zero, but the acceleration is nonzero because the spring force, and hence acceleration, is maximum at these points. The object can have nonzero velocity and nonzero acceleration simultaneously. This occurs between the amplitude points and equilibrium where both velocity and acceleration are nonzero. The object cannot have zero velocity and, simultaneously, zero acceleration. It's because when the object has zero velocity at the extreme points, the acceleration is max due to the restoring spring force. At the equilibrium, where the acceleration is zero, the velocity is maximum.
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A block of mass 0.254 kg is placed on top of a light, vertical spring of force constant 5 100 N/m and pushed downward so that the spring is compressed by 0.093 m. After the block is released from rest, it travels upward and then leaves the spring. To what maximum height above the point of release does it rise? (Round your answer to two decimal places.)
Answer:
8.86 m
Explanation:
According to the law of conservation of energy, the elastic potential energy initially stored in the spring will be converted into gravitational potential energy of the block when it is at its maximum height:
[tex]\frac{1}{2}kx^2 = mgh[/tex]
where
k = 5100 N/m is the spring constant
x = 0.093 m is the spring compression
m = 0.254 kg is the mass of the block
g = 9.8 m/s^2 is the acceleration due to gravity
h is the maximum height of the block
Solving the equation for h, we find
[tex]h=\frac{kx^2}{2mg}=\frac{(5100 N/m)(0.093 m)^2}{2(0.254 kg)(9.8 m/s^2)}=8.86 m[/tex]
An object, located 80.0 cm from a concave lens, forms an image 39.6 cm from the lens on the same side as the object. What is the focal length of the lens?
Answer:
-78.4 cm
Explanation:
We can solve the problem by using the lens equation:
[tex]\frac{1}{f}=\frac{1}{p}+\frac{1}{q}[/tex]
where
p is the distance of the object from the lens
q is the distance of the image from the lens
f is the focal length
Here we have
p = 80.0 cm
q = -39.6 cm (negative, because the image is on the same side as the object , so it is a virtual image)
Substituting, we find f:
[tex]\frac{1}{f}=\frac{1}{80.0 cm}+\frac{1}{-39.6 cm}=-0.0128 cm^{-1}[/tex]
[tex]f=\frac{1}{0.0128 cm^{-1}}=-78.4 cm[/tex]
can someone help me?!!!!!
Answer:
93 m
Explanation:
We need to find the resultant displacement.
Let's take north as positive y-direction and east as positive direction. We have
- Displacement 1 is 51 m to the north, so the two components are
[tex]x_1 =0\\y_1 = 51 m[/tex]
- Displacement 2 is 45 m [tex]60^{\circ}[/tex] north of east, so the two components are
[tex]x_1 = (45)cos 60^{\circ}=22.5 m\\y_1 = (45) sin 60^{\circ}=39.0 m[/tex]
So to find the resultant displacement we have to sum the components along each direction:
[tex]x=x_1 + x_2 = 0+22.5 m = 22.5 m\\y = y_1 + y_2 = 51 m +39.0 m = 90.0 m[/tex]
And the magnitude of the resultant displacement is
[tex]m=\sqrt{x^2+y^2}=\sqrt{(22.5 m)^2+(90.0 m)^2}=92.8 m \sim 93 m[/tex]
Ultraviolet (UV) radiation is a part of the electromagnetic spectrum that reaches the Earth from the Sun. It has wavelengths shorter than those of visible light, making it invisible to the naked eye. These wavelengths are classified as UVA, UVB, or UVC, with UVA the longest of the three at 317 nm to 400 nm. Both the U.S. Department of Health and Human Services and the World Health Organization have identified UV as a proven human carcinogen. Many experts believe that, especially for fair-skinned people, UV radiation frequently plays a key role in melanoma, the deadliest form of skin cancer, which kills more than 8000 Americans each year. UVB has a wavelength between 280 nm and 317 nm. Determine the frequency ranges of UVA and UVB. UVA Hz (smaller value) Hz (larger value) UVB Hz (smaller value) Hz (larger value)
Final answer:
To determine the frequency ranges for UVA and UVB, we use the equation f = c/λ. The UVA frequency range is 7.5 × 10^14 Hz to 9.375 × 10^14 Hz, and the UVB frequency range is 9.375 × 10^14 Hz to 1.0345 × 10^15 Hz.
Explanation:
The frequency of electromagnetic radiation, including UV radiation, can be calculated using the formula c = λf, where c is the speed of light (approximately 3 × 10^8 m/s), λ is the wavelength, and f is the frequency. To find the frequency ranges for UVA and UVB, we can rearrange the equation to f = c/λ.
Calculating UVA Frequency
For UVA with a wavelength range of 320-400 nm (or 3.2 × 10^-7 m - 4 × 10^-7 m), we use the formula to calculate the frequency as:
• Lower frequency limit: f = (3 × 10^8 m/s) / (4 × 10^-7 m) = 7.5 × 10^14 Hz
• Upper frequency limit: f = (3 × 10^8 m/s) / (3.2 × 10^-7 m) = 9.375 × 10^14 Hz
Calculating UVB Frequency
For UVB with a wavelength range of 290-320 nm (or 2.9 × 10^-7 m - 3.2 × 10^-7 m), the frequency range is:
• Lower frequency limit: f = (3 × 10^8 m/s) / (3.2 × 10^-7 m) = 9.375 × 10^14 Hz
• Upper frequency limit: f = (3 × 10^8 m/s) / (2.9 × 10^-7 m) = 1.0345 × 10^15 Hz
What vertical distance Δy does a free-falling particle travel from the moment it starts to the moment it reaches a speed of 7.9 m/s if it starts from rest? Work out your solution using one of the equations for vertical motion with constant acceleration, specifically, v2f=v2i+2aΔy, where vi and vf are the particle’s initial and final speeds, respectively, and a is the particle’s acceleration.
Answer:
3.2 m
Explanation:
The equation to use to solve this problem is:
[tex]v_f^2 = v_i^2 + 2 a \Delta y[/tex]
where
[tex]v_f[/tex] is the final velocity
[tex]v_i[/tex] is the initial velocity
a is the acceleration
[tex]\Delta y[/tex] is the distance covered
For the particle in free-fall in this problem, we have
[tex]v_i = 0[/tex] (it starts from rest)
[tex]v_f = 7.9 m/s[/tex]
[tex]g=9.8 m/s^2[/tex] (acceleration due to gravity)
By re-arranging the equation, we can find the distance travelled:
[tex]\Delta y = \frac{v_f^2 -v_i^2}{2a}=\frac{(7.9 m/s)^2-0^2}{2(9.8 m/s^2)}=3.2 m[/tex]
CP Global Positioning System (GPS). The GPS network consists of 24 satellites, each of which makes two orbits around the earth per day. Each satellite transmits a 50.0-W (or even less) sinusoidal electromagnetic signal at two frequencies, one of which is 1575.42 MHz. Assume that a satellite transmits half of its power at each frequency and that the waves travel uni- formly in a downward hemisphere. (a) What average intensity does a GPS receiver on the ground, directly below the satellite, receive? (Hint: First use Newton’s laws to find the altitude of the satellite.) (b) What are the amplitudes of the electric and magnetic fields at the GPS receiver in part (a), and how long does it take the signal to reach the receiver? (c) If the receiver is a square panel 1.50 cm on a side that absorbs all of the beam, what average pres- sure does the signal exert on it? (d) What wavelength must the receiver be tuned to?
(a) [tex]9.66\cdot 10^{-15} W/m^2[/tex]
First of all, we need to find the altitude of the satellite. The gravitational attraction between the Earth and the satellite is equal to the centripetal force that keeps the satellite in circular motion, so
[tex]G\frac{mM}{r^2}=m\omega^2 r[/tex] (1)
where
G is the gravitational constant
m is the satellite's mass
M is the earth's mass
r is the distance of the satellite from the Earth's centre
[tex]\omega[/tex] is the angular frequency of the satellite
The satellite here makes two orbits around the Earth per day, so its frequency is
[tex]\omega = \frac {2 \frac{rev}{day}}{24 \frac{h}{day} \cdot 60 \frac{min}{h}} \cdot \frac{2\pi rad/rev}{s/min}=1.45\cdot 10^{-4} rad/s[/tex]
And by solving eq.(1) for r, we find
[tex]r=\sqrt[3]{\frac{GM}{\omega^2}}=\sqrt[3]{\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}kg)}{(1.45\cdot 10^{-4} rad/s)^2}}=2.67\cdot 10^{7} m[/tex]
The radius of the Earth is
[tex]R=6.37\cdot 10^6 m[/tex]
So the altitude of the satellite is
[tex]h=r-R=2.67\cdot 10^7 m-6.37\cdot 10^6m=2.03\cdot 10^7 m[/tex]
The average intensity received by a GPS receiver on the Earth will be given by
[tex]I=\frac{P}{A}[/tex]
where
P = 50.0 W is the power
A is the area of a hemisphere, which is:
[tex]A=4\pi h^2 = 4 \pi (2.03\cdot 10^7 m)^2=5.18\cdot 10^{15} m^2[/tex]
So the intensity is
[tex]I=\frac{50.0 W}{5.18\cdot 10^{15}m^2}=9.66\cdot 10^{-15} W/m^2[/tex]
(b) [tex]2.70\cdot 10^{-6} V/m, 9.0\cdot 10^{-15}T[/tex], 0.068 s
The relationship between average intensity of an electromagnetic wave and amplitude of the electric field is
[tex]I=\frac{1}{2}c\epsilon_0 E^2[/tex]
where
c is the speed of light
[tex]\epsilon_0[/tex] is the vacuum permittivity
E is the amplitude of the electric field
Solving for E, we find
[tex]E=\sqrt{\frac{2I}{c\epsilon_0}}=\sqrt{\frac{2(9.66\cdot 10^{-15} W/m^2)}{(3\cdot 10^8 m/s)(8.85\cdot 10^{-12}F/m)}}=2.70\cdot 10^{-6} V/m[/tex]
Instead, the amplitude of the magnetic field is given by:
[tex]B=\frac{E}{c}=\frac{2.70\cdot 10^{-6} V/m}{3\cdot 10^8 m/s}=9.0\cdot 10^{-15}T[/tex]
The signal travels at the speed of light, so the time it takes to reach the Earth is the distance covered divided by the speed of light:
[tex]t=\frac{h}{c}=\frac{2.03\cdot 10^7 m}{3\cdot 10^8 m/s}=0.068 s[/tex]
(c) [tex]3.22\cdot 10^{-23}Pa[/tex]
In case of a perfect absorber (as in this case), the radiation pressure exerted by an electromagnetic wave on a surface is given by
[tex]p=\frac{I}{c}[/tex]
where
I is the average intensity
c is the speed of light
In this case, we have
[tex]I=9.66\cdot 10^{-15} W/m^2[/tex]
So the average pressure is
[tex]p=\frac{9.66\cdot 10^{-15} W/m^2}{3\cdot 10^8 m/s}=3.22\cdot 10^{-23}Pa[/tex]
(d) 0.190 m
The wavelength of the receiver must be tuned to the same wavelength as the transmitter (the satellite), which is given by
[tex]\lambda=\frac{c}{f}[/tex]
where
c is the speed of light
f is the frequency of the signal
For the satellite in the problem, the frequency is
[tex]f=1575.42 MHz=1575.42\cdot 10^6 Hz[/tex]
So the wavelength of the signal is:
[tex]\lambda=\frac{3.0\cdot 10^8 m/s}{1575.42 \cdot 10^6 Hz}=0.190 m[/tex]
A runner is moving at a constant speed of 8.00 m/s around a circular track. If the distance from the runner to the center of the track is 28.2 m, what is the centripetal acceleration of the runner?
1.13 m/s2
0.284 m/s2
3.53 m/s2
2.27 m/s2
Answer: Last option
2.27 m/s2
Explanation:
As the runner is running at a constant speed then the only acceleration present in the movement is the centripetal acceleration.
If we call a_c to the centripetal acceleration then, by definition
[tex]a_c =w^2r = \frac{v^2}{r}[/tex]
in this case we know the speed of the runner
[tex]v =8.00\ m/s[/tex]
The radius "r" will be the distance from the runner to the center of the track
[tex]r = 28.2\ m[/tex]
[tex]a_c = \frac{8^2}{28.2}\ m/s^2[/tex]
[tex]a_c = 2.27\ m/s^2[/tex]
The answer is the last option
A single resistor is connected across the terminals of a battery. Which one or more of the following changes in voltage and current leaves unchanged the electric power dissipated in the resistor? (A) Doubling the voltage and reducing the current by a factor of two. (B) Doubling the voltage and increasing the resistance by a factor of four. (C) Doubling the current and reducing the resistance by a factor of four.
Answer:
All of the choices are correct
Explanation:
The power dissipated in a single resistor connected to a battery is given by:
[tex]P=VI = I^2 R=\frac{V^2}{R}[/tex]
where
V is the voltage
I is the current
R is the resistance
Let's analyze each case:
A) Doubling the voltage (V'=2V) and reducing the current by a factor of 2 (I'=I/2). The new power dissipated is:
[tex]P'=V'I'=(2V)(\frac{I}{2})=VI=P[/tex] --> the power is unchanged
B) Doubling the voltage (V'=2V) and increasing the resistance by a factor of 4 (R'=4R). The new power dissipated is:
[tex]P'=\frac{V'^2}{R'}=\frac{(2V)^2}{4R}=\frac{V^2}{R}[/tex] --> the power is unchanged
C) Doubling the current (I'=2I) and reducing the resistance by a factor of four (R'=R/4). The new power dissipated is:
[tex]P'=I'^2 R'=(2I)^2(\frac{R}{4})=I^2 R[/tex] --> the power is unchanged
Final answer:
Option (B), which involves doubling the voltage and increasing the resistance by a factor of four, is the correct scenario where the electric power dissipated in the resistor remains unchanged, explained by the formulas P = V^2/R and P = I^2R.
Explanation:
The question addresses how voltage, current, and resistance affect electric power dissipation in a resistor. Electric power (P) dissipated by a resistor can be calculated using the formula P = V2/R or P = I2R, where V is the voltage across the resistor, I is the current through the resistor, and R is the resistance. To keep the power dissipated unchanged, we need to ensure that any changes in voltage, current, or resistance occur in a way that does not change the final calculation of P.
Option (A) suggests doubling the voltage and reducing the current by a factor of two. This will not keep power dissipated the same due to the direct relationship in the formula P = I2R.
Option (B), doubling the voltage and increasing the resistance by a factor of four, will keep the power dissipated in the resistor unchanged because the power can also be expressed as P = V2/R, and thus doubling V while quadrupling R does not change the P.
Option (C) involves doubling the current and reducing the resistance by a factor of four. According to P = I2R, this change will not keep the power dissipated unchanged as the increase in current will significantly raise power dissipation due to its squared relationship in the formula.
Therefore, Option (B) correctly describes a method to change voltage and resistance in such a way that the power dissipated in the resistor remains unchanged.
Which path will the car follow when it leaves the table? A B C D
Answer:
C
Explanation:
Objects in free-fall (also known as projectiles) follow a parabolic curve. So the answer is C.
The rate of change of atmospheric pressure P with respect to altitude h is proportional to P, provided that the temperature is constant. At a specific temperature the pressure is 102.1 kPa at sea level and 87.8 kPa at h = 1,000 m. (Round your answers to one decimal place.) (a) What is the pressure at an altitude of 4500 m? kPa (b) What is the pressure at the top of a mountain that is 6165 m high?
The rate of change of atmospheric pressure with respect to altitude is proportional to the current pressure. Using this information, we can calculate the pressure at different altitudes.
Explanation:To solve this problem, we can use the fact that the rate of change of atmospheric pressure with respect to altitude is proportional to the current pressure. We can set up a proportion using the given information to find the constant of proportionality. Then, we can use this constant to find the pressure at different altitudes.
(a) Let's use the given information to find the constant of proportionality. We have P = kP, where k is the constant of proportionality. Using the values at sea level and 1000m, we can set up the proportion 102.1/87.8 = k. Solving for k, we find k ≈ 1.16.
Now, we can use this constant to find the pressure at an altitude of 4500m. We set up the proportion 102.1/x = 1.16, where x is the pressure at 4500m. Solving for x, we find x ≈ 122.0 kPa.
(b) We can use the same constant of proportionality to find the pressure at the top of a mountain that is 6165m high. We set up the proportion 102.1/x = 1.16, where x is the pressure at the top of the mountain. Solving for x, we find x ≈ 89.2 kPa.
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Zero, a hypothetical planet, has a mass of 5.3 x 1023 kg, a radius of 3.3 x 106 m, and no atmosphere. A 10 kg space probe is to be launched vertically from its surface. (a) If the probe is launched with an initial kinetic energy of 5.0 x 107 J, what will be its kinetic energy when it is 4.0 x 106 m from the center of Zero? (b) If the probe is to achieve a maximum distance of 8.0 x 106 m from the center of Zero, with what initial kinetic energy must it be launched from the surface of Zero?
(a) [tex]3.1\cdot 10^7 J[/tex]
The total mechanical energy of the space probe must be constant, so we can write:
[tex]E_i = E_f\\K_i + U_i = K_f + U_f[/tex] (1)
where
[tex]K_i[/tex] is the kinetic energy at the surface, when the probe is launched
[tex]U_i[/tex] is the gravitational potential energy at the surface
[tex]K_f[/tex] is the final kinetic energy of the probe
[tex]U_i[/tex] is the final gravitational potential energy
Here we have
[tex]K_i = 5.0 \cdot 10^7 J[/tex]
at the surface, [tex]R=3.3\cdot 10^6 m[/tex] (radius of the planet), [tex]M=5.3\cdot 10^{23}kg[/tex] (mass of the planet) and m=10 kg (mass of the probe), so the initial gravitational potential energy is
[tex]U_i=-G\frac{mM}{R}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{3.3\cdot 10^6 m}=-1.07\cdot 10^8 J[/tex]
At the final point, the distance of the probe from the centre of Zero is
[tex]r=4.0\cdot 10^6 m[/tex]
so the final potential energy is
[tex]U_f=-G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{4.0\cdot 10^6 m}=-8.8\cdot 10^7 J[/tex]
So now we can use eq.(1) to find the final kinetic energy:
[tex]K_f = K_i + U_i - U_f = 5.0\cdot 10^7 J+(-1.07\cdot 10^8 J)-(-8.8\cdot 10^7 J)=3.1\cdot 10^7 J[/tex]
(b) [tex]6.3\cdot 10^7 J[/tex]
The probe reaches a maximum distance of
[tex]r=8.0\cdot 10^6 m[/tex]
which means that at that point, the kinetic energy is zero: (the probe speed has become zero):
[tex]K_f = 0[/tex]
At that point, the gravitational potential energy is
[tex]U_f=-G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{8.0\cdot 10^6 m}=-4.4\cdot 10^7 J[/tex]
So now we can use eq.(1) to find the initial kinetic energy:
[tex]K_i = K_f + U_f - U_i = 0+(-4.4\cdot 10^7 J)-(-1.07\cdot 10^8 J)=6.3\cdot 10^7 J[/tex]
A compact disc (CD) stores music in a coded pattern of tiny pits 10−7m deep. The pits are arranged in a track that spirals outward toward the rim of the disc; the inner and outer radii of this spiral are 25.0 mm and 58.0 mm, respectively. As the disc spins inside a CD player, the track is scanned at a constant linear speed of 1.25 m/s. Part A What is the angular speed of the CD when scanning the innermost part of the track? B What is the angular speed of the CD when scanning the outermost part of the track? C The maximum playing time of a CD is 74.0 min. What would be the length of the track on such a maximum-duration CD if it were stretched out in a straight line? D What is the average angular acceleration of a maximum-duration CD during its 74.0-min playing time? Take the direction of rotation of the disc to be positive.
(a) 50 rad/s
The angular speed of the CD is related to the linear speed by:
[tex]\omega=\frac{v}{r}[/tex]
where
[tex]\omega[/tex] is the angular speed
v is the linear speed
r is the distance from the centre of the CD
When scanning the innermost part of the track, we have
v = 1.25 m/s
r = 25.0 mm = 0.025 m
Therefore, the angular speed is
[tex]\omega=\frac{1.25 m/s}{0.025 m}=50 rad/s[/tex]
(b) 21.6 rad/s
As in part a, the angular speed of the CD is given by
[tex]\omega=\frac{v}{r}[/tex]
When scanning the outermost part of the track, we have
v = 1.25 m/s
r = 58.0 mm = 0.058 m
Therefore, the angular speed is
[tex]\omega=\frac{1.25 m/s}{0.058 m}=21.6 rad/s[/tex]
(c) 5550 m
The maximum playing time of the CD is
[tex]t =74.0 min \cdot 60 s/min = 4,440 s[/tex]
And we know that the linear speed of the track is
v = 1.25 m/s
If the track were stretched out in a straight line, then we would have a uniform motion, therefore the total length of the track would be:
[tex]d=vt=(1.25 m/s)(4,440 s)=5,550 m[/tex]
(d) [tex]-6.4\cdot 10^{-3} rad/s^2[/tex]
The angular acceleration of the CD is given by
[tex]\alpha = \frac{\omega_f - \omega_i}{t}[/tex]
where
[tex]\omega_f = 21.6 rad/s[/tex] is the final angular speed (when the CD is scanned at the outermost part)
[tex]\omega_i = 50.0 rad/s[/tex] is the initial angular speed (when the CD is scanned at the innermost part)
[tex]t=4440 s[/tex] is the time elapsed
Substituting into the equation, we find
[tex]\alpha=\frac{21.6 rad/s-50.0 rad/s}{4440 s}=-6.4\cdot 10^{-3} rad/s^2[/tex]
Answer: (a) 50 rad/s, (b) 21.6 rad/s, (c) 5550 m, (d) [tex]-6.4\cdot 10^{-3} rad/s^2[/tex]
Explanation:A compact disc (CD) stores music in a coded pattern of tiny pits 10−7m deep. The pits are arranged in a track that spirals outward toward the rim of the disc; the inner and outer radii of this spiral are 25.0 mm and 58.0 mm, respectively. As the disc spins inside a CD player, the track is scanned at a constant linear speed of 1.25 m/s
Part
A What is the angular speed of the CD when scanning the innermost part of the track?
B What is the angular speed of the CD when scanning the outermost part of the track?
C The maximum playing time of a CD is 74.0 min. What would be the length of the track on such a maximum-duration CD if it were stretched out in a straight line?
D What is the average angular acceleration of a maximum-duration CD during its 74.0-min playing time?
(a) 50 rad/s
[tex]\omega=\frac{v}{r}[/tex]
where
[tex]\omega[/tex] is the angular speed
v is the linear speed, v = 1.25 m/s
r is the distance from the centre of the CD, r = 25.0 mm = 0.025 m
Therefore, the angular speed
[tex]\omega=\frac{1.25 m/s}{0.025 m}=50 rad/s[/tex]
(b) 21.6 rad/s
The angular speed of the CD is
[tex]\omega=\frac{v}{r}[/tex]
When scanning the outermost part of the track
v = 1.25 m/s
r = 58.0 mm = 0.058 m
Therefore, the angular speed is
[tex]\omega=\frac{1.25 m/s}{0.058 m}=21.6 rad/s[/tex]
(c) 5550 m
[tex]t =74.0 min \cdot 60 s/min = 4,440 s[/tex]
the linear speed of the track is v = 1.25 m/s
the total length of the track would be:
d=vt=(1.25 m/s)(4,440 s)=5,550 m
(d) [tex]-6.4\cdot 10^{-3} rad/s^2[/tex]
The angular acceleration of the CD is given by[tex]\alpha = \frac{\omega_f - \omega_i}{t}[/tex]
where
[tex]\omega_f = 21.6 rad/s[/tex] is the final angular speed (when the CD is scanned at the outermost part)
[tex]\omega_i = 50.0 rad/s[/tex] is the initial angular speed (when the CD is scanned at the innermost part)
[tex]t=4440 s[/tex] is the time elapsed
Substituting into the equation, we find
[tex]\alpha=\frac{21.6 rad/s-50.0 rad/s}{4440 s}=-6.4\cdot 10^{-3} rad/s^2[/tex]
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A ball has a mass of 0.0935 kg just before it strikes the Earth after being dropped from a building 39.8 m tall. What is its de Broglie wavelength? The acceleration of gravity is 9.8 m/s 2 and Planck’s constant is 6.62607 × 10−34 J · s. Answer in units of m.
Answer:
[tex]2.54\cdot 10^{-34}m[/tex]
Explanation:
First of all, we need to find the final velocity of the ball just before it reaches the ground. Since the ball is in free-fall motion, its final velocity is given by
[tex]v^2 = u^2 +2gh[/tex]
where
v is the final velocity
u = 0 is the initial velocity
g = 9.8 m/s^2 is the acceleration due to gravity
h = 39.8 m is the height of the building
Solving for v,
[tex]v=\sqrt{2gh}=\sqrt{2(9.8 m/s^2)(39.8 m)}=27.9 m/s[/tex]
Now we can calculate the ball's momentum:
[tex]p=mv=(0.0935 kg)(27.9 m/s)=2.61 kg m/s[/tex]
And now we can calculate the De Broglie's wavelength of the ball:
[tex]\lambda = \frac{h}{p}=\frac{6.63\cdot 10^{-34} Js}{2.61 kg m/s}=2.54\cdot 10^{-34}m[/tex]
Consider an electron with charge −e and mass m orbiting in a circle around a hydrogen nucleus (a single proton) with charge +e. In the classical model, the electron orbits around the nucleus, being held in orbit by the electromagnetic interaction between itself and the protons in the nucleus, much like planets orbit around the sun, being held in orbit by their gravitational interaction. When the electron is in a circular orbit, it must meet the condition for circular motion: The magnitude of the net force toward the center, Fc, is equal to mv 2/r. Given these two pieces of information, deduce the velocity v of the electron as it orbits around the nucleus.
Final answer:
In the classical model of the hydrogen atom, the velocity of the electron as it orbits around the nucleus can be deduced using the equation for the centripetal force. The magnitude of the net force towards the center, Fc, is equal to mv^2/r, and solving for v gives the velocity of the electron.
Explanation:
In the classical model of the hydrogen atom, the electron orbits the proton in a circular path. The magnitude of the net force towards the center, also known as the centripetal force, is equal to the mass of the electron times its velocity squared divided by the radius of the orbit (Fc = mv^2 / r). In this case, we have the mass of the electron, the radius of the orbit, and we need to find the velocity of the electron as it orbits around the nucleus.
Using the given equation of the centripetal force, we can rearrange it to solve for the velocity (v). So, v = sqrt(Fc * r / m). Plugging in the values for the mass of the electron, the radius of the orbit, and the known value of the centripetal force, we can calculate the velocity.
For example, if the radius of the orbit is 5.28 x 10^-11 m and the mass of the electron is 9.11 x 10^-31 kg, the velocity of the electron would be v = sqrt(Fc * r / m) = sqrt(m * v^2 / r * r / m) = v = sqrt(v^2) = v = 2.18 x 10^6 m/s.
Nail tips exert tremendous pressures when they are hit by hammers because they exert a large force over a small area. What force (in N) must be exerted on a nail with a circular tip of 1.15 mm diameter to create a pressure of 2.63 ✕ 109 N/m2? (This high pressure is possible because the hammer striking the nail is brought to rest in such a short distance.)
Answer:
2780 N
Explanation:
Pressure is defined as the ratio between the force applied and the area of the surface:
[tex]p=\frac{F}{A}[/tex]
Here we know the pressure:
[tex]p=2.63 \cdot 10^9 N/m^2[/tex]
we also know the diameter of the tip, d = 1.15 mm, so we can calculate the radius
[tex]r=\frac{1.15 mm}{2}=0.58 mm = 5.8\cdot 10^{-4} m[/tex]
and so the area
[tex]A=\pi r^2 = \pi (5.8\cdot 10^{-4} m)^2=1.057\cdot 10^{-6} m^2[/tex]
And so we can re-arrange the equation to find the force:
[tex]F=pA=(2.63\cdot 10^9 N/m^2)(1.057\cdot 10^{-6} m^2)=2780 N[/tex]
We can calculate the necessary force to drive the nail home by first calculating the area of the nail tip using its given diameter then using the given pressure and the formula for pressure (P = F/A) to solve for the force.
Explanation:To calculate the force required to drive the nail we need to use the equation for pressure which is pressure = force/area. Pressure (P) is given as 2.63 x 109 N/m2. The area (A) can be calculated using the formula for the area of a circle which is A = πr2. The radius (r) of the nail tip can be calculated from its diameter (1.15 mm divided by 2) making sure to convert the units to meters.
After you derive the area, you will use it to calculate the force (F). Rearranging the formula for pressure to solve for force gives F = P * A. This will give the force necessary to drive the nail home exerting the stated pressure.
Learn more about Physics of Pressure here:https://brainly.com/question/33912587
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A spring of constant k hangs from a ceiling with a mass m attached at the other end. The natural length of the hanging spring without the mass is L. a) Draw this scenario, including forces and known or calculable displacements. b) What is the equilibrium length of the spring with the mass attached? c) Suppose the spring is pulled down to create a displacement y from this equilibrium position and then released. Write an expression y (t) describing its position as a function of time t. Let "up" be the positive vertical direction. d) What is the period of oscillation for this simple harmonic oscillator? e) Suppose you cut the spring in half. What would be the spring constant of each half-spring? (The two are identical, so they will have the same spring constant.) To determine this, redraw your diagram as two springs of equal length connected "in series," i.e., top to bottom. Then indicate the forces acting on each of the three interesting bodies: the mass m and each halfspring of constant k’. Apply Hooke’s Law to each half-spring individually and then to the two springs together. f) Take the two half-springs from part e) and connect them "in parallel" (that is, side by side, hanging from the ceiling and each connected to the mass m). Draw the new diagram and repeat parts a) - d)
a) See part a of the attached picture.
There are two forces involved here:
- The weight of the block attached to the spring, of magnitude
W = mg
where m is the mass of the block and g = 9.8 m/s^2 is the acceleration due to gravity, pointing downward
- The restoring force of the spring,
F = kx
with k being the spring constant and x the displacement of the spring with respect to the equilibrium position, pointing upward
b) [tex]L' = L+\frac{mg}{k}[/tex]
The spring is in equilibrium when the restoring force of the spring is equal to the weigth of the block:
[tex]kx = mg[/tex]
where
x is the displacement of the spring with respect to the natural length of the spring, L. Solving for x,
[tex]x=\frac{mg}{k}[/tex]
And since the natural length is L, the equilibrium length of the spring is
[tex]L' = L+x=L+\frac{mg}{k}[/tex]
c) [tex]y(t) = y cos(\sqrt{\frac{k}{m}} t + \pi)[/tex]
Let's assume that
y = 0
corresponds to the equilibrium position of the spring (which is stretched by an amount [tex]L+\frac{mg}{k}[/tex]). If doing so, the vertical position of the mass at time t is given by
[tex]y(t) = y cos(\omega t + \pi)[/tex]
where
[tex]\omega = \sqrt{\frac{k}{m}}[/tex] is the angular frequency
t is the time
y is the initial displacement with respect to the equilibrium position
[tex]\pi[/tex] is the phase shift, so that the position at time t=0 is negative:
y(0) = -y
So rewriting the angular frequency:
[tex]y(t) = y cos(\sqrt{\frac{k}{m}} t + \pi)[/tex]
d) [tex]T=2 \pi \sqrt{\frac{m}{k}}[/tex]
The period of oscillation is given by
[tex]T=\frac{2\pi}{\omega}[/tex]
where
[tex]\omega = \sqrt{\frac{k}{m}}[/tex] is the angular frequency
Substituting [tex]\omega[/tex], we find an expression for the period
[tex]T=2 \pi \sqrt{\frac{m}{k}}[/tex]
e) k' = 2k (see part e of attached picture)
Here the two half springs of spring constant k' are connected in series, so the sum of the stretchings of the two springs is equal to the total stretching of the spring, x:
[tex]x=x_1 + x_2[/tex]
Also, since the two springs are identical, their stretching will be the same:
[tex]x_1 = x_2 = x'[/tex]
so we have
[tex]x=x'+x'=2x'[/tex]
[tex]x'=\frac{x}{2}[/tex]
Substituting x find in part (b),
[tex]x'=\frac{mg}{2k}[/tex] (1)
Hooke's law for each spring can be written as
[tex]F=k' x'[/tex] (2)
where
F = mg (3)
is still the weight of the block
Using (1), (2) and (3) together, we find an expression for the spring constant k' of each spring
[tex]mg=k'\frac{mg}{2k}\\k' =2k[/tex]
So, the spring constant of each half-spring is twice the spring constant of the original spring.
fa) See part f) of attached picture
This time we have the block hanging from both the two half-springs, each with spring constant k'. So at equilibrium, the weight of the block is equal to the sum of the restoring forces of the two springs:
[tex]k' x_1 + k' x_2 = mg[/tex]
fb) [tex]x'=\frac{L}{2}+\frac{mg}{2k'}[/tex]
For the two springs in parallel, the sum of the restoring forces of the two springs must be equal to the weight of the block:
[tex]F_1 + F_2 = mg\\k_1 x_1 + k_2 x_2 = mg[/tex]
The two springs are identical, so they have same spring constant:
[tex]k_1 = k_2 = k'[/tex]
So (1) can be rewritten as
[tex]k' (x_1 + x_2) = mg[/tex]
And since the two springs are identical, their stretching x' is the same:
[tex]x_1 = x_2 = x'[/tex]
so we can rewrite this as
[tex]k' (2x') = mg[/tex]
[tex]x'=\frac{mg}{2k'}[/tex]
and so the equilibrium length of each spring will be
[tex]x'=\frac{L}{2}+\frac{mg}{2k'}[/tex]
fc) [tex]y(t) = y cos(\sqrt{\frac{2k'}{m}} t + \pi)[/tex]
The system of two springs in parallel can be treated as a system of a single spring with equivalent spring constant given by
[tex]k_{eq}=k_1 + k_2 = 2k'[/tex]
where k' is the spring constant of each spring.
So, let's assume again that
y = 0
corresponds to the equilibrium position as calculated in the previous part. If doing so, the vertical position of the mass at time t is given by:
[tex]y(t) = y cos(\omega t + \pi)[/tex]
where this time we have
[tex]\omega = \sqrt{\frac{k_{eq}}{m}}[/tex] is the angular frequency of the system
t is the time
y is the initial displacement with respect to the equilibrium position
[tex]\pi[/tex] is the phase shift, which we put so that the position at time t=0 is negative:
y(0) = -y
If we rewrite the angular frequency,
[tex]\omega = \sqrt{\frac{2k'}{m}}[/tex]
the position of the mass is
[tex]y(t) = y cos(\sqrt{\frac{2k'}{m}} t + \pi)[/tex]
fd) [tex]T=2 \pi \sqrt{\frac{m}{2k'}}[/tex]
Similarly to part d), the period of oscillation is
[tex]T=\frac{2\pi}{\omega}[/tex]
where
[tex]\omega = \sqrt{\frac{k_{eq}}{m}}=\sqrt{\frac{2k'}{m}}[/tex] is the angular frequency
Substituting [tex]\omega[/tex], we find
[tex]T=2 \pi \sqrt{\frac{m}{2k'}}[/tex]
Spider-Man and Ned were testing the distance he could shoot his web depending on the angle at which he points his web shooter. He tested this 3 times at each angle.
0°: 18.2m, 16.7m, and 17.9m
45°: 79.6m, 74.3m, and 76.2m
75°: 62.4m, 61.5m, and 62.7m
What is the Independent variable?
Question 1 options:
The distance spider-man’s web shoots
The angle at which spider-man points his web shooter
Testing each angle 3 times
The location where they are testing
What is the dependent variable?
Question 2 options:
The distance spider-man’s web shoots
The angle at which spider-man points his web shooter
Testing each angle 3 times
The location where they are testing
Look at the 1st graph below, which location has the warmest temperatures in July?
Question 3 options: See Picture below
Barrow, Ak
Christchurch, NZ
Quito, Equador
Madison, WI
According to the 1st graph below, which location does not experience seasonal temperature fluctuations?
Question 4 options:
Barrow, AK
Quito, Equador
Christchurch, NZ
Madison, WI
Use the 2nd chart below to answer the following question. If cool stars are red and hot stars are blue, where would a bright (highly luminous) blue star be found on a Hertzsprung-Russell diagram?
Question 5 options:
Upper right
Lower right
Upper left
Lower left
A group of scientists has developed a computer program that will predict the phase that a star will be after a certain number of years. The scientists used data from real stars to teach the computer how the universe works. Now, it can be used to quickly run a simulation of what would likely happen over billions of years in the universe.
A group of students is wondering the rate at which a star progresses through life. They plan two experiments using computer simulations. They ran each simulation from the protostar phase 100 times and recorded the outcomes.
In Experiment 1 they used one size of star (1 solar mass) and compared seven different lengths of time. In Experiment 2 they used one length of time and 3 different sizes of stars (1, 15, and 30 solar masses). The results of both experiments are below in Figure 2
According to the 3rd data chart in which is the best conclusion of the changes occurring from 5-10 billion years?
Question 6 options:
Main sequence stars are becoming red giants.
Protostars are becoming red giants.
The stars are becoming black holes.
The stars are exploding.
Answer:
3. Madison, WI
4. Quito, Equador
6. Main Sequence stars are becoming red giants
Explanation:
Answer:
1) The angle at which spider-man points his web shooter
2) The distance spider-man’s web shoots
3) Madison, WI
4) Quito, Equador
5) Upper left
6) Main sequence stars are becoming red giants.
Explanation:
1) The experimenters are choosing the angle at which to point the web shooter. This means the angle at which they shoot is independent of the other variables.
2) The distance the web shoots is not chosen by the experimenters and depends on the angle at which it is shot.
3) July is the 7th month and the graph for Madison, WI shows temperatures exceeding all other locations at this time.
4) Fluctuations refer to a changing of temperatures over seasons. Since Quito, Equador has a flat graph, this indicates no fluctuations.
5) This graph plots the hottest (blue) stars on the left and the highest luminosity (bright) stars at the top. Thus, bright blue stars are plotted in the top left.
6) At 5 billion years there are 0 giants and 100 main sequence stars. At 10 billions years there are 94 giants and 6 main sequence stars. This shows that the main sequence stars are becoming giants.