Answer:
2.45 N/m
Explanation:
We can find the spring constant by using Hooke's law:
[tex]F=kx[/tex]
where
F is the force applied to the spring
k is the spring constant
x is the stretching of the spring
Here, the force applied is the weight of the mass hanged on the spring. The mass is
m = 10 g = 0.010 kg
So the weight is
[tex]F=mg=(0.010 kg)(9.8 m/s^2)=0.098 N[/tex]
while the stretching is
x = 4 cm = 0.04 m
So the spring constant is
[tex]k=\frac{F}{x}=\frac{0.098 N}{0.04 m}=2.45 N/m[/tex]
A proton moves with a speed of 4.00 106 m/s horizontally, at a right angle to a magnetic field. What magnetic field strength is required to just balance the weight of the proton and keep it moving horizontally? (The mass and charge of the proton are 1.67 ✕ 10−27 kg and 1.60 ✕ 10−19 C, respectively.)
Answer:
Magnetic field, [tex]B=2.55\times 10^{-14}\ T[/tex]
Explanation:
It is given that,
Speed of proton, [tex]v=4\times 10^6\ m/s[/tex]
Mass of the proton, [tex]m=1.67\times 10^{-27}\ kg[/tex]
Charge on proton, [tex]q=1.6\times 10^{-19}\ C[/tex]
We need to find the magnetic field strength required to just balance the weight of the proton and keep it moving horizontally.
The Lorentz force is given by :
[tex]F=q(v\times B)=qvB\ sin90[/tex].............(1)
The weight of proton,
[tex]W=mg[/tex]..............(2)
From equation (1) and (2), we get :
[tex]mg=qvB[/tex]
[tex]B=\dfrac{mg}{qv}[/tex]
[tex]B=\dfrac{1.67\times 10^{-27}\ kg\times 9.8\ m/s^2}{1.6\times 10^{-19}\ C\times 4\times 10^6\ m/s}[/tex]
[tex]B=2.55\times 10^{-14}\ T[/tex]
Hence, this is the required solution.
Final answer:
The magnetic field strength required to balance the weight of a proton and keep it moving horizontally is 3.07 x 10^-4 Tesla, calculated by setting the magnetic force (qvB) equal to the gravitational force (mg) and solving for B with given values for q, v, m, and g.
Explanation:
To calculate the magnetic field strength required to balance the weight of a proton and keep it moving horizontally, we can use the relationship between the magnetic force and the gravitational force acting on the proton. The magnetic force that will balance the weight of the proton is given by FB = qvB sin(θ), where q is the charge of the proton, v is its velocity, B is the magnetic field strength, and θ is the angle between the velocity and the magnetic field. Because the proton is moving at a right angle to the magnetic field, sin(θ) will be 1.
The weight of the proton (gravitational force) is given by W = mg, where m is the mass of the proton and g is the acceleration due to gravity. Setting the magnetic force equal to the weight gives us:
FB = W
qvB = mg
B = mg/qv
Substituting the given values for mass m = 1.67 x 10-27 kg, charge q = 1.60 x 10-19 C, speed v = 4.00 x 106 m/s, and the acceleration due to gravity g = 9.81 m/s2, we can find the magnetic field strength:
B = (1.67 x 10-27 kg * 9.81 m/s2) / (1.60 x 10-19 C * 4.00 x 106 m/s)
B = 3.07 x 10-4 T
The magnetic field strength required is 3.07 x 10-4 Tesla.
A hypothetical atom has three energy levels: the ground-state level and levels 1.50 eV and 5.00 eV above the ground state. What is the longest wavelength in the line spectrum for this atom? Let Planck's constant h = 4.136 x 10^-15 ev s, and the speed of light c = 3.00 x 10^8 m/s. (a)355 nm (b)780 nm (c)882 nm (d) 827 nm
Answer:
option (d)
Explanation:
E1 = 1.5 eV = 1.5 x 1.6 x 10^-19 J, E2 = 5 eV = 5 x 1.6 x 10^-19 J, c = 3 x 10^8 m/s, h = 6.62 x 10^-34 Js
Wavelength associated with 1.5 eV is λ1.
E1 = h c / λ1
λ1 = h c / E1
λ1 = (6.62 x 10^-34 x 3 x 10^8) / (1.5 x 1.6 x 10^-19)
λ1 = 8.275 x 10^-7 m = 827 nm
Wavelength associated with 5 eV is λ2.
E2 = h c / λ2
λ2 = h c / E2
λ2 = (6.62 x 10^-34 x 3 x 10^8) / (5 x 1.6 x 10^-19)
λ2 = 2.4825 x 10^-7 m = 248 nm
So, the longest wavelength is 827 nm
A team of astronauts is on a mission to land on and explore a large asteroid. In addition to collecting samples and performing experiments, one of their tasks is to demonstrate the concept of the escape speed by throwing rocks straight up at various initial speeds. With what minimum initial speed Vesc will the rocks need to be thrown in order for them never to "fall" back to the asteroid? Assume that the asteroid is approximately spherical, with an average density p 3.84 x108 g/m3 and volume V 2.17 x 1012 m3 Recall that the universal gravitational constant is G 6.67 x 10-11 N m2/kg2
Explanation:
The escape velocity [tex]V_{esc}[/tex] is given by the following equation:
[tex]V_{esc}=\sqrt{\frac{2GM}{R}}[/tex] (1)
Where:
[tex]G[/tex] is the Gravitational Constant and its value is [tex]6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}[/tex]
[tex]M[/tex] is the mass of the asteroid
[tex]R[/tex] is the radius of the asteroid
On the other hand, we know the density of the asteroid is [tex]\rho=3.84(10)^{8}g/m^{3}[/tex] and its volume is [tex]V=2.17(10)^{12}m^{3}[/tex].
The density of a body is given by:
[tex]\rho=\frac{M}{V}[/tex] (2)
Finding [tex]M[/tex]:
[tex]M=\rhoV=(3.84(10)^{8} g/m^{3})(2.17(10)^{12}m^{3})[/tex] (3)
[tex]M=8.33(10)^{20}g=8.33(10)^{17}kg[/tex] (4) This is the mass of the spherical asteroid
In addition, we know the volume of a sphere is given by the following formula:
[tex]V=\frac{4}{3}\piR^{3}[/tex] (5)
Finding [tex]R[/tex]:
[tex]R=\sqrt[3]{\frac{3V}{4\pi}}[/tex] (6)
[tex]R=\sqrt[3]{\frac{3(2.17(10)^{12}m^{3})}{4\pi}}[/tex] (7)
[tex]R=8031.38m[/tex] (8) This is the radius of the asteroid
Now we have all the necessary elements to calculate the escape velocity from (1):
[tex]V_{esc}=\sqrt{\frac{2(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(8.33(10)^{17}kg)}{8031.38m}}[/tex] (9)
Finally:
[tex]V_{esc}=117.626m/s[/tex] This is the minimum initial speed the rocks need to be thrown in order for them never return back to the asteroid.
To calculate the minimum speed to escape the gravitational pull of the asteroid, you'll first determine the asteroid's mass using its density and volume. Find its radius assuming it's a sphere. Plug these into the escape speed equation sqrt(2*G*M/R) to determine the velocity.
Explanation:The escape velocity of an object from another object’s gravitational pull can be determined by the formula: V_esc = sqrt(2*G*M/R). Here, V_esc is the escape speed, G is the universal gravitational constant, M is the mass of the object (in this case, the asteroid), and R is the radius of the object. To find M in this case, we can multiply the given density (p) of the asteroid by its volume (V), and convert this to kg. Once we know the mass, we can find the radius of the asteroid (assuming it is spherical), using the formula for the volume of a sphere (V = 4/3 * pi * R^3).
Once we have determined all the values above, we can substitute them into the escape speed formula to find the minimum initial speed a rock would need to be thrown in order never to fall back to the asteroid.
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The tension in the string of a simple pendulum is:
a. constant
b. Maximum in the extreme position
c. Zero in the mean position
d. None of the above
Answer:
D. None of the above
Explanation:
There are only two forces acting on a pendulum:
- The force of gravity (downward)
- The tension in the string
We can consider the axis along the direction of the string: here we have the tension T, acting towards the pivot, and the component of the weight along this direction, acting away from the pivot. Their resultant must be equal to the centripetal force, so we can write:
[tex]T-mg cos \theta = m\frac{v^2}{r}\\T=m\frac{v^2}{r}+mg cos \theta[/tex]
where
T is the tension in the string
[tex]\theta[/tex] is the angle between the tension and the vertical
m is the mass
g is the acceleration of gravity
v is the speed of the pendulum
r is the length of the string
From the formula we see that the value of the tension, T, depends only on the value of v (the speed) and [tex]\theta[/tex], the angle. We notice that:
- Since [tex]\theta[/tex] and v constantly change, T must change as well
- At [tex]\theta=0^{\circ}[/tex] (equilibrium position), [tex]cos \theta=1[/tex] (maximum value), and also the speed v is maximum, so the tension has the maximum value at the equilibrium position
- For [tex]\theta[/tex] increasing, the [tex]cos \theta[/tex] decreases and the speed v decreases as well, so the tension T decreases: this means that the value of the tension will be minimum in the extreme positions.
So the correct answer is D. None of the above
A long, straight wire with 2 A current flowing through it produces magnetic field strength 1 T at its surface. If the wire has a radius R, where within the wire is the field strength equal to 84 % of the field strength at the surface of the wire? Assume that the current density is uniform throughout the wire. (μ 0 = 4π × 10-7 T · m/A)
Answer:
[tex]r = 3.36 \times 10^{-7} m[/tex]
Explanation:
As per Ampere's law of magnetic field we know that
line integral of magnetic field along closed ampere's loop is equal to the product of current enclosed and magnetic permeability of medium
So it is given as
[tex]\int B. dl = \mu_0 i_{en}[/tex]
here we can say that enclosed current is given as
[tex]i_{en} = \frac{i}{\pi R^2} (\pi r^2)[/tex]
now from ampere'e loop law for any point inside the wire we will have
[tex]B.(2\pi r) = \mu_o (\frac{ir^2}{R^2}[/tex]
[tex]B = \frac{\mu_0 i r}{2\pi R^2}[/tex]
now we know that magnetic field inside the wire is 84% of the field at its surface
so we will have
[tex]0.84 \frac{\mu_o i}{2\pi R} = \frac{\mu_o i r}{2\pi R^2}[/tex]
so we have
[tex]r = 0.84 R[/tex]
now we know
[tex]\frac{\mu_o i}{2\pi R} = 1[/tex]
here i = 2 A
[tex]R = 2\times 10^{-7} m[/tex]
so now we have
[tex]r = 3.36 \times 10^{-7} m[/tex]
The point ( r ) within the wire where the field strength equals 84% of the field strength at the wire surface is : 0.84 R
Given data :
Radius of wire = R
current in the wire = 2A
magnetic field strength = 1 T
Determine the area within the wire where the field strength equals 84%we will apply Ampere's law
i) Ampere's law applied inside the wire
B₁ (2πr ) = μ₀I ( r² / R² )
ii) Ampere's law applied at the surface
B₂ ( 2πr ) = μ₀ I
Resolving equations above
Therefore : B₁ / B₂ = 0.84 also r / R = 0.84
Hence ( r ) = 0.84 R
Therefore we can conclude that The point ( r ) within the wire where the field strength equals 84% of the field strength at the wire surface is : 0.84 R
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A sailboat is traveling to the right when a gust of wind causes the boat to accelerate leftward at 2.5m/s2 for 4s. After the wind stops, the sailboat is traveling to the left with a velocity of 3.0m/s.Assuming the acceleration from the wind is constant, what was the initial velocity of the sailboat before the gust of wind?Answer using a coordinate system where rightward is positive.
Answer:
+7.0 m/s
Explanation:
Let's take rightward as positive direction.
So in this problem we have:
a = -2.5 m/s^2 acceleration due to the wind (negative because it is leftward)
t = 4 s time interval
v = -3.0 m/s is the final velocity (negative because it is leftward)
We can use the following equation:
v = u + at
Where u is the initial velocity
We want to find u, so if we rearrange the equation we find:
[tex]u = v - at = (-3.0 m/s) - (-2.5 m/s^2)(4 s)=+7.0 m/s[/tex]
and the positive sign means the initial direction was rightward.
An Australian emu is running due north in a straight line at a speed of 13.0 m/s and slows down to a speed of 10.5 m/s in 4.20 s. (a) What is the magnitude and direction of the bird’s acceleration? (b) Assuming that the acceleration remains the same, what is the bird’s velocity after an additional 1.50 s has elapsed?
Explanation:
It is given that,
Initial velocity of the bird, u = 13 m/s
Final speed of the bird, v = 10.5 m/s
Time taken, t = 4.20 s
(a) Acceleration of the bird is given by :
[tex]a=\dfrac{v-u}{t}[/tex]
[tex]a=\dfrac{10.5\ m/s-13\ m/s}{4.20\ s}[/tex]
[tex]a=-0.59\ m/s^2[/tex]
So, The direction of acceleration is opposite to the direction of motion.
(b) We need to find the bird’s velocity after an additional 1.50 s has elapsed i.e. t = 4.2 + 1.5 = 5.7 s. Let v' is the new final velocity.
It can be calculated using first equation of motion as :
[tex]a=\dfrac{v'-u}{t}[/tex]
v' = u + at
[tex]v'=(-0.59)\times 5.7+13[/tex]
v' = 9.64 m/s
Hence, this is the required solution.
Point charge A is located at point A and point charge B is at point B. Points A and B are separated by a distance r. To determine the electric potential at the mid-point along a line between points A and B, which of the following mathematical approaches is correct? a. The difference in the absolute value (the sign of the charges does not enter into the calculation) of the two electric potentials is determined at a distance r/2 from each of the charges. b. The algebraic sum of the two electric potentials is determined at a distance r/2 from each of the charges, making sure to include the signs of the charges. c. The electric potential due to each charge is determined at a distance r/2 from each of the charges and an average is taken of the two values. d. The vector sum of the two electric potentials determines the total electric potential at a distance r/2 from each of the charges
Answer:
B. The algebraic sum of the two electric potentials is determined at a distance r/2 from each of the charges, making sure to include the signs of the charges.
Explanation:
Total electric potential is the sum of all the electric potential. And because electric potential is a scalar quantity you have to account for the signs.
A 64.0-kg ice skater is moving at 4.04 m/s when she grabs the loose end of a rope, the opposite end of which is tied to a pole. She then moves in a circle of radius 0.890 m around the pole. (a) Determine the magnitude of the force exerted by the horizontal rope on her arms. kN (b) Compare this force with her weight. Frope W =
We have that for the Question it can be said that the magnitude of the force exerted by the horizontal rope on her arms and the ratio of the Force to the weight is
F=1150.561NF/W=1.8325
From the question we are told
A 64.0-kg ice skater is moving at 4.04 m/s when she grabs the loose end of a rope, the opposite end of which is tied to a pole. She then moves in a circle of radius 0.890 m around the pole. (a) Determine the magnitude of the force exerted by the horizontal rope on her arms. kN (b) Compare this force with her weight. F-rope W =
Generally the equation for the force applied is mathematically given as
[tex]F=\frac{( mv^2)}{R}\\\\Therefore\\\\F=\frac{( mv^2)}{R}\\\\F=\frac{( (64)(4.0)^2)}{0.890}\\\\[/tex]
F=1150.561N
b)
Generally the equation for the Weight is mathematically given as
W=mg
Therefore
W=64*9.81
W=627.84N
Therefore
The Force to weight ratio is
[tex]F/W=1150.561N/627.84N[/tex]
F/W=1.8325
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The force exerted by the rope on the skater's arms as she moves in a circular path is 1.167 kN. This force is about 1.860 times her weight, which is 627.2 N.
Explanation:The skater is experiencing centripetal force exerted by the rope, which causes her to move in a circular path. The magnitude F of this force can be calculated using the formula F = mv²/r, where m is the skater's mass (64.0 kg), v is her velocity (4.04 m/s), and r is the radius of her circular path (0.890 m).
By substituting the given numbers into this formula, we get: F = (64.0 kg)(4.04 m/s)² / 0.890 m = 1166.67 N. In kilonewtons, this force is 1.167 kN.
To compare this force with her weight, we can calculate the weight (W) using the formula W = mg, where g is the acceleration due to gravity (around 9.8 m/s²). Substituting the given mass into this formula gives us: W = (64.0 kg)(9.8 m/s²) = 627.2 N.
Comparing these two forces shows that the force exerted by the rope on her arms is about 1.860 times her weight.
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A particle's position is given by x = 7.00 - 9.00t + 3t2, in which x is in meters and t is in seconds. (a) What is its velocity at t = 1 s? (b) Is it moving in the positive or negative direction of x just then? (c) What is its speed just then? (d) Is the speed increasing or decreasing just then? (Try answering the next two questions without further calculation.) (e) Is there ever an instant when the velocity is zero? If so, give the time t; if not, answer "0". (f) Is there a time after t = 3 s when the particle is moving in the negative direction of x? If so, give the time t; if not, answer "0".
[tex]x(t)=7.00\,\mathrm m-\left(9.00\dfrac{\rm m}{\rm s}\right)t+\left(3\dfrac{\rm m}{\mathrm s^2}\right)t^2[/tex]
a. The particle has velocity at time [tex]t[/tex],
[tex]\dfrac{\mathrm dx(t)}{\mathrm dt}=-9.00\dfrac{\rm m}{\rm s}+\left(6\dfrac{\rm m}{\mathrm s^2}\right)t[/tex]
so that after [tex]t=1\,\mathrm s[/tex] it will have velocity [tex]\boxed{-3.00\dfrac{\rm m}{\rm s}}[/tex].
b. The sign of the velocity is negative, so it's moving in the negative [tex]x[/tex] direction.
c. Its speed is 3.00 m/s.
d. The particle's velocity changes according to
[tex]\dfrac{\mathrm d^2x(t)}{\mathrm dt^2}=6\dfrac{\rm m}{\mathrm s^2}[/tex]
which is positive and indicates the velocity/speed of the particle is increasing.
e. Yes. The velocity is increasing at a constant rate. Solving for [tex]\dfrac{\mathrm dx(t)}{\mathrm dt}=0[/tex] is trivial; this happens when [tex]\boxed{t=1.50\,\mathrm s}[/tex].
f. No, the velocity is positive for all [tex]t[/tex] beyond 1.50 s.
Which of the following is an example of the conclusion phase of the scientific method?
a scientist decides on a question to explore
a scientist collects data
a scientists creates graphs and performs calculations
a scientist examines the results and answers the lab question
Answer:
a scientist examines the results and answers the lab question- last choice
Answer:
D. a scientist examines the results and answers the lab question
An isotope of Uranium, Z = 92 and A = 235, decays by emitting an alpha particle. Calculate the number of neutrons in the nucleus left behind after the radioactive decay.
Answer:
141
Explanation:
The atomic number (Z) corresponds to the number of protons:
Z = p
while the mass number (A) corresponds to the number of protons+neutrons:
A = p + n
So the number of neutrons in a nucleus is equal to the difference between mass number and atomic number:
n = A - Z
For the initial nucleus of Uranium, Z = 92 and A = 235, so the initial number of neutrons is
n = 235 - 92 = 143
An alpha particle carries 2 protons and 2 neutrons: so, when the isotope of Uranium emits an alpha particle, it loses 2 neutrons. Therefore, the number of neutrons after the decay will be
n = 143 - 2 = 141
Four point charges of magnitude 3.6 mu or micro CC are at the corners of a square of side 4 m. (a) Find the electrostatic potential energy if all of the charges are negative.
The electrostatic potential energy if all of the charges are negative is
U = -1.026 x [tex]10^{-5}[/tex] J.
We have,
The electrostatic potential energy U of a system of point charges can be calculated using the formula:
U = 1/4π∈ [tex]\sum_{i = 1}^n \sum_{j > i}^n q_iq_j/r_{ij}[/tex]
Now,
n is the number of charges.
q are the magnitudes of charges.
r is the distance between the charges.
∈ is the vacuum permittivity
∈ = 8.85 x [tex]10^{-12}[/tex] C² / N - m²
Now,
Given that all charges are negative.
[tex]q_i[/tex] = - 3.6 mu
And they are at the corners of a square with a side of 4m, the distances between charges are all diagonals of the square is [tex]r_{ij} = 4\sqrt{2}m[/tex]
Substituting the values.
U = 1/4π∈ [tex]\sum_{i = 1}^n \sum_{j > i}^n q_iq_j/r_{ij}[/tex]
U = -1.026 x [tex]10^{-5}[/tex] J
Thus,
The electrostatic potential energy if all of the charges are negative is
U = -1.026 x [tex]10^{-5}[/tex] J.
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A ball is thrown upward in the air, and its height above the ground after t seconds is H ( t ) = 57 t − 16 t 2 feet. Find the time t when the ball will be traveling upward at 14.25 feet per second.
Answer:
1.34 seconds
Explanation:
h(t) = 57t - 16t²
Velocity is the derivative of position with respect to time:
v(t) = dh/dt
v(t) = 57 - 32t
When v = 14.25:
14.25 = 57 - 32t
32t = 42.75
t = 1.34
Two point charges of +2.0 μC and -6.0 μC are located on the x-axis at x = -1.0 cm and x = +2.0 cm respectively. Where should a third charge of +3.0-μC be placed on the +x-axis so that the potential at the origin is equal to zero? (k = 1/4πε0 = 8.99 × 109 N · m2/C2)
Answer:
see attachment
Explanation:
The electric potential varies inversely with the distance. So, the third charge should be placed at a distance of 3 cm from the origin on the x-axis.
What is electric potential?The work done on an electric charge to shift it from infinity to a point is known as electric potential at that point. And its expression is,
[tex]V = \dfrac{kq}{r}[/tex]
here, k is the coulomb's constant.
Given data:
The magnitude of two point charges are, [tex]+2.0 \;\rm \mu C[/tex] and [tex]-6.0 \;\rm \mu C[/tex].
The location of each charge on the x-axis is -1.0 cm and +2.0 cm.
Let the third charge ( [tex]+3.0 \;\rm \mu C[/tex] ) be placed at a distance of x. Then the electric potential at origin is,
[tex]V = \dfrac{k \times 2.0}{0.01} +\dfrac{k \times -6.0}{0.02} +\dfrac{k \times 3.0}{x}[/tex]
Since, potential at origin is zero (V = 0). Then,
[tex]0 = \dfrac{k \times 2.0}{0.01} +\dfrac{k \times -6.0}{0.02} +\dfrac{k \times 3.0}{x}\\\\\dfrac{k \times 6.0}{0.02} = \dfrac{k \times 2.0}{0.01} +\dfrac{k \times 3.0}{x}\\\\\dfrac{6.0}{0.02} = \dfrac{2.0}{0.01} +\dfrac{3.0}{x}\\\\x = 0.03 \;\rm m =3 \;\rm cm[/tex]
Thus, we can conclude that the third charge should be placed at a distance of 3 cm from the origin on the x-axis.
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From Center Station, a train departs every 30 minutes on the Fast Line and a train departs every 50 minutes on the State Line. If two trains depart from Center Station at 8:00 A.M., one on each of the two lines, what is the next time that two trains, one on each line, will depart at the same time?
Explanation:
You need to find the least common multiple (LCM) of 30 and 50. First, write the prime factorization of each:
30 = 2×3×5
50 = 2×5²
The LCM must contain all the factors of both, so:
LCM = 2×3×5²
LCM = 150
It will take 150 minutes (or 2 hours and 30 minutes) before two trains depart at the same time again.
The next time that two trains, one on each line, will depart from Center Station at the same time is at 10:30 A.M. This is calculated by finding the least common multiple of the two train schedules.
Explanation:This question requires finding the least common multiple (LCM) of the two train schedules, which represents the time duration when both trains will depart again at the same time. The LCM of 30 and 50 is 150 minutes. This means, from 8:00 A.M., it will be the next 150 minutes or 2 hours and 30 minutes when both trains will depart from Center Station at the same time. Therefore, the answer is 10:30 A.M.
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If two metal balls each have a charge of -10^-6 C and the repulsive force between them is 1 N, how far apart are they? (Coulomb's constant is k = 9.0x 10^9 N-m^2/C^2? 9.5 m 9.0 mm 9.5 cm 0.9 m .
Answer:
The distance between the charges is 9.5 cm
(c) is correct option
Explanation:
Given that,
Charge [tex]q= 10^{-6}\ C[/tex]
Force F = 1 N
We need to calculate the distance between the charges
Using Coulomb's formula
[tex]F = \dfrac{kq_{1}q_{2}}{r^2}[/tex]
Where, q = charge
r = distance
F = force
Put the value into the formula
[tex]1=\dfrac{9.0\times10^{9}\times(-10^{-6})^2}{r^2}[/tex]
[tex]r=\sqrt{9\times10^{9}\times(-10^{-6})^2}[/tex]
[tex]r=0.095\ m[/tex]
[tex]r= 9.5\ cm[/tex]
Hence, The distance between the charges is 9.5 cm
A 75 kg skier rides a 2830-m-long lift to the top of a mountain. The lift makes an angle of 14.6°with the horizontal. What is the change in the skier's gravitational potential energy?
Answer:
Gravitational potential energy = 524.85 kJ
Explanation:
Refer the figure.
Gravitational potential energy = mgh
Mass, m = 75 kg
Acceleration due to gravity, g = 9.81 m/s²
We have
[tex]sin14.6=\frac{h}{2830}\\\\h=713.36m[/tex]
Gravitational potential energy = 75 x 9.81 x 713.36 = 524854.62 J = 524.85 kJ
An experimenter measures the frequency, f, of an electromagnetic wave Its wavelength in free space is a) c/f b) cf c) f/c d) independent of the frequency.
Answer:
Wavelength, [tex]\lambda=\dfrac{c}{f}[/tex]
Explanation:
In an electromagnetic wave both electric and magnetic field propagate simultaneously. Radio waves, microwaves, gamma rays etc are some of the examples of electromagnetic waves.
If f is the frequency of electromagnetic wave, c is the speed of light, then the relationship between the frequency f, wavelength and the speed is given by :
[tex]\lambda=\dfrac{c}{f}[/tex]
Hence, the correct option that shows the wavelength of electromagnetic wave in free space is(a) " c/f ".
A bike with 15cm diameter wheels accelerates uniformly from rest to a speed of 7.1m/s over a distance of 35.4m. Determine the angular acceleration of the bike's wheels.
Answer:
9.47 rad/s^2
Explanation:
Diameter = 15 cm, radius, r = diameter / 2 = 7.5 cm = 0.075 m, u = 0, v = 7.1 m/s,
s = 35.4 m
let a be the linear acceleration.
Use III equation of motion.
v^2 = u^2 + 2 a s
7.1 x 7.1 = 0 + 2 x a x 35.4
a = 0.71 m/s^2
Now the relation between linear acceleration and angular acceleration is
a = r x α
where, α is angular acceleration
α = 0.71 / 0.075 = 9.47 rad/s^2
In 1610 Galileo made a telescope and used it to study the planet Jupiter. He discovered four moons. One of them was Ganymede. The mean radius of the orbit of Ganymede around Jupiter is 10.7 × 108m and the period of the orbit is 7.16 days. i) Determine the mass of Jupiter.
Answer: [tex]1.893(10)^{27}kg [/tex]
Explanation:
This problem can be solved by the Third Kepler’s Law of Planetary motion, which states:
“The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.
In other words, this law stablishes a relation between the orbital period [tex]T[/tex] of a body (moon, planet, satellite) orbiting a greater body in space with the size [tex]a[/tex] of its orbit.
This Law is originally expressed as follows:
[tex]T^{2}=\frac{4\pi^{2}}{GM}a^{3}[/tex] (1)
Where;
[tex]T=7.16days=618624s[/tex] is the period of the orbit Ganymede describes around Jupiter
[tex]G[/tex] is the Gravitational Constant and its value is [tex]6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}[/tex]
[tex]M[/tex] is the mass of Jupiter (the value we need to find)
[tex]a=10.7(10)^{8}m[/tex] is the semimajor axis of the orbit Ganymede describes around Jupiter (assuming it is a circular orbit, the semimajor axis is equal to the radius of the orbit)
If we want to find [tex]M[/tex], we have to express equation (1) as written below and substitute all the values:
[tex]M=\frac{4\pi^{2}}{GT^{2}}a^{3}[/tex] (2)
[tex]M=\frac{4\pi^{2}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(618624s)^{2}}(10.7(10)^{8}m)^{3}[/tex] (3)
Finally:
[tex]M=1.8934(10)^{27}kg[/tex] This is the mass of Jupiter
A wheelhas a radius of 4.8 m. Howfar (path length) does a point on the circumference travel if thewheel is rotated through angles of 30°, 30 rad, and 30 rev,respectively?
Answer:
(a) 2.512 m
(b) 144 m
(c) 904.32 m
Explanation:
radius, r = 4.8 m
(a) for 30 degree
As we know that in 360 degree it rotates a complete round that means circumference.
In 360 degree, it rotates = 2 x π x r
in 30 degree, it rotates = 2 x π x r x 30 / 360
= 2 x 3.14 x 4.8 x 30 / 360
= 2.512 m
(b) for 30 rad
As we know that in one complete rotation, it rotates by 2π radian.
so,
for 2π radian it rotates = 2 x π x r
for 30 radian, it rotates = 2 x π x r x 30 / 2 π = 144 m
(c) For 30 rev
In one complete revolution, it travels = 2 x π x r
in 30 rev, it travels = 2 x π x r x 30 = 2 x 3.14 x 4.8 x 30 = 904.32 m
Astronomers discover an exoplanet (a planet of a star other than the Sun) that has an orbital period of 3.87 Earth years in its circular orbit around its sun, which is a star with a measured mass of 3.59 x 1030 kg. Find the radius of the exoplanet's orbit.
Answer: [tex]4.487(10)^{11}m[/tex]
Explanation:
This problem can be solved using the Third Kepler’s Law of Planetary motion:
“The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.
This law states a relation between the orbital period [tex]T[/tex] of a body (the exoplanet in this case) orbiting a greater body in space (the star in this case) with the size [tex]a[/tex] of its orbit:
[tex]T^{2}=\frac{4\pi^{2}}{GM}a^{3}[/tex] (1)
Where:
[tex]T=3.87Earth-years=122044320s[/tex] is the period of the orbit of the exoplanet (considering [tex]1Earth-year=365days[/tex])
[tex]G[/tex] is the Gravitational Constant and its value is [tex]6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}[/tex]
[tex]M=3.59(10)^{30}kg[/tex] is the mass of the star
[tex]a[/tex] is orbital radius of the orbit the exoplanet describes around its star.
Now, if we want to find the radius, we have to rewrite (1) as:
[tex]a=\sqrt[3]{\frac{T^{2}GM}{4\pi^{2}}}[/tex] (2)
[tex]a=\sqrt[3]{\frac{(122044320s)^{2}(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(3.59(10)^{30}kg)}{4\pi^{2}}}[/tex] (3)
Finally:
[tex]a=4.487(10)^{11}m[/tex] This is the radius of the exoplanet's orbit
Given the orbital period and mass of the exoplanet, its radius of orbit is 4.488 × 10¹¹m.
Given the data in the question;
Orbital period;[tex]T = 3.87 \ Earth\ years = [ 3.87yrs*365days*24hrs*60min*60sec = 122044320s[/tex] Mass of the Planet; [tex]M = 3.59*10^{30}kg[/tex]Radius of the exoplanet's orbit; [tex]r= \ ?[/tex]To determine the radius of the exoplanet's orbit, we use the equation from Kepler's Third Law:
[tex]T^2 = \frac{4\pi^2 }{GM}r^3\\[/tex]
Where, T is the period of the orbit of the exoplanet, G is the Gravitational Constant, M is the mass of the star and r is orbital radius.
We make "r", the subject of the formula
[tex]r = \sqrt[3]{\frac{T^2GM}{4\pi ^2} }[/tex]
We substitute our given values into the equation
[tex]r = \sqrt[3]{\frac{(122044320s)^2*(6.67430 * 10^{-11} m^3/kg s^2)*(3.59*10^{30}kg)}{4*\pi ^2} } \\\\r = \sqrt[3]{\frac{(1.48948*10^{16}s^2)*(6.67430 * 10^{-11} m^3/kg s^2)*(3.59*10^{30}kg)}{4*\pi ^2} }\\\\r = \sqrt[3]{\frac{3.5689*10^{36}m^3}{4*\pi ^2} }\\\\r = \sqrt[3]{9.04*10^{34}m^3}\\\\r = 4.488*10^{11}m[/tex]
Therefore, given the orbital period and mass of the exoplanet, its radius of orbit is 4.488 × 10¹¹m.
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44. A rescue helicopter is hovering over a person whose boat has sunk. One of the rescuers throws a life preserver straight down to the victim with an initial velocity of 1.40 m/s and observes that it takes 1.8 s to reach the water. (a) List the knowns in this problem. (b) How high above the water was the preserver released? Note that the downdraft of the helicopter reduces the effects of air resistance on the falling life preserver, so that an acceleration equal to that of gravity is reasonable.
Answer:
18.4 m
Explanation:
(a)
The known variables in this problem are:
u = 1.40 m/s is the initial vertical velocity (we take downward direction as positive direction)
t = 1.8 s is the duration of the fall
a = g = 9.8 m/s^2 is the acceleration due to gravity
(b)
The vertical distance covered by the life preserver is given by
[tex]d=ut + \frac{1}{2}at^2[/tex]
If we substitute all the values listed in part (a), we find
[tex]d=(1.40 m/s)(1.8 s)+\frac{1}{2}(9.8 m/s^2)(1.8 s)^2=18.4m[/tex]
Using the equation of motion, the calculation shows the life preserver was released from approximately 18.4 meters above the water.
Initial velocity of the life preserver (vo) = 1.40 m/s (downward)Time the life preserver takes to reach the water (t) = 1.8 sAcceleration due to gravity (a) ≈ 9.8 m/s² (downward)y = vot + 1/2(at²)
Substituting the known values, we get:
y = (1.40 m/s)(1.8 s) + 1/2(9.8 m/s²)(1.8 s)²
y = 2.52 m + 15.876 m
y = 18.396 m
The life preserver was released approximately 18.4 meters above the water.
Calculate the deflection at point C of a beanm subjected to uniformly distributed load w 275 N/mm on span AB and point load P-10 kN at C. Assume that L = 5 m and EI = 1.50 × 107 N·m2
To calculate the deflection at point C of the beam, we can use the deflection equation for beams under a uniformly distributed load and a point load.
Explanation:To calculate the deflection at point C of the beam, we will use the formula for deflection under a uniformly distributed load and a point load. The deflection equation for beams is given by
δ = (5wL⁴ - PL³) / (384EI)
where δ is the deflection, w is the uniformly distributed load, L is the span length, P is the point load, E is the modulus of elasticity, and I is the moment of inertia of the beam. Substituting the given values into the equation, we can calculate the deflection at point C.
A dielectric-filled parallel-plate capacitor has plate area A = 30.0 cm2 , plate separation d = 9.00 mm and dielectric constant k = 3.00. The capacitor is connected to a battery that creates a constant voltage V = 15.0 V . Throughout the problem, use ϵ0 = 8.85x10-12 C2/N.m2 . Find the energy U1 of the dielectric-filled capacitor.
Answer:
[tex]9.96\cdot 10^{-10}J[/tex]
Explanation:
The capacitance of the parallel-plate capacitor is given by
[tex]C=\epsilon_0 k \frac{A}{d}[/tex]
where
ϵ0 = 8.85x10-12 C2/N.m2 is the vacuum permittivity
k = 3.00 is the dielectric constant
[tex]A=30.0 cm^2 = 30.0\cdot 10^{-4}m^2[/tex] is the area of the plates
d = 9.00 mm = 0.009 m is the separation between the plates
Substituting,
[tex]C=(8.85\cdot 10^{-12}F/m)(3.00 ) \frac{30.0\cdot 10^{-4} m^2}{0.009 m}=8.85\cdot 10^{-12} F[/tex]
Now we can calculate the energy of the capacitor, given by:
[tex]U=\frac{1}{2}CV^2[/tex]
where
C is the capacitance
V = 15.0 V is the potential difference
Substituting,
[tex]U=\frac{1}{2}(8.85\cdot 10^{-12}F)(15.0 V)^2=9.96\cdot 10^{-10}J[/tex]
A disc initially at rest experiences an angular acceleration of 3.11 rad/s for a time of 15.0 s. What will the angular speed of the disc be at this time, in units of rad/s?
Answer:
The angular speed of the disc be at this time is 46.65 rad/s.
Explanation:
Given that,
Angular acceleration [tex]\alpha= 3.11\ rad/s^2[/tex]
Time t =15.0 s
We will calculate the angular speed of the disc
A disc initially at rest.
So, [tex]\omega=0[/tex]
Using rotational kinematics equation
[tex]\omega'=\omega+\alpha\ t[/tex]
Where, [tex]\omega[/tex] = initial angular speed
[tex]\omega'[/tex] =final angular speed
[tex]\alpha[/tex] = angular acceleration
Put the value in the equation
[tex]\omega'=0+3.11\times15[/tex]
[tex]\omega'=46.65\ rad/s[/tex]
Hence, The angular speed of the disc be at this time is 46.65 rad/s.
A rifle with a weight of 35 N fires a 4.5-g bullet with a speed of 240 m/s. (a) Find the recoil speed of the rifle in m/s. (b) If a 650-N man holds the rifle firmly against his shoulder, find the recoil speed of the man and rifle in m/s.
(a) 0.30 m/s
The total momentum of the rifle+bullet system before the shot is zero:
[tex]p_i = 0[/tex]
The total momentum of the system after the shot is the sum of the momenta of the rifle and of the bullet:
[tex]p_f = m_r v_r + m_b v_b[/tex]
where we have
[tex]m_r = \frac{W}{g}=\frac{35 N}{9.8 m/s^2}=3.57 kg[/tex] is the mass of the rifle
[tex]v_r[/tex] is the final velocity of the rifle
[tex]m_b = 4.5 g = 0.0045 kg[/tex] is the mass of the bullet
[tex]v_b = 240 m/s[/tex] is the final velocity of the bullet
Since the total momentum must be conserved, we have
[tex]p_i = p_f[/tex]
So
[tex]m_r v_r + m_b v_b=0[/tex]
and so we can find the recoil velocity of the rifle:
[tex]v_r = - \frac{m_b v_b}{m_r}=-\frac{(0.0045 kg)(240 m/s)}{3.57 kg}=-0.30 m/s[/tex]
And the negative sign means it travels in the opposite direction to the bullet: so, the recoil speed is 0.30 m/s.
(b) 0.016 m/s
The mass of the man is equal to its weight divided by the acceleration of gravity:
[tex]m=\frac{W}{g}=\frac{650 N}{9.8 m/s^2}=66.3 kg[/tex]
This time, we have to consider the system (man+rifle) - bullet. Again, the total momentum of the system before the shot is zero:
[tex]p_i = 0[/tex]
while the total momentum after the shot is
[tex]p_f = m_r v_r + m_b v_b[/tex]
where this time we have
[tex]m_r = 66.3 kg+3.57 kg=69.9 kg[/tex] is the mass of the rifle+person
[tex]v_r[/tex] is the final velocity of the man+rifle
[tex]m_b = 4.5 g = 0.0045 kg[/tex] is the mass of the bullet
[tex]v_b = 240 m/s[/tex] is the final velocity of the bullet
Since the total momentum must be conserved, we have
[tex]m_r v_r + m_b v_b=0[/tex]
and so we can find the recoil velocity of the man+rifle:
[tex]v_r = - \frac{m_b v_b}{m_r}=-\frac{(0.0045 kg)(240 m/s)}{66.9 kg}=-0.016 m/s[/tex]
So the recoil speed is 0.016 m/s.
The recoil speed of the rifle is 0.0031 m/s when held loosely away from the shoulder. When a 650 N man holds the rifle firmly against his shoulder, the effective mass of the rifle-man system is 28.0 kg, resulting in a recoil speed of 0 m/s.
Explanation:To calculate the recoil speed of the rifle in m/s, we use the principle of conservation of momentum. The momentum of the rifle before firing is equal to the momentum of the bullet after firing. The momentum of an object is calculated by multiplying its mass by its velocity. Given that the mass of the bullet is 4.5 g (0.0045 kg) and the velocity is 240 m/s, we can find the momentum of the bullet. Then, using the principle of conservation of momentum, we can calculate the recoil speed of the rifle.
(a) The momentum of the bullet is calculated as:
Momentum = mass x velocity = 0.0045 kg x 240 m/s = 0.108 kg·m/s
Since the momentum of the bullet before firing is equal to the momentum of the rifle after firing, we can write:
0.108 kg·m/s = mass of the rifle x recoil speed of the rifle
Rearranging the equation, we can solve for the recoil speed of the rifle:
Recoil speed of the rifle = 0.108 kg·m/s ÷ mass of the rifle = 0.108 kg·m/s ÷ 35 N = 0.0030857 m/s
(b) When a 650 N man holds the rifle firmly against his shoulder, the effective mass of the rifle-man system is 28.0 kg. To find the recoil speed of the man and rifle together, we can again use the principle of conservation of momentum. The initial momentum of the rifle-man system is zero, as they are at rest. Therefore, the final momentum of the system after firing must also be zero. We can write:
0 = (mass of the rifle + mass of the man) x recoil speed of the system
Rearranging the equation, we can solve for the recoil speed of the system:
Recoil speed of the system = 0 ÷ (mass of the rifle + mass of the man) = 0 ÷ (28 kg + 650 N ÷ 9.8 m/s²) = 0 m/s
A barrel ride at an amusement park starts from rest and speeds up to 0.520 rev/sec in 7.26 s. What is the angular acceleration during that time? (Unit = rad/s^2)
The angular acceleration of the barrel ride is 0.0714 rad/s^2.
Explanation:To find the angular acceleration, we first need to convert the rotational speed from rev/s to radians per second (rad/s) because the standard unit for angular speed and acceleration is in rad/s and rad/s² respectively. We know that 1 revolution is equal to 2π radians, therefore 0.520 rev/s equals 0.520 x 2π rad/s.
Angular acceleration (α) is calculated using the formula α = Δω / Δt, where Δω is the change in angular velocity and Δt is the time it takes for the change. As the ride started from rest, the change in angular velocity was simply its final angular velocity. Thus, by putting the values in the formula, we will get the angular acceleration during the 7.26 s.
The angular acceleration of the barrel ride at the amusement park can be calculated using the formula:
α = (Δω) / t
Plugging in the values from the question, we have:
Δω = 0.520 rev/sec
t = 7.26 s
Therefore, the angular acceleration during that time is α = (0.520 rev/sec) / (7.26 s) = 0.0714 rad/s^2.
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The angular acceleration of the barrel ride at the amusement park, which sped up from rest to 0.520 rev/sec in 7.26 s, is 0.45 rad/s².
Explanation:To answer this question the first step we should do is to convert the given rate of 0.520 revolutions per second to radians per second, as our target unit is rad/s². We know that 1 revolution is equal to 2π radians, so multiplying the revolution rate by 2π, we get: 0.520 rev/sec * 2π rad/rev = 3.27 rad/sec.
Next, we need to calculate the angular acceleration using the formula α = (ωf- ωi) / t. Where ωi is the initial angular velocity, ωf is the final angular velocity, and t is the time. The ride starts from rest, so the initial angular velocity is 0, the final angular velocity is 3.27 rad/sec and the time is 7.26 seconds.
Substituting these values into the equation, we get: α = (3.27 rad/sec - 0 rad/sec) / 7.26 s = 0.45 rad/s².
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The equation for free fall at the surface of a celestial body in outer space (s in meters, t in seconds) is sequals10.04tsquared. How long does it take a rock falling from rest to reach a velocity of 28.6 StartFraction m Over sec EndFraction on this celestial body in outer space?
Answer:
1.42 s
Explanation:
The equation for free fall of an object starting from rest is generally written as
[tex]s=\frac{1}{2}at^2[/tex]
where
s is the vertical distance covered
a is the acceleration due to gravity
t is the time
On this celestial body, the equation is
[tex]s=10.04 t^2[/tex]
this means that
[tex]\frac{1}{2}g = 10.04[/tex]
so the acceleration of gravity on the body is
[tex]g=2\cdot 10.04 = 20.08 m/s^2[/tex]
The velocity of an object in free fall starting from rest is given by
[tex]v=gt[/tex]
In this case,
g = 20.08 m/s^2
So the time taken to reach a velocity of
v = 28.6 m/s
is
[tex]t=\frac{v}{g}=\frac{28.6 m/s}{20.08 m/s^2}=1.42 s[/tex]
It takes approximately [tex]\( 1.424 \)[/tex] seconds for the rock to reach a velocity of 28.6 m/s on this celestial body.
We need to use the given equation and the relationship between position, velocity, and acceleration.
The equation for the position [tex]\( s \)[/tex] as a function of time [tex]\( t \)[/tex] is given by:
[tex]\[s = 10.04t^2\][/tex]
Step 1: Find the Acceleration
This equation is similar to the general form of the kinematic equation for free fall under constant acceleration:
[tex]\[s = \frac{1}{2} a t^2\][/tex]
Comparing the two equations:
[tex]\[10.04t^2 = \frac{1}{2} a t^2\][/tex]
We can solve for the acceleration [tex]\( a \)[/tex]:
[tex]\[10.04 = \frac{1}{2} a\][/tex]
[tex]\[a = 2 \times 10.04 = 20.08 \, \text{m/s}^2\][/tex]
Step 2: Use the Acceleration to Find the Time
The velocity [tex]\( v \)[/tex] of an object in free fall under constant acceleration is given by:
[tex]\[v = at\][/tex]
We need to find the time [tex]\( t \)[/tex] when the velocity [tex]\( v \)[/tex] is 28.6 m/s:
[tex]\[28.6 = 20.08 t\][/tex]
Solving for [tex]\( t \)[/tex]:
[tex]\[t = \frac{28.6}{20.08}\][/tex]
[tex]\[t \approx 1.424 \, \text{seconds}\][/tex]