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
A science teacher tells her class that their final project requires students to measure a specific variable and determine the velocity of a car with no more than 2.5% error. Jennifer and Johnny work hard and decide the velocity of the car is 34.87 m/s. The teacher informs them that the actual velocity is 34.15 m/s. Will Jennifer and Johnny pass their final project?
Answer:
Yes
Explanation:
The velocity measured by Jennifer and Johnny is
[tex]v_m = 34.87 m/s[/tex]
The actual velocity is
[tex]v=34.15 m/s[/tex]
We can calculate the % error of the students measurement as follows:
[tex]Err = \frac{v_m - v}{v}\cdot 100 = \frac{34.87 m/s-34.15 m/s}{34.15 m/s}\cdot 100 =0.021 \cdot 100 = 2.1 \%[/tex]
Which is lower than the 2.5% maximum error required, so the two students will pass the test.
Final answer:
Jennifer and Johnny's measurement has a percentage error of approximately 2.11%, which is less than the maximum allowed error of 2.5%. Therefore, they will pass their final project.
Explanation:
To determine if Jennifer and Johnny passed their final project, we need to calculate the percentage error of their measured velocity. The percentage error is calculated using the formula:
Percentage Error = |(Actual Value - Experimental Value) / Actual Value| × 100%
First, let's find the absolute difference between the actual velocity (34.15 m/s) and the measured velocity (34.87 m/s):
|34.15 m/s - 34.87 m/s| = |(-0.72 m/s)| = 0.72 m/s
Now, we calculate the percentage error:
Percentage Error = (0.72 m/s / 34.15 m/s) × 100% ≈ 2.11%
Since the percentage error they obtained (2.11%) is less than the maximum allowed error of 2.5%, Jennifer and Johnny will pass their final project.
Squids are the fastest marine invertebrates, using a powerful set of muscles to take in and then eject water in a form of jet propulsion that can propel them to speeds of over 11.5 m/s. What speed (in m/s) would a stationary 1.60 kg squid achieve by ejecting 0.115 kg of water (not included in the squid's mass) at 3.50 m/s? Neglect other forces, including the drag force on the squid. HINT m/s
Answer:
0.25 m/s
Explanation:
This problem can be solved by using the law of conservation of momentum - the total momentum of the squid-water system must be conserved.
Initially, the squid and the water are at rest, so the total momentum is zero:
[tex]p_i = 0[/tex]
After the squid ejects the water, the total momentum is
[tex]p_f = m_s v_s + m_w v_w[/tex]
where
[tex]m_s = 1.60 kg[/tex] is the mass of the squid
[tex]v_s[/tex] is the velocity of the squid
[tex]m_2 = 0.115 kg[/tex] is the mass of the water
[tex]v_w = 3.50 m/s[/tex] is the velocity of the water
Due to the conservation of momentum,
[tex]p_i = p_f[/tex]
so
[tex]0=m_s v_s + m_w v_w[/tex]
so we can find the final velocity of the squid:
[tex]v_s = -\frac{m_w v_w}{m_s}=-\frac{(0.115 kg)(3.50 m/s)}{1.60 kg}=-0.25 m/s[/tex]
and the negative sign means the direction is opposite to that of the water.
By applying the conservation of momentum, the squid would achieve a speed of about 0.80 m/s in the opposite direction to that of the ejected water.
Explanation:This problem is related to the principle of conservation of momentum. According to this principle, the total momentum before and after the ejection of the water should be the same, because no external force is acting on this system of the squid and the water it ejects.
The initial momentum of the system is 0, because the squid is initially stationary. When the squid ejects water, it gets a momentum in the opposite direction. Hence,
0 = momentum of squid + momentum of water
or, 0 = (mass of squid × speed of squid) + (mass of ejected water × speed of ejected water),
Solving this equation for the speed of the squid gives:
speed of squid = - (mass of ejected water × speed of ejected water) / mass of squid
Substituting the given values into this equation:
speed of squid = - (0.115 kg × 3.50 m/s) / 1.60 kg ≈ -0.80 m/s.
The negative sign indicates that the squid moves in the opposite direction to that of the ejected water.
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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
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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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 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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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.
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.
If a victim of sexual harassment asks a supervisor not report it, the supervisor should respect his or her wishes.
True
False
This is true. The victim could be too embarrassed and would not want others to know because people could bash her for that.
If a victim of sexual harassment asks a supervisor not report it, the supervisor should respect his or her wishes is False.
If a victim of sexual harassment asks a supervisor not to report it, the supervisor should not automatically respect their wishes. It is essential to prioritize the safety and well-being of the victim and address the issue appropriately.
Sexual harassment is a serious and unlawful matter, and supervisors have a legal and ethical obligation to address and report such incidents following company policies and the law.
Reporting incidents of sexual harassment is crucial for investigating the matter, providing support to the victim, and taking appropriate actions to prevent further harassment and protect the well-being of employees.
Supervisors should follow their organization's policies and procedures for handling harassment complaints and ensure that victims are treated with sensitivity and respect throughout the process. Confidentiality should be maintained to the extent possible, but it should not prevent the necessary actions to address and resolve the issue effectively.
Hence, If a victim of sexual harassment asks a supervisor not report it, the supervisor should respect his or her wishes is False.
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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.
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 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 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
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
If you were to drop a rock from a tall building, assuming that it had not yet hit the ground, and neglecting air resistance. What is its vertical displacement (in m) after 6 s?
Answer:
176.4 m
Explanation:
U = 0, t = 6s, g = 9.8 m/s^2
Use second equation of motion
H = ut + 1/2 gt^2
H = 0 + 0.5 × 9.8 × 6 × 6
H = 176.4 m
It is the displacement from the point of dropping of object.
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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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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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
An object is dropped from a tower, 400 ft above the ground. The object's height above ground x seconds after the fall is s(x)equals400minus16xsquared. About how long does it take the object to hit the ground? What is the object's velocity at the moment of impact?
1) 5 s
The vertical position of the object is given by
[tex]y(t) = h - \frac{1}{2}gt^2 = 400 - 16 t^2[/tex]
where
h=400 ft represents the initial height
g = 32 ft/s^2 is the acceleration of gravity
t is the time
We want to find the time t at which the object reaches the ground, so the time t at which
y(t) = 0
By substituting this into the equation, we find
[tex]0 = 400 - 16t^2\\t=\sqrt{\frac{400}{16}}=5 s[/tex]
2) 160 ft/s
The object is released from rest, so the initial velocity is zero
u = 0
The final vertical velocity can be found by using
[tex]v^2 - u^2 = 2ah[/tex]
where
v is the final velocity
a = 32 ft/s^2 is the acceleration of gravity
h = 400 ft is the vertical distance covered
Solving for v, we find
[tex]v=\sqrt{u^2 +2ay}=\sqrt{2(32 ft/s)(400 ft)}=160 ft/s[/tex]
Final answer:
The object takes approximately 5 seconds to hit the ground and has an impact velocity of 160 feet per second.
Explanation:
Time to Hit the Ground and Velocity at Impact
To find out how long it takes the object to hit the ground, we need to solve the equation s(x) = 400 - 16x2 for the moment when the height s(x) is zero (s(x) = 0). Setting the equation to zero, we get 0 = 400 - 16x2, which simplifies to x2 = 25 after dividing both sides by 16. Taking the square root of both sides gives us x ≈ 5 seconds, which is the time the object takes to hit the ground.
For the object's velocity at the moment of impact, we use the formula v = gt, where g is the acceleration due to gravity (32 feet per second squared), and t is the time in seconds. Thus, the velocity at impact is v = 32 * 5 = 160 feet per second.
A projectile is shot from the edge of a vertical cliff 60.0 m above the ocean. It has a speed of 100 m/s and is fired at an angle of 35.0° below the horizontal. How far from the foot of the vertical cliff does the projectile hit the water?
Answer:
79.5 m
Explanation:
Let t be the time taken to hit the surface of water and x be the horizontal distance traveled.
use II equation of motion in Y axis direction
h = uy t + 1/2 g t^2
- 60 = - 100 Sin 35 x t - 1/2 x 9.8 x t^2
-60 = - 57.35 t - 4.9 t^2
4.9 t^2 + 57.35 t - 60 = 0
[tex]t = \frac{-57.35\pm \sqrt{57.35^{2} + 4 \times 4.9 \times 60}}{2\times 4.9}[/tex]
By solving we get
t = 0.97 second
The horizontal distance traveled is
x = ux t
x = 100 Cos 35 x 0.97
x = 79.5 m
Final answer:
To find how far from the cliff a projectile hits the water, one must use projectile motion principles to calculate the time of flight based on vertical movement and then determine the horizontal distance traveled during this time.
Explanation:
The question involves calculating how far from the foot of a vertical cliff a projectile hits the water when it is shot from an elevation with a given initial speed and angle. To solve this, we need to use the concepts of projectile motion, specifically focusing on the horizontal distance traveled by a projectile. The key equations involve splitting the initial velocity into its horizontal and vertical components, calculating the time of flight based on the vertical motion, and then using this time to find the horizontal distance traveled.
Given the projectile has a speed of 100 m/s and is fired at an angle of 35.0° below the horizontal from a height of 60.0 m, the calculation involves several steps:
Determine the initial horizontal and vertical velocity components.
Calculate the time of flight using the vertical motion equations.
Finally, compute the horizontal distance traveled using the time of flight.
Due to the complexity and the need for specific formulae and calculations, a detailed step-by-step solution would be necessary to find the exact distance.
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 wheel is rotating at 30.0 rpm. The wheel then accelerates uniformly to 50.0 rpm in 10.0 seconds. Determine the – a) rate of angular acceleration in rad/s2 and –
Answer:
The angular acceleration is [tex]0.209\ rad/s^2[/tex]
Explanation:
Given that,
Angular velocity, [tex]\omega_{i} = 30.0\ rpm[/tex]
Angular velocity, [tex]\omega_{f} = 50.0\ rpm[/tex]
Time t = 10.0 sec
We need to calculate the angular acceleration
Using formula of angular acceleration
[tex]\alpha=\dfrac{\Delta \omega}{\Delta t}[/tex]
[tex]\alpha=\dfrac{\omega_{f}-\omega_{i}}{\Delta t}[/tex]
[tex]\alpha=\dfrac{50.0-30.0}{10.0}[/tex]
Now, we change the angular velocity in rad/s.
[tex]\omega=20\times\dfrac{2\pi}{60}[/tex]
[tex]\omega=2.09\ rad/s[/tex]
[tex]\alpha=\dfrac{2.09}{10.0}[/tex]
[tex]\alpha=0.209\ rad/s^2[/tex]
Hence, The angular acceleration is [tex]0.209\ rad/s^2[/tex]
Answer:
The rate of angular acceleration is 0.209 rad/s²
Explanation:
the solution is in the attached Word file
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 astronomer finds that a meteorite sample has a mass of 15.6 g along with an apparent mass of 9.2 g when submerged in water. find the density of the meteorite?
Answer:
Density of meteorite = 2.44 g/cm³
Explanation:
Apparent mass = Mass of solid - Mass of water displaced
Mass of water displaced = Mass of solid - Apparent mass
= 15.6 - 9.2 = 6.4 g
Density of water = 1 g/cm³
Volume of water displaced [tex]=1\times 6.4=6.4cm^3[/tex]
Volume of meteorite = Volume of water displaced = 6.4 cm³
[tex]\texttt{Density of meteorite}=\frac{\texttt{Mass of meteorite}}{\texttt{Volume of meteorite}}=\frac{15.6}{6.4}=2.44g/cm^3[/tex]
Density of meteorite = 2.44 g/cm³
A 2.0kg solid disk rolls without slipping on a horizontal surface so that its center proceeds to the right with a speed of 5.0 m/s. What is the instantaneous speed of the point of the disk that makes contact with the surface?
Answer:
Instantaneous speed of contact point will be ZERO
Explanation:
As we know that disc is rolling without slipping on horizontal surface
So here the speed of center of the disc is given as
v = 5 m/s
now at the contact point the tangential speed will be in reverse direction
[tex]v_t = R\omega[/tex]
now we know that net contact speed with respect to its lower surface must be zero
[tex]v_{net} = v - v_t = 0[/tex]
so net velocity of contact point with respect to its lower surface must be ZERO here
The instantaneous speed of the point of a disc that is rolling without slipping and makes contact with the surface is zero. This is because that point is momentarily at rest relative to the surface at that instant.
Explanation:The instantaneous speed of the point of the disc that makes contact with the surface is zero. This is because for a disc rolling without slipping on a surface, the point of contact at any instant is momentarily at rest relative to the surface. This can be understood by imagining the point of contact as the 'pivot' or point about which the disc rotates while rolling. As the disc rolls, the pivot point changes, but whichever point is in contact with the surface at a given instant is not moving relative to the surface. Hence, the instantaneous speed of that point is zero.
Learn more about Rolling Disc Instantaneous Speed here:https://brainly.com/question/17661499
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A battery has an emf of 15.0 V. The terminal voltage of the battery is 12.2 V when it is delivering 26.0 W of power to an external load resistor R. (a) What is the value of R? Ω (b) What is the internal resistance of the battery?
Answer:
(a) 5.725 Ω
(b) 1.3 Ω
Explanation:
(a)
E = emf of the battery = 15.0 Volts
V = terminal voltage of the battery = 12.2 Volts
P = Power delivered to external load resistor "R" = 26.0 W
R = resistance of external load resistor
Power delivered to external load resistor is given as
[tex]P = \frac{V^{2}}{R}[/tex]
26.0 = 12.2²/R
R = 5.725 Ω
(b)
r = internal resistance of the battery
i = current coming from the battery
Power delivered to external load resistor is given as
P = i V
26.0 = i (12.2)
i = 2.13 A
Terminal voltage is given as
V = E - ir
12.2 = 15 - (2.13) r
r = 1.3 Ω
Final answer:
The external load resistor R is calculated to be 5.72 Ohms using the power and terminal voltage, and the internal resistance of the battery is 1.31 Ohms as determined from the emf, terminal voltage, and the current flowing through R.
Explanation:
To solve for external load resistor R and the internal resistance of the battery, we can use the formulas related to electric power and the relation between emf, terminal voltage, and internal resistance.
(a) Value of R:
The power delivered to the resistor (P) is given by:
P = V2 / R
Where V is the terminal voltage and R is the resistance.
Substituting the given values:
26.0 W = (12.2 V)2 / R
Hence, R = (12.2 V)2 / 26.0 W = 5.72 Ω (Ohms)
(b) Internal resistance of the battery:
We know that terminal voltage V is emf - (current * internal resistance).
The current I flowing through R can be calculated using the power:
I = P / V = 26.0 W / 12.2 V = 2.13 A
Now, using the emf (E) of the battery and terminal voltage (V):
E = V + Ir
15.0 V = 12.2 V + (2.13 A × r)
We solve for r, the internal resistance:
r = (15.0 V - 12.2 V) / 2.13 A = 1.31 Ω (Ohms)
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.
If a light bulb's power rating is 75W, it means it consumesjoule of electric energy in 1 second of time
Answer:
Electrical energy consumed, E = 1.95 × 10⁻⁵ kWh
Explanation:
It is given that,
Power rating on the light bulb, P = 75 W
We need to find the electrical energy consumed by the bulb in 1 second of time i.e in 0.00026 hours.
[tex]P=\dfrac{energy\ consumed}{time}[/tex]
Energy consumed, E = P × t
E = 75 W × 0.00026 h
E = 0.0195 W-h
Energy consumed is calculated in kilo watt hour. Since, 1 watt = 0.001 kW
So, E = 1.95 × 10⁻⁵ kWh
Hence, this is the required solution.
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.
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]