Two small objects each with a net charge of +Q exert a force of magnitude F on each other. We replace one of the objects with another whose net charge is + 4Q. We move the +Q and +4Q charges to be 3 times as far apart as they were. What is the magnitude of the force on the +4Q charge ?A. F
B. 4F
C. 4F/3
D. 4F/9
E. F/3
Answers
Answer:
F'= 4F/9
Explanation:
Two small objects each with a net charge of +Q exert a force of magnitude F on each other. If r is the distance between them, then the force is given by :
...(1)
Now, if one of the objects with another whose net charge is + 4Q is replaced and also the distance between +Q and +4Q charges is increased 3 times as far apart as they were. New force is given by :
.....(2)
Dividing equation (1) and (2), we get :
Hence, the correct option is (d) i.e. " 4F/9"
Final answer:
The magnitude of the force on the +4Q charge, after replacing one of the original +Q charges and moving the charges three times farther apart, is calculated to be 4F/9 using Coulomb's Law. Therefore, the correct answer is D.
Explanation:
The magnitude of the electrostatic force between two charges can be described by Coulomb's Law, which states that F = k × (q1 × q2) / r^2, where F is the force between the charges, k is Coulomb's constant, q1 and q2 are the magnitudes of the charges, and r is the distance between the centers of the two charges. Originally, two objects each with charge +Q exert a force of magnitude F on each other. After one charge is replaced with a +4Q charge and they are moved to be three times as far apart, the force on the +4Q charge can be calculated using the modified version of Coulomb's Law that takes into account the new charges and distance.
Using the original scenario as a reference, where F = k × (Q × Q) / r^2, when the charge is replaced and the distance is tripled, the new force F' = k × (Q × 4Q) / (3r)^2 = 4kQ^2 / 9r^2. By comparing F' with F, we find that F' = (4/9)F. Thus, the magnitude of the force on the +4Q charge is 4F/9.
Related Questions
Tarzan, whose mass is 96 kg, is hanging at rest from a tree limb. Then he lets go and falls to the ground. Just before he lets go, his center of mass is at a height 2.1 m above the ground and the bottom of his dangling feet are at a height 1.3 above the ground. When he first hits the ground he has dropped a distance 1.3, so his center of mass is (2.1 - 1.3) above the ground. Then his knees bend and he ends up at rest in a crouched position with his center of mass a height 0.4 above the ground.Consider the point particle system. What is the speed v at the instant just before Tarzan's feet touch the ground?
In a charging process, 4 × 1013 electrons are removed from one small metal sphere and placed on a second identical sphere. Initially, both metal spheres were neutral. After the charging process, the electrical potential energy associated with the two spheres is found to be −0.063 J. What is the distance between the two spheres?
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The planet earth orbits around the sun and also spins around its own axis. 33% part (a) calculate the angular momentum of the earth in its orbit around the sun in kg • m2/s
B. What is the on-axis magnetic field strength 4.70 cm from the ring?
Answers
Answer:
0.000314 Am²
6.049*10^-7 T
Explanation:
A
From the definitions of magnetic dipole moment, we can establish that
= , where
= the magnetic dipole moment in itself
= Current, 100 A
= Area, πr² (r = diameter divided by 2). Converting to m², we have 0.000001 m²
On solving, we have
= ,
= 100 * 3.14 * 0.000001
= 0.000314 Am²
B
=(0)/4* 2/³, where
(0) = constant of permeability = 1.256*10^-6
z = 4.7 cm = 0.047 m
B = 1.256*10^-6 / 4*3.142 * [2 * 0.000314/0.047³]
B = 1*10^-7 * 0.000628/1.038*10^-4
B = 1*10^-7 * 6.049
B = 6.049*10^-7 T
Final answer:
The magnetic dipole moment of the superconducting ring is 0.000314 Ampere*m². The on-axis magnetic field strength 4.70 cm from the ring is 6.56 μT.
Explanation:
The magnetic dipole moment (μ) of a current (I) circulating around a loop of radius (r) is given by the formula μ = Iπr². Substituting the given values in SI units (I=100 Ampere, r=1.00 mm = 0.001 m), we get μ = 100 * π * (0.001)² = 0.000314 Ampere*m².
To find out the on-axis magnetic field strength (B) at a certain distance from the ring, we use the formula B = μ0/(4π) * (2μ/r³), where μ0 represents the permeability of free space. Plugging the values in SI units (μ0 = 4π × 10-7 T*m/A, r=4.70 cm = 0.047 m), The magnetic field is calculated to be B = (4π × 10-7 T*m/A) /(4π) * (2 * 0.000314 m² / 0.0473m³) = 6.56 × 10-6 T or 6.56 μT.
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Answers
A.The time taken for the car to stop is 8.75 s
B.The distance travelled when the brakes were applied till the car stops is 136.89 m
A. Determination of the time taken for the car to stop.
- We'llbegin bycalculatingthedecelerationof thecar
Initial velocity (u) = 70 mph = 0.447 × 70 = 31.29 m/s
Final velocity (v) = 30 mph = 0.447 × 30 = 13.41 m/s
Time (t) = 5 s
Deceleration (a) =?
a = –3.576 m/s²
- Finally,we shall determine the time taken for the car to stop.
Initial velocity (u) = 31.29 m/s
Final velocity (v) = 0 m/s
Deceleration (a) = –3.576 m/s²
Time (t) =?
t = 8.75 s
Thus, the time taken for the car to stop is 8.75 s
B.Determination of the total distance travelled when the brakes were applied.
Initial velocity (u) = 31.29 m/s
Final velocity (v) = 0 m/s
Deceleration (a) = –3.576 m/s²
Distance (s) =?
s = 136.89 m
Therefore, the total distance travelled by the car when the brakes were applied is 136.89 m
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Answer:8.75 s,
136.89 m
Explanation:
Given
Initial velocity
velocity after 5 s is
Therefore acceleration during these 5 s
therefore time required to stop
v=u+at
here v=final velocity =0 m/s
initial velocity =31.29 m/s
(b)total distance traveled before stoppage
s=136.89 m
Answers
Answers
Answer:
Explanation:
solution below
Final answer:
The quantum of energy for one atomic oscillator in tungsten, given the effective interatomic spring stiffness of 360 N/m, the mass of one tungsten atom as 3.074 x 10^-25 kg, and the reduced Planck's constant of 1.0546 x 10^-34 J · s, can be calculated to be approximately 1.33 x 10^-21 J.
Explanation:
To calculate the quantum of energy for one atomic oscillator in tungsten, we will consider the model of an atom being connected to two springs, both having an effective interatomic spring stiffness of four times the given value (90 N/m). This value thus becomes 360 N/m.
One mole of tungsten has a mass of 0.185 kg, thus the mass of one atom can be determined by dividing this value by Avogadro's number (6.0221 x 10^23 molecules/mole), which gives approximately 3.074 x 10^-25 kg.
The quantum of energy, or the energy of one quantum (the smallest possible energy increment), is given by the formula E = ħω, where ħ is the reduced Planck's constant (1.0546 x 10^-34 J · s) and ω is the angular frequency, given by sqrt(k/m), where k is the spring constant and m is the mass.
Substituting the known values into these equations gives ω= sqrt((360)/(3.074 x 10^-25)) and E= (1.0546 x 10^-34) x sqrt((360)/(3.074 x 10^-25)), which results in a quantum of energy of approximately 1.33 x 10^-21 J.
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Answers
Final answer:
This Physics problem involves balancing the forces and torques acting on a 3.6-m-long pole. By applying the principles of equilibrium and calculations of torque, we find that 114 N of force is needed to keep the pole in a horizontal position.
Explanation:
This is a physics problem related to the concepts of equilibrium and torque. From the details provided, we know that the pole has a mass of 21 kg and it's 3.6 meters long. The center of gravity (cg) of the pole, since it's uniform, is at the middle, which is at 1.8 m from either end of the pole. We are then told that you are holding the pole 35 centimeters (or 0.35 meters) from its tip.
To keep the pole horizontal in equilibrium, the downward force due to the weight of the pole at its center of mass (which is equal to the mass of the pole times gravity, or 21*9.8 = 205.8 N) needs to be balanced by the sum of the torques produced by the forces you are applying at the end you are holding and the force exerted by the fence post at the other end.
Let the force you apply be F1 and the force the fence post exerts be F2. We have F2 at 0.35 m from one end (the pivot point), and F1 at 3.6 - 0.35 = 3.25 m from the pivot. Given that the torque (t) equals to Force (F) times the distance from the pivot (d), and that the net torque should equal zero in equilibrium, we have:
0.35*F2 = 3.25*F1 (1)
Because the net force should also be zero in equilibrium, we have:
F1 + F2 = 205.8 (2)
Solving these two equations, we'll be able to calculate that the force you must exert to keep the pole motionless in a horizontal position, F1, is approximately 114 N.
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Final answer:
To balance the 3.6m-long, 21 kg pole and keep it horizontally motionless, a force of approximately 114N is required
Explanation:
The subject question is a classic example of Torque problem specific to Physics, which involves the concepts of force, weight, and distance. To keep the pole motionless and horizontally balanced, the force you exert must counterbalance the torque due to the pole's weight. Assuming the pole is uniform, its center of gravity (cg) is at its midpoint, 1.8m from each end. The weight of the pole acts downward at this midpoint, providing a clockwise torque about the point of support, which is the fence post.
This torque is calculated as Torque = r * F = 1.8m (distance from fence post to cg) * Weight of pole = 1.8m * 21kg * 9.8m/s² (gravitational acceleration) = ~370 N.m. As the pole is motionless, the total torque about any point must be zero. Hence, the counter-clockwise torque provided by the force you exert is equal to the clockwise torque due to the weight of the pole. Using the distance from the point of your hold to the fence post (3.25m) we can calculate the force you need to exert: Force = Torque/distance = 370 N.m/3.25m = ~114N.
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-Calculate the net work Wnet in J done on the bucket of water by the two forces F1 and Fg.
Answers
Final answer:
To calculate the work done by gravity on the bucket of water as it is lifted up the well, multiply the weight of the bucket by the lifting distance. The net work done on the bucket by the force applied by the farmer and gravity is the sum of the work done by both forces. The net work is represented by the equation Wnet = W1 + Wg.
Explanation:
To calculate how much work gravity does on the bucket filled with water as the farmer lifts it up the well, we need to multiply the force of gravity (weight) by the vertical distance the bucket is lifted. The equation for work is W = Fd, where W is the work done, F is the force, and d is the distance. In this case, the force of gravity is the weight of the bucket, which can be calculated by multiplying the mass by the acceleration due to gravity (9.8 m/s^2).
So, the work done by gravity (Wg) on the bucket is Wg = Fg * d = (m * g) * d = (3.9 kg * 9.8 m/s^2) * d = 38.22 d Joules.
To calculate the net work done on the bucket by the two forces, we can use the equation Wnet = W1 + Wg, where W1 is the work done by force F1 and Wg is the work done by gravity. Since force F1 and the displacement (lifting distance) are both vertical, the work done by F1 is given by W1 = F1 * d.
Therefore, the net work done on the bucket by forces F1 and gravity is Wnet = F1 * d + Fg * d = (57.5 N) * d + (3.9 kg * 9.8 m/s^2) * d = (57.5 N + 38.22 d) Joules.
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