Calculate the momentum for the 5 kg bowling ball moving at 6 m/s.

Two charges, qA and qB, are separated by a distance, d, and exert a force, F, on each other. Analyze Coulomb's law and answer th

maria [59]

Answer: a. F doubled

b. F reduced by one-quarter i.e

1/4*(F)

c. 1/9*(F)

d. F increased by a factor of 4 i.e 4*F

e. F reduces 3/4*(F)

Explanation: Coulombs law states the force F of attraction/repulsion experience by two charges qA and qB is directly proportional to thier product and inversely proportional to the square of distance d between them. That is

F = k*(qA*qB)/d²

a. If qA is doubled therefore the force is doubled since they are directly proportional.

b. If qA and qB are half, that means thier new product would be qA/2)*qB/2 =qA*qB/4

Which means the product of charge is divided by 4 so the force would be divided by 4 too since they are directly proportional.

c. If d is tripped that is multiplied by 3. From the formula new d would be (3*d)²=9d² but force is inversely proportional to d² so instead of multiplying by 9 the force will be divided by 9

d. If d is cut into half that is divided by 2. The new d would be (d/2)²=d²/4. So d² is divided by 4 so the force would be multiplied by 4

e. If qA is tripled that is multiplied by 3. F would be multiplied by 3 also, if at the same time d is doubled (2*d)²= 4*d² . Force would be divided by 4 at same time. So we have,

3/4*F

6 0

4 years ago

Which of these planets has a giant red spot? saturn jupiter uranus neptune

ki77a [65]

Jupiter has a giant red spot

4 0

3 years ago

Read 2 more answers

A 100 g ball collides elastically with a 300 g ball that is at rest. If the 100 g ball was traveling

sammy [17]

Answer:

The magnitude of the velocities of the two balls after the collision is 3.1 m/s (each one).

Explanation:

We can find the velocity of the two balls after the collision by conservation of linear momentum and energy:

$P_{1} = P_{2}$

$m_{1}v_{1_{i}} + m_{2}v_{2_{i}} = m_{1}v_{1_{f}} + m_{2}v_{2_{f}}$

Where:

m₁: is the mass of the ball 1 = 100 g = 0.1 kg

m₂: is the mass of the ball 2 = 300 g = 0.3 kg

$v_{1_{i}}$ : is the initial velocity of the ball 1 = 6.20 m/s

$v_{2_{i}}$ : is the initial velocity of the ball 2 = 0 (it is at rest)

$v_{1_{f}}$ : is the final velocity of the ball 1 =?

$v_{2_{f}}$ : is the initial velocity of the ball 2 =?

$m_{1}v_{1_{i}} = m_{1}v_{1_{f}} + m_{2}v_{2_{f}}$

$v_{1_{f}} = v_{1_{i}} - \frac{m_{2}v_{2_{f}}}{m_{1}}$ (1)

Now, by conservation of kinetic energy (since they collide elastically):

$\frac{1}{2}m_{1}v_{1_{i}}^{2} = \frac{1}{2}m_{1}v_{1_{f}}^{2} + \frac{1}{2}m_{2}v_{2_{f}}^{2}$

$m_{1}v_{1_{i}}^{2} = m_{1}v_{1_{f}}^{2} + m_{2}v_{2_{f}}^{2}$ (2)

By entering equation (1) into (2) we have:

$m_{1}v_{1_{i}}^{2} = m_{1}(v_{1_{i}} - \frac{m_{2}v_{2_{f}}}{m_{1}})^{2} + m_{2}v_{2_{f}}^{2}$

$0.1 kg*(6.20 m/s)^{2} = 0.1 kg*(6.2 m/s - \frac{0.3 kg*v_{2_{f}}}{0.1 kg})^{2} + 0.3 kg(v_{2_{f}})^{2}$

By solving the above equation for $v_{2_{f}}$ :

$v_{2_{f}} = 3.1 m/s$

Now, $v_{1_{f}}$ can be calculated with equation (1):

$v_{1_{f}} = 6.20 m/s - \frac{0.3 kg*3.1 m/s}{0.1 kg} = -3.1 m/s$

The minus sign of $v_{1_{f}}$ means that the ball 1 (100g) is moving in the negative x-direction after the collision.

Therefore, the magnitude of the velocities of the two balls after the collision is 3.1 m/s (each one).

I hope it helps you!

5 0

3 years ago

Magnetic fields are produced either by electric currents or time-varying electric fields.

Fynjy0 [20]

Answer:

option (a)

Explanation:

The magnetic field is produced by the electric current, and the direction of magnetic field is given by the Maxwells' right hand rule.

The magnetic field is also be produced by the time varying magnetic fields.

6 0

3 years ago

If the mass of a wheel is increased by a factor of 7 and the radius is increased by a factor of 15, by what factor is the moment

inn [45]

Answer:

The moment of inertia will be increased by a factor of 1575

Explanation:

Let the moment of inertia of a wheel be given as;

$I = \frac{1}{2} mr^2\\\\$

where;

I is the moment of inertia of the wheel

m is mass of the wheel

r is radius of the wheel

when the mass is increased by a factor of 7 and the radius is increased by a factor of 15

$I_1 = \frac{1}{2}mr^2\\ \\I_2 = \frac{1}{2}(7m)(15r)^2\\\\I_2 = \frac{1}{2}(7m)(225r^2)\\\\I_2 = 7\times 225[\frac{1}{2}(mr^2)]\\\\I_2 = 1575[\frac{1}{2}(mr^2)]\\\\Recall, \frac{1}{2}(mr^2) = I_1 \\\\I_2 = 1575 I_1$

Thus, the moment of inertia will be increased by a factor of 1575

3 0

3 years ago