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Arturiano [62]
3 years ago
14

4. Describe how the velocity of an object changes if it undergoes uniformly acceleration motion. Can its direction change?

Physics
1 answer:
valentinak56 [21]3 years ago
4 0

Answer:

n the case of linear motion, the change occurs in the magnitude of the velocity, the direction remaining constant.

In the case of circular motion, the magnitude of the velocity remains constant, the change in its direction occurring.

Explanation:

Velocity is a vector therefore it has magnitude and direction, a change in either of the two is the consequence of an acceleration on the system.

In the case of linear motion, the change occurs in the magnitude of the velocity, the direction remaining constant.

              a_{t} = (v₂-v₁)/Δt

In the case of circular motion, the magnitude of the velocity remains constant, the change in its direction occurring.

                    a_{c} = v2/R

In the general case, both the module and the address change

             a = Ra (  a_{t}^2 +   a_{c}^2)  

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A charge of -3.02 μC is fixed in place. From a horizontal distance of 0.0377 m, a particle of mass 9.43 x 10^-3 kg and charge -9
Andreyy89

Answer:

d = 0.0306 m

Explanation:

Here we know that for the given system of charge we have no loss of energy as there is no friction force on it

So we will have

U + K = constant

\frac{kq_1q_2}{r_1} + \frac{1}{2}mv_1^2 = \frac{kq_1q_2}{r_2} + \frac{1}{2}mv_2^2

now we know when particle will reach the closest distance then due to electrostatic repulsion the speed will become zero.

So we have

\frac{(9 \times 10^9)(3.02 \mu C)(9.78 \mu C)}{0.0377} + \frac{1}{2}(9.43 \times 10^{-3})(80.4)^2 = \frac{(9 \times 10^9)(3.02 \mu C)(9.78 \mu C)}{r} + 0

7.05 + 30.5 = \frac{0.266}{r}

r = 7.08 \times 10^{-3} m

so distance moved by the particle is given as

d = r_1 - r_2

d = 0.0377 - 0.00708

d = 0.0306 m

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3 years ago
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