All categories

2 years ago

Object's velocity change

Physics

1 answer:

marshall27 [118]2 years ago

3 0

the definition of acceleration in kinematics, allows to find that the correct answer is:

2. speed up ( acceleration)

The kinematics study the movement of the body, the acceleration is defined as the change of the speed by the time in the interval

a = Δv / t

Where the bold letters indicate vectors, a is the acceleration, v the velocity and t the time

Analyzing this expression we see that for there to be a change in velocity, there must be an acceleration of the body.

Let's analyze the different claims

1. True. If it is stopped and you start to move there is an acceleration, therefore there is a change in speed, but after you are moving the acceleration becomes zero and there is no change in speed

2. True. Whenever there is acceleration there is a change in speed

3. False. Moving slowly does not change the acceleration, therefore there is no change in speed

4. True. If you are moving and you stop at this moment there is an acceleration, therefore there is a change in speed, but after being stopped the acceleration is zero and there is no change in speed

5. False. If you change the direction at the instant of change there is an acceleration but after you go in the opposite direction there is not acceleration therefore there is no change in speed.

In conclusion using the definition of acceleration in kinematics, we can find the answer the condition for a change in velocity, the correct statement is:

2. speed up ( acceleration)

Learn more here: brainly.com/question/5063616

You might be interested in

13 points and brainlyest if possible. Thanks.

nikdorinn [45]

Most likely it would be C not completely sure

3 0

3 years ago

Read 2 more answers

Please answer the following question!

Semmy [17]

Answer:

Explanation:

the momentum will always be 0 when it is at rest because the object isnt moving!

Hope this helped!

6 0

2 years ago

A stone is dropped into a river from a bridge at a height h above the water. Another stone is thrown vertically down at a time t

Mumz [18]

Answer:

$v_{y_0} = \frac{\frac{g}{2}t(t - 2\sqrt{\frac{2h}{g}})}{\sqrt{\frac{2h}{g}} - t}$

Explanation:

We will apply the equations of kinematics to both stones separately.

First stone:

Let us denote the time spent after the second stone is thrown as 'T'.

$y - y_0 = v_{y_0}(t+T) + \frac{1}{2}a(t+T)^2\\0 - h = 0 + \frac{1}{2}(-g)(t+T)^2\\(t+T)^2 = \frac{2h}{g}\\T = \sqrt{\frac{2h}{g}}-t$

Second stone:

$y - y_0 = v_{y_0}T + \frac{1}{2}aT^2\\0 - h = v_{y_0}T -\frac{1}{2}gT^2\\-h = v_{y_0}(\sqrt{\frac{2h}{g}} - t) - \frac{g}{2}(\sqrt{\frac{2h}{g}} - t)^2\\-h = v_{y_0}(\sqrt{\frac{2h}{g}} - t) - \frac{g}{2}(\frac{2h}{g} + t^2 - 2t\sqrt{\frac{2h}{g}})\\-h = v_{y_0}\sqrt{\frac{2h}{g}} - v_{y_0}t - h -\frac{g}{2}t^2 + gt\sqrt{\frac{2h}{g}}\\v_{y_0}(\sqrt{\frac{2h}{g}} - t) = \frac{g}{2}t^2 - gt\sqrt{\frac{2h}{g}}\\v_{y_0} = \frac{\frac{g}{2}t(t - 2\sqrt{\frac{2h}{g}})}{\sqrt{\frac{2h}{g}} - t}$

6 0

3 years ago

Read 2 more answers

In a nuclear physics experiment, a proton (mass 1.67×10^(−27)kg, charge +e=+1.60×10^(−19)C) is fired directly at a target nucleu

Arte-miy333 [17]

The given question is incomplete. The complete question is as follows.

In a nuclear physics experiment, a proton (mass $1.67 \times 10^(-27)$ kg, charge +e = $+1.60 \times 10^(-19)$ C) is fired directly at a target nucleus of unknown charge. (You can treat both objects as point charges, and assume that the nucleus remains at rest.) When it is far from its target, the proton has speed $2.50 \times 10^6$ m/s. The proton comes momentarily to rest at a distance $5.31 \times 10^(-13)$ m from the center of the target nucleus, then flies back in the direction from which it came. What is the electric potential energy of the proton and nucleus when they are $5.31 \times 10^{-13}$ m apart?