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2 years ago

The center-seeking change in velocity of an object moving in a circle is:

Physics

2 answers:

NeTakaya2 years ago

5 0

The center-seeking change in velocity of an object moving in a circle is the centripetal acceleration.

So, by Newton's laws, we know that an object moving with a given velocity will remain in constant motion with a constant velocity until we apply an acceleration.

So we define acceleration as the rate of change of the velocity, also remember that velocity is a vector (has magnitude and direction), so, if there is a change the direction of the velocity, we have an acceleration that causes that.

In circular motion, the velocity vector is always perpendicular to the radius of the circle, and it can only be possible if the velocity direction is changing constantly. This will happen because of something called centripetal acceleration.

This acceleration points radially inwards (to the center of the circle) so is also perpendicular to the velocity of the moving object, and this is what causes the constant change in the direction of the velocity of the moving object.

Just to give an example, if you have a string with a mass on one end, and with your hand, you rotate the mass (from the string), the tension of the string would be the centripetal acceleration.

If you want to learn more about circular motion, you can read:

brainly.com/question/2285236

Helen [10]2 years ago

4 0

Answer:

centripetal acceleration.

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madam [21]

Answer:

Assumption: the air resistance on this ball is negligible. Take $g = 10\; \rm m \cdot s^{-2}$ .

a. The momentum of the ball would be approximately $60\;\rm kg \cdot m \cdot s^{-1}$ two seconds after it is tossed into the air.

b. The momentum of the ball would be approximately $\rm \left(-45\; \rm kg \cdot m \cdot s^{-1}\right)$ three seconds after it reaches the highest point (assuming that it didn't hit the ground.) This momentum is smaller than zero because it points downwards.

Explanation:

The momentum $p$ of an object is equal its mass $m$ times its velocity $v$ . That is: $\vec{p} = m \cdot \vec{v}$ .

Assume that the air resistance on this ball is negligible. If that's the case, then the ball would accelerate downwards towards the ground at a constant $g \approx -10\; \rm m \cdot s^{-2}$ . In other words, its velocity would become approximately $10\; \rm m \cdot s^{-1}$ more negative every second.

The initial velocity of the ball is $60\; \rm m \cdot s^{-1}$ . After two seconds, its velocity would have become $60\;\rm m \cdot s^{-1} + 2\; \rm s \times \left(-10\;\rm m \cdot s^{-1}\right) = 40\; \rm m \cdot s^{-1}$ . The momentum of the ball at that time would be around $p = m \cdot v \approx 60\; \rm kg \cdot m \cdot s^{-1}$ .

When the ball is at the highest point of its trajectory, the velocity of the ball would be zero. However, the ball would continue to accelerate downwards towards the ground at a constant $g \approx -10\; \rm m \cdot s^{-2}$ . That's how the ball's velocity becomes negative.

After three more seconds, the velocity of the ball would be $0\; \rm m \cdot s^{-1} + 3\; \rm s \times \left(-10\; \rm m \cdot s^{-2}\right) = -30 \; \rm m \cdot s^{-1}$ . Accordingly, the ball's momentum at that moment would be $p = m \cdot v \approx \left(-45\; \rm kg \cdot m \cdot s^{-1}\right)$ .

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