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2 years ago

A golf ball rolls up a hill toward a miniature-golf hole. Assume the direction toward the hole is positive.

Physics

1 answer:

umka21 [38]2 years ago

4 0

Answer:

a) 1m/s

b) -1m/s

Explanation:

So the ball starts rolling up the hill with a speed of 2m/s (v0=2m/s) and it decelerates at a constant rate of 0.5 m/s2 (a = -0.5 m/s², minus because it decelerates).

Now, the equation for calculating velocity is:

v = v0 + a • t, where t represents time

So, ball's velocity after 2 seconds is:

v = 2m/s + (-0.5m/s²• 2s)

v = 2m/s + (-1m/s)

v = 1m/s

We use the same formula for finding ball's velocity after 6 seconds:

v = v0 + a • t

v = 2m/s + (-0.5m/s² • 6s)

v = 2m/s + (-3m/s)

v = -1m/s

Of course, velocity can't be negative, this only means that the ball moves in 1m/s velocity, only in opposite direction (downhill).

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How much energy (in kW-h) does a 900 Watt stove use in a week if it is used for 1.5 hours each day?

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Answer:

9.45 kWh

Explanation:

Energy = Power × time

E = 900 W × (1.5 h/day × 7 day)

E = 9450 Wh

E = 9.45 kWh

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2 years ago

The earth's interior heat and pressure drive a process of heat transfer in the mantle called _____.

Galina-37 [17]

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2 years ago

Read 2 more answers

Suppose a baseball pitcher throws the ball to his catcher.

amm1812

a) Same

b) Same

c) Same

d) Throw the ball takes longer

e) F is larger when the ball is catched

Explanation:

The change in speed of an object is given by:

$\Delta v = |v-u|$

where

u is the initial velocity of the object

v is the final velocity of the object

The change in speed is basically the magnitude of the change in velocity (because velocity is a vector, while speed is a scalar, so it has no direction).

In this problem:

- In situation 1 (pitcher throwing the ball), the initial velocity is

u = 0 (because the ball starts from rest)

while the final velocity is v, so the change in speed is

$\Delta v=|v-0|=|v|$

- In situation 2 (catcher receiving the ball), the initial velocity is now

u = v

while the final velocity is now zero (ball coming to rest), so the change in speed is

$\Delta v =|0-v|=|-v|$

Which means that the two situations have same change in speed.

The change in momentum of an object is given by

$\Delta p = m \Delta v$

where

m is the mass of the object

$\Delta v$ is the change in velocity

If we want to compare only the magnitude of the change in momentum of the object, then it is given by

$|\Delta p|=m|\Delta v|$

- In situation 1 (pitcher throwing the ball), the change in momentum is

$\Delta p = m|\Delta v|=m|v|=mv$

- In situation 2 (catcher receiving the ball), the change in momentum is

$\Delta p = m\Delta v = m|-v|=mv$

So, the magnitude of the change in momentum is the same (but the direction is opposite)

The impulse exerted on an object is equal to the change in momentum of the object:

$I=\Delta p$

where

I is the impulse

$\Delta p$ is the change in momentum

As we saw in part b), the change in momentum of the ball in the two situations is the same, therefore the impulse exerted on the ball will also be the same, in magnitude.

However, the direction will be opposite, as the change in momentum has opposite direction in the two situations.

To compare the time of impact in the two situations, we have to look closer into them.

- When the ball is thrown, the hand "moves together" with the ball, from back to ahead in order to give it the necessary push. We can verify therefore that the time is longer in this case.

- When the ball is cacthed, the hand remains more or less "at rest", it doesn't move much, so the collision lasts much less than the previous situation.

Therefore, we can say that the time of impact is longer when the ball is thrown, compared to when it is catched.

The impulse exerted on an object can also be rewritten as the product between the force applied on the object and the time of impact:

$I=F\Delta t$