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Maru [420]
3 years ago
10

A kicked ball rolls across the grass and eventually comes to a stop in 4.0 sec. When the ball was kicked, its initial velocity w

as 20 mi/ hr. What is the acceleration of the ball as it rolls across the grass?
Physics
1 answer:
kotegsom [21]3 years ago
6 0

Answer:

-2.24 m/s²

Explanation:

Given:

v₀ = 20 mi/hr = 8.94 m/s

v = 0 m/s

t = 4.0 s

Find: a

v = v₀ + at

0 m/s = 8.94 m/s + a (4.0 s)

a = -2.24 m/s²

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A car is stopped for a traffic signal. When the light turns green, the car accelerates, increasing its speed from 0 to 5.40 m/s
tekilochka [14]

Answer:

a) 378Ns

b) 477.27N

Explanation:

Impulse is the defined as the product of the applied force and time taken. This is expressed according to the formula

I = Ft = m(v-u)

m is the mass = 70kg

v is the final velocity = 5.4m/s

u is the initial velocity = 0m/s

Get the impulse

I = m(v-u)

I = 70(5.4-0)

I = 70(5.4)

I = 378Ns

b) Average total force is expressed as

F = ma (Newton's second law)

F = m(v-u)/t

F = 378/0.792

F = 477.27N

Hence the average total force experienced by a 70.0-kg passenger in the car during the time the car accelerates is 477.27N

3 0
3 years ago
a foul ball is hit into the stands at a baseball game. the ball rises to a height of 38 meters and is caught on its way down by
lisov135 [29]

The velocity of the ball when it was caught is 12.52 m/s.

<em>"Your question is not complete it seems to be missing the following, information"</em>,

find the velocity of the ball when it was caught.

The given parameters;

maximum height above the ground reached by the ball, H = 38 m

height above the ground where the ball was caught, h = 30 m

The height traveled by the ball when it was caught is calculated as follows;

y = H - h

y = 38 - 30 = 8 m

The velocity of the ball when it was caught is calculated as;

v_f^2 = v_0 + 2gh\\\\v_f^2 = 0 + (2\times 9.8 \times 8)\\\\v_f^2 = 156.8\\\\v_f = \sqrt{156.8} \\\\v_f = 12.52 \ m/s

Thus, the velocity of the ball when it was caught is 12.52 m/s.

Learn more here: brainly.com/question/14582703

4 0
3 years ago
Help please I don't understand this
BARSIC [14]
It's either staying there or is going at the same pace
4 0
4 years ago
Using 6400 km as the radius of Earth, calculate how high above Earth’s surface you would have to be in order to weigh 1/16th of
FrozenT [24]
Gravity obeys the inverse square law.  At 6400 km above the center of the Earth (Earth's surface) you weigh x.  Twice that reduces your weight to 1/4th.  Four times that height reduces your weight to 1/16th.  4 times 6400 km is 25,600 km.  But that is above the center of the earth, and the question requests the height above the surface, so we deduct 6400 km to arrive at our final answer:  19,200 km.

Incidentally, it doesn't exactly work the opposite way.  At the center of the Earth the mass would be equally distributed around you, and you would therefore be weightless.
6 0
3 years ago
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