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Goryan [66]
4 years ago
9

A ball is dropped from somewhere above a window that is 2.00 m in height. As it falls, it is visible to a person boxing through

the window for 200 ms as it passes by the 2.00 m height of the window. From what height above the top of the window was the ball dropped?
Physics
1 answer:
Irina18 [472]4 years ago
5 0

Answer:

4.14 m

Explanation:

In the last leg of the journey the ball covers 2 m in 2ms or 0.2 s .

Let in this last leg , u be the initial velocity.

s = ut + 1/2 g t²

2 = .2 u + .5 x 9.8 x .04

u = 9.02 m /s .

Let v be the final velocity in this leg

v² = u² + 2 g s

v² = (9.02)² + 2 x 9.8 x 2

= 81.36 +39.2

v = 10.97 m / s

Now consider the whole height from where the ball dropped . Let it be h.

Initial velocity u = 0

v² = u² +2gh

(10.97 )² = 2 x 9.8 h

h = 6.14 m

Height from window

= 6.14 - 2m

= 4.14 m

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8 0
3 years ago
A 72 kg swimmer dives horizontally off a raft floating in a lake. The diver's speed immediately after leaving the raft is 3.8 m/
juin [17]

Answer:

F = 1094.4 N

Explanation:

From impulse - momentum theorem, we now that ;

Impulse = momentum

Where;

Formula for impulse = force (F) × time(t)

Momentum = mass(m) × velocity(v)

Now, we are given;

Mass of swimmer; m = 72 kg

Speed; v = 3.8 m/s

Time; t = 0.25 s

Thus;

F × t = mv

F = mv/t

F = (72 × 3.8)/0.25

F = 1094.4 N

This value of force is the magnitude of the average horizontal force by diver on the raft.

4 0
3 years ago
A bullet traveling at 350 m/s hits a tree, penetrating 0.125 m into the tree before stopping. The mass of the bullet is 0.200 kg
Sloan [31]

Answer:

-4.9\cdot 10^5 m/s^2

Explanation:

The motion of the bullet is a uniformly accelerated motion, therefore we can find its acceleration by using the following suvat equation:

v^2-u^2=2as

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance covered

For the bullet in this problem:

u = 350 m/s is the initial velocity of the bullet

v = 0 is the final velocity (the bullet comes to a stop)

s = 0.125 m is the stopping distance of the bullet

Therefore, by solving the equation for a, we find its acceleration:

a=\frac{v^2-u^2}{2s}=\frac{0^2-350^2}{2(0.125)}=-4.9\cdot 10^5 m/s^2

And the negative sign tells that the direction of the acceleration is opposite to that of the velocity.

7 0
4 years ago
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Answer:

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6 0
3 years ago
The barrel of a rifle has a length of 0.855 m. A bullet leaves the muzzle of a rifle with a speed of 553 m/s. What is the accele
Novay_Z [31]

Answer:

Acceleration of the bullet will be 1778835.6 m/sec^2      

Explanation:

We have given length of the barrel refile s= 0.855 m

When the bullet leaves the muzzle its velocity is 553 m/sec

So final velocity v = 553 m/sec

Initial velocity will be 0 that is u = 0 m/sec

According to third equation of motion v^2=u^2+2as

553^2=0^2+2\times a\times 0.855

a=178835.6m/sec^2

5 0
4 years ago
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