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Triss [41]
3 years ago
8

38.A student pushes a 0.15 kg box down against a spring doing 25 J of work on the spring. The student releases the box which lau

nches the box into the air. What is the maximum height reached by the box assuming negligible frictional forces
Physics
1 answer:
Gnoma [55]3 years ago
3 0

Answer:

Explanation:

Potential energy stored in the spring = 25 J

This energy is converted into gravitational potential energy . If h be the height attained

gravitational potential energy = mgh

mgh = 25

.15 x 9.8 x h = 25

h = 17 m

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A rock is thrown vertically upward with a speed of 18.0 m/s from the roof of a building that is 50.0 m above the ground. Assume
Andreyy89

Answer:

(a) 5.7 s

(b) 39 m/s

Explanation:

(a) u = 18 m/s

At the maximum height, the final velocity of ball is zero. lte teh time taken by the ball to go from 50 m height to maximum height is t.

use first equation of motion.

v = u + g t

0 = 18 - 10 x t

t = 1.8 s

Let the maximum height attained by the ball when it thrown from 50 m height is h'.

Use third equation of motion

v^2 = u^2 + 2 g h'

0 = 18^2 - 2 x 10 x h'

h' = 16.2 m

Total height from the ground H = h + h' = 50 + 16.2 = 76.2 m

Let t' be the time taken by the ball to hit the ground as it falls from maximum height.

use third equation of motion

H = ut + 1/2 x g t'^2

76.2 = 0 + 1/2 x 10 x t'^2

t' = 3.9 s

Total time taken by the ball to hit the ground = T = t + t' = 1.8 + 3.9 = 5.7 s

(b) Let v be the velocity with which the ball strikes the ground.

v^2 = u^2 + 2 g H

v^2 = 0 + 2 x 10 x 76.2

v = 39 m/s

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4 years ago
A direct injection (DI) fuel injector system sprays high-pressure fuel, up to _______ psi, into the combustion chamber as the pi
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4 years ago
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What height will the object reach? 12 points. Will give brainliest.
umka21 [38]

Answer:

12.7 m

Explanation:

The following data were obtained from the question:

Initial velocity (u) = 56.7 Km/hr

Maximum height (h) =..?

First, we shall convert 56.7 Km/hr to m/s. This can be obtained as follow:

Initial velocity (m/s) = 56.7 x 1000/3600

Initial velocity (m/s) = 15.75 m/s

Next, we shall determine the time taken to get to the maximum height. This can be obtained as follow:

Initial velocity (u) = 15.75 m/s

Final velocity (v) = 0 m/s

Acceleration due to gravity (g) = 9.8 m/s²

Time (t) =?

v = u – gt (since the ball is going against gravity)

0 = 15.75 – 9.8 × t

Rearrange

9.8 × t = 15.75

Divide both side by 9.8

t = 15.75/9.8

t = 1.61 secs.

Finally, we shall determine the maximum height as follow

h = ½gt²

Acceleration due to gravity (g) = 9.8 m/s²

Time (t) = 1.61 secs.

Height (h) =..?

h = ½gt²

h = ½ × 9.8 × 1.61²

h = 4.9 x 1.61²

h = 12.7 m

Therefore, the maximum height reached by the ball is 12.7 m

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3 years ago
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Answer:

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