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Nostrana [21]
3 years ago
11

A steel ball is dropped from a building's roof and passes a window, taking 0.12 s to fall from the top to the bottom of the wind

ow, a distance of 1.20 m. It then falls to a sidewalk and bounces back past the window, moving from bottom to top in 0.12 s. Assume that the upward flight is an exact reverse of the fall. The time the ball spends below the bottom of the window is 1.76 s. How tall is the building?
Physics
1 answer:
Dvinal [7]3 years ago
8 0
The top of the WINDOW is 19 meters from the ground. also steel balls dont bounce.
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A factory worker pushes a 32.0 kg crate a distance of 7.0 m along a level floor at constant velocity by pushing horizontally on
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Answer:

(a) 81.54 N

(b) 570.75 J

(c) - 570.75 J

(d) 0 J, 0 J

(e) 0 J  

Explanation:

mass of crate, m = 32 kg

distance, s = 7 m

coefficient of friction = 0.26

(a) As it is moving with constant velocity so the force applied is equal to the friction force.

F = 0.26 x m x g = 0.26 x 32 x 9.8 = 81.54 N

(b) The work done on the crate

W = F x s = 81.54 x 7 = 570.75 J

(c) Work done by the friction

W' = - W = - 570.75 J

(d) Work done by the normal force

W'' = m g cos 90 = 0 J

Work done by the gravity

Wg = m g cos 90 = 0 J

(e) The total work done is

Wnet = W + W' + W'' + Wg = 570.75 - 570.75 + 0 = 0 J  

6 0
3 years ago
____________ are used to calculate the distance a continent has moved in a year. a. Space satellites c. Laser beams b. Mirrors d
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<span> Space satellites, laser beams, mirrors</span> are used to calculate the distance a continent has moved in a year.

Therefore, your correct answer would be "all of the above".
4 0
3 years ago
**20points**
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B) a new element is formed
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Machines makes work easier by increasing the amount of force that is applied, and changing the direction in which the force is applied !! Hope it helped (p.s. I had this same question)
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2 * 1.5 * (.850/2)^2A small ball with mass 1.50 kg is mounted on one end of a rod 0.850 m long and of negligible mass. The syste
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Answer

given,

mass of the rod = 1.50 Kg

length of rod = 0.85 m

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now calculating the rotational inertia of the system.

I = m L^2        

where L is the length of road, we will take whole length of rod because mass is at  the end of it.      

I = 1.5 \times 0.85^2  

I = 1.084 kg.m²                        

hence, the rotational inertia the system is equal to I = 1.084 kg.m²

8 0
3 years ago
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