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Mice21 [21]
3 years ago
13

A student, standing on a scale in an elevator at rest, sees that his weight is 997 N. As the elevator rises, the scale reads 122

6 N and then returns to normal. When the elevator slows to a stop at the top floor, the scale reads 650 N and then returns to normal. Determine the magnitude of the acceleration when the elevator is slowing to a stop. (no negative numbers)
Physics
1 answer:
Artyom0805 [142]3 years ago
3 0
The total force F on the student is given by:

F = N - mg = ma

N normal force of the scale on to the student(scale readout)
m mass of the student
g acceleration due to gravity
a acceleration of the elevator

997 = mg
650 = m(g - a)

ma = 997 - 650
a = g(997 - 650)/997 
a = 3,4

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Naily [24]

Answer:

I will answer in English.

Here we will use the relation

Velocity*time = distance

So:

a) velocity = 3m/s

time = 2s

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b) velocity = 2m/s

time = 3.5s

Distance = 2m/s*3.5s = 7m

c) velocity = 10m/s

time = 0.5s

Distance = 10m/s*0.5s = 5m

d) velocity = 4m/s

time = 2.5s

Distance = 4m/s*2.5s = 9m

e) velocity = 1.5m/s

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8 0
3 years ago
In the laboratory, a ball is dropped onto a force-sensing platform several times, each time hitting a different surface (foam, f
NeX [460]

Answer:

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Explanation:

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           I = F t

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Starting point. Higher

          Em₀ = U = mgh

Lower end point, just before hitting the scale

          Em_{f} = K = ½ m v²

in the path in the air there is no friction

          Em₀ = Em_{f}

          m g h = ½ m v²

          v = \sqrt{2gh}

this height is different for the descent and ascent of the ball, so we have two moments

         Δp = p_{f} - p₀

         Δp = m (v_{f} -v₀)

         

therefore we have the relationship

         

         I = Δp

Graphing the momentum against the change in moment yields a linear relationship.

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3 years ago
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ira [324]

Answer:

1,836J

Explanation:

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myrzilka [38]

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Answer:

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