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Suppose the elevator starts from rest and maintains a constant upward acceleration of 2.90 m/s2 . A bolt in the elevator ceiling 3.00 m above the elevator floor works loose and falls. It falls out the instant the elevator begins to move. a.)How long does it take for the bolt to reach the floor of the elevator? b.)Just as it reaches the floor, how fast is the bolt moving according to an observer in the elevator? c.)Just as it reaches the floor, how fast is the bolt moving according to an observer standing on the floor landings of the building? d.)According to an observer in the elevator, how far has the bolt traveled between the ceiling and floor of the elevator? e.)According to an observer standing on the floor landings of the building, how far has the bolt traveled between the ceiling and floor of the elevator?

Respuesta :

The correct answer is 2.26 m.

Initial velocity of elevator u =0

Constant upward acceleration of the elevator a= 2.90 m/s²

Height of the bolt from the floor of the elevator h=3.00 m

Initial velocity of the bolt in the elevator u = 0 m/s

Elevator is moves upward with a constant acceleration 2.90 m/s²

so the net acceleration a' = (a+g) m/s²

=( 2.90 + 9.8) m/s²

a' = 12.7 m/s²

Initial velocity of the bolt u = 0 m/s

Distance traveled by the bolt h = 3.00 m

Substitute these values in the equation

h=ut+ 1/2 ( a )t², we get

3.00 = (0)x+1/2 (12.7)(t^2)

3.00 = 6.35 t²

t²= 3.00/6.35

t = 0.68 s

i.e., after 0.68 s the bolt reaches the floor of the elevator.

(b)For the observer in the elevator acceleration

a'=(a+g) m/s²

=( 2.90 + 9.8) m/s²

a'= 12.7 m/s²

Time taken bolt to reach the floor of the elevator t = 0.68 s From the equation v=u+a't we get

v=(0)+(12.7) (0.68)

v = 8.636 m/s

Velocity of the bolt when it reaches the floor of the elevator according to an observer into elevator 8.636 m/s

ii) An observer standing on the floor landings of the buildings: for observer standing on the floor landing, the acceleration of the

bolt

appears a = g m/s²

= 9.8m/s²

Time taken the bolt to reach the floor t = 0.68 s

Initial velocity of the bolt u = 0 m/s

Substitute these values in the equation

v=u+at we get

v=0+(9.8) (0.68 s)

v= 8.636 m/s

Velocity of the bolt when it reaches the floor of the elevator according to

an observer standing on the floor landing = 8.636  m/s

c) i) According to an observer in the elevator in this situation acceleration a= 12.7 m/s²

time t= 0.68 s initial velocity u = 0

substitute these values in the equation 1

s = ut +1/2 at²

We get

s =(0) (0.659)+1/2(12.7 ) (0.68)²

= 2.93 m

Distance traveled by the bolt according to the observer on the elevator 2.93 m

iii)According to an observer standing on the floor landing in this situation

acceleration a=g m/s²

= 9.8m/s²

Time t= 0.68 s

Initial velocity u = 0 m/s Substitute these values in the equation

1 ut + 1/2 at ²,

s=ut+1/2at²

we get

s=(0)+1/2(9.8) (0.68)²

= 2.26 m

Distance traveled by the bolt according to the observer standing on the floor landing 2.26 m.

To learn more about acceleration refer the link:

https://brainly.com/question/18623334

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