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1 answer:
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Answer:
Minimum time interval (t2)=0.90 SECONDS
Explanation:
- coefficient of friction for employees footwear = 0.5
- coefficient of friction for typical athletic shoe = 0.810
- frictional force = coefficient of friction X acceleration due to gravity X mass of body
- Acceleration due to gravity is a constant = 9.81 m/s
- Let frictional force for employee footwear = FF1
- Let frictional force for athletic footwear =FF2
FF1 = O.5 X 9.81 X mass of body
= 4.905 x mass of body
FF2 = 0.810 X 9.81 X mass of body
= 7.9461 x mass of body
The body started from rest there by making the initial velocity zero ( u = 0)
From d= ut + 1/2 a x
- d =
x a x
.....................................i
where d= distance and it is given as 3.25m
- F =ma ...................................ii
making acceleration subject of the formula from equation ii
- a =
Making t subject of formula from equation (i)
- t=
where
-
= 4.905
=7.9461
Let
- t1 = minimum time taken for frictional force for employee foot wear
- t1 =
=1.15 seconds
- t2 =
= 0.90 seconds
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The correct answer is
<span>C) -10.7 m/s
In fact, the first rock is moving upward with velocity +4.5 m/s, while the second rock is moving downward with velocity -6.2 m/s, with respect to a fixed reference frame. In the reference frame of the first rock, instead, the second rock is moving with velocity equal to its velocity in the fixed frame minus the velocity of the reference frame of the first rock:
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<span>
</span>
<span>C) -10.7 m/s
In fact, the first rock is moving upward with velocity +4.5 m/s, while the second rock is moving downward with velocity -6.2 m/s, with respect to a fixed reference frame. In the reference frame of the first rock, instead, the second rock is moving with velocity equal to its velocity in the fixed frame minus the velocity of the reference frame of the first rock:
</span>
</span>