A bathroom scale is compressed 0.11 in. when a 110-lb person stands on it. Assuming that the scale behaves like a spring that o
1 answer:
Answer:
(a) 220 lb
(b) 10.75 in-lb
Explanation:
(a) We know that if a 110 lb person compresses the spring 0.11 inches, then a person weighing double that amount will compress the spring double, i.e. the person would weigh 220 lb.
(b) The amount of work done in compressing the spring 0.11 inch is given by the integral:
110/.11=1210
=10.75 in-lb
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Explanation:
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As we can see from above that v is higher when h ishigher.
The final speed of the orange is 7.35 m/s
Explanation:
The motion of the orange is a free fall motion, since there is only the force of gravity acting on it. Therefore, it is a uniformly accelerated motion with constant acceleration towards the ground. So we can use the following suvat equation:
where
v is the final velocity
u is the initial velocity
a is the acceleration
t is the time elapsed
For the orange in this problem, we have
u = 0 (it is dropped from rest)
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Substituting t = 0.75 s, we find the final velocity (and speed) of the orange:
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