Answer:
0.07756 m
Explanation:
Given mass of object =0.20 kg
spring constant = 120 n/m
maximum speed = 1.9 m/sec
We have to find the amplitude of the motion
We know that maximum speed of the object when it is in harmonic motion is given by
where A is amplitude and
is angular velocity
Angular velocity is given by
where k is spring constant and m is mass
So 

I have three problems with this question.
#1). If you copied the question exactly the way it's written,
then the question is written very badly. The wording is
misleading, and the more you try to think about it and
puzzle it out, the more it'll damage your understanding
of Physics.
There is no relationship between the force exerted on an
elevator and the distance the elevator is lifted.
-- If the force is anything more than the weight of the elevator ...
even one ounce more ... then it'll lift the elevator as high as
you want.
-- If the force is anything less than the weight of the elevator ...
even one ounce less, then that elevator is headed for the bottom.
#2). You didn't post any graph below, so if we need the graph
to answer the question, then we can't answer the question.
#3). I guess that's OK, because you didn't ask any question.
Every one has a reason in life. You may not know what it is just yet, but you do.
Answer: The magnitude of the force exerted on the roof is 490522.5 N.
Explanation:
The given data is as follows.
Below the roof,
= 0 m/s
At top of the roof,
= 39 m/s
We assume that
is the pressure at lower surface of the roof and
be the pressure at upper surface of the roof.
Now, according to Bernoulli's theorem,


= ![0.5 \times 1.29 \times [(39)^{2} - (0)^{2}]](https://tex.z-dn.net/?f=0.5%20%5Ctimes%201.29%20%5Ctimes%20%5B%2839%29%5E%7B2%7D%20-%20%280%29%5E%7B2%7D%5D)
= 
= 981.045 Pa
Formula for net upward force of air exerted on the roof is as follows.
F = 
= 
= 490522.5 N
Therefore, we can conclude that the magnitude of the force exerted on the roof is 490522.5 N.
Answer:
51.85m/s
Explanation:
Given parameters:
Mass of ball = 0.0459kg
Force = 2380N
Time taken = 0.001s
Unknown:
Speed of the ball afterwards = ?
Solution:
To solve this problem, we use Newton's second law of motion:
F = m x
F is the force
m is the mass
v is the final velocity
u is the initial velocity
t is the time taken
2380 = 0.0459 x
0.0459v = 2.38
v = 51.85m/s