A) 
The total energy of the system is equal to the maximum elastic potential energy, that is achieved when the displacement is equal to the amplitude (x=A):
(1)
where k is the spring constant.
The total energy, which is conserved, at any other point of the motion is the sum of elastic potential energy and kinetic energy:
(2)
where x is the displacement, m the mass, and v the speed.
We want to know the displacement x at which the elastic potential energy is 1/3 of the kinetic energy:

Using (2) we can rewrite this as

And using (1), we find

Substituting
into the last equation, we find the value of x:

B) 
In this case, the kinetic energy is 1/10 of the total energy:

Since we have

we can write

And so we find:

Answer:
Nerve and muscle function and fluid balance
Explanation:
:)
Answer:
nitrogen
Explanation:
because I also had this in exam and I was correct
Answer:
you would be better off if the car bounced backwards
Explanation:
because if the hood was dismembered than you have a high chance of very bad injury but if it is just bounced back you would have less chance of getting hurt if properly sitting and seat belted.
The grandfather clock will now run slow (Option A).
<h3>What is Time Period of an oscillation?</h3>
- The time period of an oscillation refers to the time taken by an object to complete one oscillation.
- It is the inverse of frequency of oscillation; denoted by "T".
Now,
, where L is the length and g is the gravitational constant, is the formula for a pendulum's period. - The period will increase as one climbs a very tall mountain because g will slightly decrease.
- Due to this and the previous issue, the clock runs slowly and it seems that one second is longer than it actually is.
Hence, the grandfather clock will now run slow (Option A).
To learn more about the time period of an oscillation, refer to the link: brainly.com/question/26449711
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