Answer:
The same amount of energy is required to either stretch or compress the spring.
Explanation:
The amount of energy required to stretch or compress a spring is equal to the elastic potential energy stored by the spring:

where
k is the spring constant
is the stretch/compression of the spring
In the first case, the spring is stretched from x=0 to x=d, so

and the amount of energy required is

In the second case, the spring is compressed from x=0 to x=-d, so

and the amount of energy required is

so we see that the amount of energy required is the same.
Answer:
The the linear speed (in m/s) of a point on the rim of this wheel at an instant=0.418 m/s
Explanation:
We are given that
Angular acceleration, 
Diameter of the wheel, d=21 cm
Radius of wheel,
cm
Radius of wheel, 
1m=100 cm
Magnitude of total linear acceleration, a=
We have to find the linear speed of a at an instant when that point has a total linear acceleration with a magnitude of 1.7 m/s2.
Tangential acceleration,


Radial acceleration,
We know that

Using the formula

Squaring on both sides
we get






Hence, the the linear speed (in m/s) of a point on the rim of this wheel at an instant=0.418 m/s
These are characteristics of a wave. The amplitude is how high and low the waves go. Crests are high points on the wave, and troughs are low points on the wave.
Answer:
In odd nuclei, the left out proton or neutron will contribute to the spin of the nucleus.
Explanation:
The meaning of odd nuclei is atomic mass is odd.
A=odd number.
A=Z+n
Here, Z is proton either it will odd or n will odd which is neutron.
Now according to the shell model the left out proton or neutron will contribute to the spin and parity.
For example,
Take the case of isotope of nitrogen-15.
Here Z is 7, and n is 8 will not contribute in spin.
Now, for Z=7.

Here,

and, L=1.
Fort parity,

Put the value of L.
Parity will be -1.
Now, spin will be
.
Answer:
Stars create new elements in their cores by squeezing elements together in a process called nuclear fusion.
Explanation: