<span>Crust. The thin solid outermost layer of Earth. ...Asthenosphere. The lower layer of the crust. ...Lithosphere.Plasticity: is solid but still being able to. flow without being a liquid.The cool, rigid outermost layer of the Earth. ...<span>the solid part of the earth consisting of the crust and outer mantle.</span></span>
Answer:
19.3m/s
Explanation:
Use third equation of motion

where v is the velocity at halfway, u is the initial velocity, g is gravity (9.81m/s^2) and h is the height at which you'd want to find the velocity
insert values to get answer
![v^2-0^2=2(9.81m/s^2)(38/2)\\v^2=9.81m/s^2 *38\\v^2=372.78\\v=\sqrt[]{372.78} \\v=19.3m/s](https://tex.z-dn.net/?f=v%5E2-0%5E2%3D2%289.81m%2Fs%5E2%29%2838%2F2%29%5C%5Cv%5E2%3D9.81m%2Fs%5E2%20%2A38%5C%5Cv%5E2%3D372.78%5C%5Cv%3D%5Csqrt%5B%5D%7B372.78%7D%20%5C%5Cv%3D19.3m%2Fs)
<span>when it returns to its original level after encountering air resistance, its kinetic energy is
decreased.
In fact, part of the energy has been dissipated due to the air resistance.
The mechanical energy of the ball as it starts the motion is:
</span>

<span>where K is the kinetic energy, and where there is no potential energy since we use the initial height of the ball as reference level.
If there is no air resistance, this total energy is conserved, therefore when the ball returns to its original height, the kinetic energy will still be 100 J. However, because of the presence of the air resistance, the total mechanical energy is not conserved, and part of the total energy of the ball has been dissipated through the air. Therefore, when the ball returns to its original level, the kinetic energy will be less than 100 J.</span>
<span>Wedges is your answer please mark brainliest </span>
Answer:
0.22 b
Explanation:
Quadrupole moment of the nucleon is,

And also,

And, 
Now,

For Bismuth
and A is 209.

Therefore, the expected value of quadrupole is 0.22 b which is quite related with experimental value which is 0.37 b