Answer:
The neutron loses all of its kinetic energy to nucleus.
Explanation:
Given:
Mass of neutron is 'm' and mass of nucleus is 'm'.
The type of collision is elastic collision.
In elastic collision, there is no loss in kinetic energy of the system. So, total kinetic energy is conserved. Also, the total momentum of the system is conserved.
Here, the nucleus is still. So, its initial kinetic energy is 0. So, the total initial kinetic energy will be equal to kinetic energy of the neutron only.
Now, final kinetic energy of the system will be equal to the initial kinetic energy.
Now, as the nucleus was at rest initially, so the final kinetic energy of the nucleus will be equal to the initial kinetic energy of the neutron.
Thus, all the kinetic energy of the neutron will be transferred to the nucleus and the neutron will come to rest after collision.
Therefore, the neutron loses all of its kinetic energy to nucleus.
The answer to your question is true.
Answer:
Power_input = 85.71 [W]
Explanation:
To be able to solve this problem we must first find the work done. Work is defined as the product of force by distance.

where:
W = work [J] (units of Joules)
F = force [N] (units of Newton)
d = distance [m]
We need to bear in mind that the force can be calculated by multiplying the mass by the gravity acceleration.
Now replacing:
![W = (80*10)*3\\W = 2400 [J]](https://tex.z-dn.net/?f=W%20%3D%20%2880%2A10%29%2A3%5C%5CW%20%3D%202400%20%5BJ%5D)
Power is defined as the work done over a certain time. In this way by means of the following formula, we can calculate the required power.

where:
P = power [W] (units of watts)
W = work [J]
t = time = 40 [s]
![P = 2400/40\\P = 60 [W]](https://tex.z-dn.net/?f=P%20%3D%202400%2F40%5C%5CP%20%3D%2060%20%5BW%5D)
The calculated power is the required power. Now as we have the efficiency of the machine, we can calculate the power that is introduced, to be able to do that work.
![Effic=0.7\\Effic=P_{required}/P_{introduced}\\P_{introduced}=60/0.7\\P_{introduced}=85.71[W]](https://tex.z-dn.net/?f=Effic%3D0.7%5C%5CEffic%3DP_%7Brequired%7D%2FP_%7Bintroduced%7D%5C%5CP_%7Bintroduced%7D%3D60%2F0.7%5C%5CP_%7Bintroduced%7D%3D85.71%5BW%5D)
Solution:
Let the slope of the best fit line be represented by '
'
and the slope of the worst fit line be represented by '
'
Given that:
= 1.35 m/s
= 1.29 m/s
Then the uncertainity in the slope of the line is given by the formula:
(1)
Substituting values in eqn (1), we get
= 0.03 m/s