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.
I believe it’s the mass of the box but I don’t no if I’m right
Hope this helped
Answer:
a) 5.5×10^17 Hz
b) visible light
Explanation:
Since the wavelength of the electromagnetic radiation must be about the size of the about itself, this implies that;
λ= 5.5 × 10^-10 m
Since;
c= λ f and c= 3×10^8 ms-1
f= c/λ
f= 3×10^8/5.5 × 10^-10
f= 5.5×10^17 Hz
The electromagnetic wave is visible light
Training everyday
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1) At the moment of being at the top, the piston will not only tend to push the penny up but will also descend at a faster rate at which the penny can reach in 'free fall', in that short distance. Therefore, at the highest point, the penny will lose contact with the piston. Therefore the correct answer is C.
2) To solve this problem we will apply the equations related to the simple harmonic movement, hence we have that the acceleration can be defined as
![a = -\omega^2 A](https://tex.z-dn.net/?f=a%20%3D%20-%5Comega%5E2%20A)
Where,
a = Acceleration
A = Amplitude
= Angular velocity
From a reference system in which the downward acceleration is negative due to the force of gravity we will have to
![a = -g](https://tex.z-dn.net/?f=a%20%3D%20-g)
![-\omega^2 A = -g](https://tex.z-dn.net/?f=-%5Comega%5E2%20A%20%3D%20-g)
![\omega = \sqrt{\frac{g}{A}}](https://tex.z-dn.net/?f=%5Comega%20%3D%20%5Csqrt%7B%5Cfrac%7Bg%7D%7BA%7D%7D)
From the definition of frequency and angular velocity we have to
![\omega = 2\pi f](https://tex.z-dn.net/?f=%5Comega%20%3D%202%5Cpi%20f)
![f = \frac{1}{2\pi} \sqrt{\frac{g}{A}}](https://tex.z-dn.net/?f=f%20%3D%20%5Cfrac%7B1%7D%7B2%5Cpi%7D%20%5Csqrt%7B%5Cfrac%7Bg%7D%7BA%7D%7D)
![f = \frac{1}{2\pi} \sqrt{\frac{9.8}{4*10^{-2}}}](https://tex.z-dn.net/?f=f%20%3D%20%5Cfrac%7B1%7D%7B2%5Cpi%7D%20%5Csqrt%7B%5Cfrac%7B9.8%7D%7B4%2A10%5E%7B-2%7D%7D%7D)
![f = 2.5Hz](https://tex.z-dn.net/?f=f%20%3D%202.5Hz)
Therefore the maximum frequency for which the penny just barely remains in place for the full cycle is 2.5Hz