Complete Question
The complete question is is shown on the first uploaded
Answer:
The elastic potential energy at point B is
The kinetic energy at point D is 
Explanation:
Looking at the given point we can observe that mechanically energy(i.e potential and kinetic energy ) is conserved and it value is 
So at point B


KE at point B is 50J
So 
Now at point D

at point D is 25J
So 
To solve the problem it is necessary to apply the Torque equations and their respective definitions.
The Torque is defined as,

Where,
I=Inertial Moment
Angular acceleration
Also Torque with linear equation is defined as,

Where,
F = Force
d= distance
Our dates are given as,
R = 30 cm = 0.3m
m = 1.5 kg
F = 20 N
r = 4.0 cm = 0.04 m
t = 4.0s
Therefore matching two equation we have that,

For a wheel the moment inertia is defined as,
I= mR2, replacing we have





Then the velocity of the wheel is

Therefore the correct answer is D.
Answer:3.87*10^-4
Explanation:
What is the decrease in mass, delta mass Xe , of the xenon nucleus as a result of this deca
We have been given the wavelength of the gamma ray, find the frequency using c = freq*wavelength.
C=f*lambda
3*10^8=f*3.44*10^-12
F=0.87*10^20 hz
Then with the frequency, find the energy emitted using equation
E=hf E = freq*Plank's constant
E=.87*10^20*6.62*10^-34
E=575.94*10^(-16)
With this energy, convert into MeV from joules.
With the energy in MeV, use E=mc^2 using c^2 = 931.5 MeV/u.
Plugging and computing all necessary numbers gives you
3.87*10^-4 u.
Answer:
The speed of ejection is 
Solution:
As per the question:
Magnetic field density, B = 0.4 T
Density of the material in the sunspot, 
Now,
To calculate the speed of ejection of the material, v:
The magnetic field energy density is given by:

This energy density equals the kinetic energy supplied by the field.
Thus


where
m = mass of the sunspot in
= 

