Answer:
Elastic Collision
Inelastic Collision
The total kinetic energy is conserved. The total kinetic energy of the bodies at the beginning and the end of the collision is different.
Momentum does not change. Momentum changes.
No conversion of energy takes place. Kinetic energy is changed into other energy such as sound or heat energy.
Highly unlikely in the real world as there is almost always a change in energy. This is the normal form of collision in the real world.
An example of this can be swinging balls or a spacecraft flying near a planet but not getting affected by its gravity in the end.
Answer:
d) 2Fr
Explanation:
We know that the work done in moving the charge from the right side to the left side in the k shell is W = ∫Fdr from r = +r to -r. F = force of attraction between nucleus and electron on k shell. F = qq'/4πε₀r² where q =charge on electron in k shell -e and q' = charge on nucleus = +e. So, F = -e × +e/4πε₀r² = -e²/4πε₀r².
We now evaluate the integral from r = +r to -r
W = ∫Fdr
= ∫(-e²/4πε₀r²)dr
= -∫e²dr/4πε₀r²
= -e²/4πε₀∫dr/r²
= -e²/4πε₀ × -[1/r] from r = +r to -r
W = e²/4πε₀[1/-r - 1/+r] = e²/4πε₀[-2/r} = -2e²/4πε₀r.
Since F = -e²/4πε₀r², Fr = = -e²/4πε₀r² × r = = -e²/4πε₀r and 2Fr = -2e²/4πε₀r.
So W = -2e²/4πε₀r = 2Fr.
So, the amount of work done to bring an electron (q = −e) from right side of hydrogen nucleus to left side in the k shell is W = 2Fr
Answer:
First, let’s correct the question. Acceleration is the rate of change in velocity. Its unit therefore is ft/sec/sec. If S is the distance traveled for a given duration, S = Vot + (1/2)at^2 where Vo is the initial velocity, a is the acceleration and t is the time. For Vo = 0, a = 6m/sec/sec and t = 3 sec. The distance traveled is S = 0 + (1/2) x 6 x 3^2 = 27 meters
Answer:
5. 9GmM/(10R)
Explanation:
m is the mass of the satellite
M is the mass of the earth
W is the energy required to launch the satellite
Energy at earth surface = Potential energy (PE) + W
W = Energy at earth surface - Potential energy (PE)
But PE = 
Therefore: W = Energy at earth surface - 
Energy at earth surface (E) at an altitude of 5R = 
But 
Therefore: 
W = E - PE
Answer:
q = 2.65 10⁻⁶ C
Explanation:
For this exercise we use Coulomb's law
F =
In this case they indicate that the load is of equal magnitude
q₁ = q₂ = q
the force is attractive because the signs of the charges are opposite
F = 
q = 
we calculate
q = 
q = 
Ra 7 10-12
q = 2.65 10⁻⁶ C
