Answer:
We can conclude that there is a decrease in kinetic energy of the particles due to their elastic collision, since kinetic energy is directly proportional to squared velocity of the particles.
Explanation:
Given:
initial velocity of particle A, Ua = 5m/s
initial velocity of particle B, Ub = 10 m/s
final velocity of particle A, Va = 4m/s
final velocity of particle B, Vb = 7m/s
For particle A:
The final velocity is 1 less than the initial velocity.
For particle B:
The final velocity is 3 less than the initial velocity.
We can conclude that there is a loss in kinetic energy due to elastic collision of the two particles, since kinetic energy is directly proportional to squared velocity of the particles. A decrease in velocity means decrease in kinetic energy.
Answer:
The gravitational potential energy of a system is -3/2 (GmE)(m)/RE
Explanation:
Given
mE = Mass of Earth
RE = Radius of Earth
G = Gravitational Constant
Let p = The mass density of the earth is
p = M/(4/3πRE³)
p = 3M/4πRE³
Taking for instance,a very thin spherical shell in the earth;
Let r = radius
dr = thickness
Its volume is given by;
dV = 4πr²dr
Since mass = density* volume;
It's mass would be
dm = p * 4πr²dr
The gravitational potential at the center due would equal;
dV = -Gdm/r
Substitute (p * 4πr²dr) for dm
dV = -G(p * 4πr²dr)/r
dV = -G(p * 4πrdr)
The gravitational potential at the center of the earth would equal;
V = ∫dV
V = ∫ -G(p * 4πrdr) {RE,0}
V = -4πGp∫rdr {RE,0}
V = -4πGp (r²/2) {RE,0}
V = -4πGp{RE²/2)
V = -4Gπ * 3M/4πRE³ * RE²/2
V = -3/2 GmE/RE
The gravitational potential energy of the system of the earth and the brick at the center equals
U = Vm
U = -3/2 GmE/RE * m
U = -3/2 (GmE)(m)/RE
Answer:
why should we do , do by your own , no sense
Explanation:
Answer:
0 kg m/s before and after collision
Explanation:
Let m, v be the mass and speed of the 2 balls, respectively, before the collision. Since they have the same mass and same speed but in opposite direction, the total momentum of the system would be:
P = mv - mv = 0 kg m/s
As the collision is elastic. The total momentum after the collision is the same as the total momentum before the collision, which is 0.
Answer:
Explanation:
Potential due to a charged metallic sphere having charge Q and radius r on its surface will be
v = k Q / r . On the surface and inside the metallic sphere , potential is the same . Outside the sphere , at a distance R from the centre potential is
v = k Q / R
a ) On the surface of the shell , potential due to positive charge is
V₁ = 
On the surface of the shell , potential due to negative charge is
V₁ = 
Total potential will be zero . they will cancel each other.
b ) On the surface of the sphere potential
= 
= 22.5 x 10⁵ V
On the surface of the sphere potential due to outer shell
= 
= -9 x 10⁵
Total potential
=( 22.5 - 9 ) x 10⁵
= 13.5 x 10⁵ V
c ) In the space between the two , potential will depend upon the distance of the point from the common centre .
d ) Inside the sphere , potential will be same as that on the surface that is
13.5 x 10⁵ V.
e ) Outside the shell , potential due to both positive and negative charge will cancel each other so it will be zero.
