Answer:
Explanation:
Given
mass of rod is M
Distance between rails is L
If the direction of magnetic field is directed upwards
If rod is moving right then,
Area enclosed is increasing therefore a Clockwise current will induce
and if rod is moving left then enclosed area is decreasing therefore an anticlockwise current will setup in the circuit
(b)Induced EMF will be
where v=velocity of rod
and current is
(c)Force necessary to move rod with constant velocity v is
External force.
Air resistance, friction, grass rubbing on a rolling ball, etc.
If there is no external force on a moving object, it keeps going.
Forever !
Answer:
1.622m/s²
Explanation:
Given the following
mass of 7.36 × 10²² kg
Radius r = 1.74×10^6m
Gravitational constant G = 6.67×10^-11
Acceleration due to gravity us expressed as:
g = GM/r²
Substitute the given values into the formula
g = 6.67×10^-11×7.36 × 10²²/(1.74×10^6)²
g = 49.0912×10¹¹/3.0276×10¹²
g = 16.22×10^{11-12}
g = 16.22×10^-1
g = 1.622m/s²
Hence the acceleration due to gravity on the Moon is 1.622m/s²
Refer to the diagram shown below.
The initial KE (kinetic energy) of the system is
KE₁ = (1/2)mu²
After an inelastic collision, the two masses stick together.
Conservation of momentum requires that
m*u = 2m*v
Therefore
v = u/2
The final KE is
KE₂ = (1/2)(2m)v²
= m(u/2)²
= (1/4)mu²
= (1/2) KE₁
The loss in KE is
KE₁ - KE₂ = (1/2) KE₁.
Conservation of energy requires that the loss in KE be accounted for as thermal energy.
Answer: 1/2
Answer:
a)P₂ =4 bar
b)W= - 1482.48 KJ
It means that work done on the system.
c)S₂ - S₁ = 3.42 KJ/K
Explanation:
Given that
T₁ = 300 K ,V₁ = 3 m³ ,P₁=2 bar
T₂ = 600 K ,V₂=V₁ 3 m³
Given that tank is rigid and insulated.It means that volume of the gas will remain constant.
Lets take the final pressure = P₂
For ideal gas P V = m R T
P₂ =4 bar
Internal energy
ΔU = m Cv ΔT
Cv=0.71 KJ/kg.k for air
m= 6.96 kg
ΔU= 6.96 x 0.71 x (600 - 300)
ΔU=1482.48 KJ
From first law
Q= ΔU + W
Q= 0 Insulated
W = - ΔU
W= - 1482.48 KJ
It means that work done on the system.
Change in the entropy
S₂ - S₁ = 3.42 KJ/K