Answer:
The minimum speed the car must have at the top of the loop to not fall = 35 m/s
Explanation:
Anywhere else on the loop, the speed needed to keep the car in the loop is obtained from the force that keeps the body in circular motion around the loop which has to just match the force of gravity on the car. (Given that frictional force = 0)
mv²/r = mg
v² = gr = 9.8 × 25 = 245
v = 15.65 m/s
But at the top, the change in kinetic energy of the car must match the potential energy at the very top of the loop-the-loop
Change in kinetic energy = potential energy at the top
Change in kinetic energy = (mv₂² - mv₁²)/2
v₁ = velocity required to stay in the loop anywhere else = 15.65 m/s
v₂ = minimum velocity the car must have at the top of the loop to not fall
And potential energy at the top of the loop = mgh (where h = the diameter of the loop)
(mv₂² - mv₁²)/2 = mgh
(v₂² - v₁²) = 2gh
(v₂² - (15.65)²) = 2×9.8×50
v₂² - 245 = 980
v₂² = 1225
v₂ = 35 m/s
Hence, the minimum speed the car must have at the top of the loop to not fall = 35 m/s
Answer:D
Explanation:
Given
mass of A is twice the mass of B half the velocity of B
Suppose 
and 
be the average force exerted on A and B respectively
and According to Newton third law of motion Force on the body A is equal to Force on body B but opposite in direction as they are action and reaction force.
Thus 
and option d is correct
Explanation:
We know that,
Where,
h = Height at which pressure is to be calculated
Answer:
0.025V + (0.000218V/s³) t³
Explanation:
Parameters given:
Radius of coil, r = 3.85 cm = 0.0385 m
Number of turns, N = 450
Magnetic field, B = ( 1.20×10^(−2) T/s )t + (2.60×10^(−5) T/s4 )t^4.
The magnitude of Induced EMF is given as:
E = N * A * dB/dt
Where A is the area of the coil
First, we differentiate the magnetic field with respect to time:
dB/dt = 0.012 + 0.000104t³
Therefore, EMF will be:
E = 450 * 3.142 * (0.012 + 0.000104t³)
E = 2.096(0.012 + 0.000104t³)
E = 0.025V + (0.000218V/s³)t³
Answer:
0.4455 m
Explanation:
g = Acceleration due to gravity = 9.81 m/s²
Total mass is
Here the spring constant is not given so let us assume it as 
Here, the forces are balanced
The springs are compressed by 0.4455 m
