V = 1/3 Bh v = 1/3 (13 ac)(43560ft^2/ac)(481ft) v = 90793560 ft^3 * 0.3048m/ft * 0.3048m/ft * 0.3048m/ft = 2570987m^3
Answer
given,
mass of copper rod = 1 kg
horizontal rails = 1 m
Current (I) = 50 A
coefficient of static friction = 0.6
magnetic force acting on a current carrying wire is
F = B i L
Rod is not necessarily vertical


the normal reaction N = mg-F y
static friction f = μ_s (mg-F y )
horizontal acceleration is zero


B_w = B sinθ
B_d = B cosθ
iLB cosθ= μ_s (mg- iLB sinθ)





B = 0.1 T
This is A.) lake. A river is a small amount of water that isn't always fresh water. A stream is too small. And an ocean is made of salt water.
The correct answer to the question is : 9375 N.
CALCULATION:
As per the question, the mass of the car m = 1500 Kg.
The diametre of the circular track D = 200 m.
Hence, the radius of the circular path R = 
= 
= 100 m.
The velocity of the truck v = 25 m/s.
When a body moves in a circular path, the body needs a centripetal force which helps the body stick to the orbit. It acts along the radius and towards the centre.
Hence, the force acting on the car is centripetal force.
The magnitude of the centripetal force is calculated as -
Force F = 
= 
= 9375 N. [ANS}
The centripetal force is provided to the car in two ways. It is the friction which provides the necessary centripetal force. Sometimes friction is not sufficient. At that time, the road is banked to some extent which provides the necessary centripetal force.