Answer:
The correct answer is option A.
Explanation:
Force is defined as push or pull on an object with mass due to which change velocity occurs (acceleration).


if ,v = constant
Net force on the vehilce
F = Normal - Weight
0 = N = W
N = W
So, the force experienced by the object will be due to its mass, and higher the mass more will be the experienced by an object.So, large truck will experience larger net force.
Answer:
The magnitude of the induced voltage in the loop is 20 mV.
Explanation:
given;
length of loop, L = 0.43 m
width of loop,w = 0.43 m
velocity of moved loop, v = 0.15m/s
magnetic field strength,B = 0.31 T
To determine the magnitude of the induced voltage in the loop, we apply Faraday's law;
magnitude induced E.M.F = BLv
magnitude induced E.M.F = 0.31 x 0.43 x 0.15 = 0.02 V = 20 mV
Therefore, the magnitude of the induced voltage in the loop is 20 mV.
Answer:
Here is the complete question:
https://www.chegg.com/homework-help/questions-and-answers/magnetic-field-372-t-achieved-mit-francis-bitter-national-magnetic-laboratory-find-current-q900632
a) Current for long straight wire 
b) Current at the center of the circular coil 
c) Current near the center of a solenoid 
Explanation:
⇒ Magnetic Field due to long straight wire is given by (B),where

Plugging the values,
Conversion
,and 

⇒Magnetic Field at the center due to circular coil (at center) is given by,
So 
⇒Magnetic field due to the long solenoid,
Then
So the value of current are
,
and
respectively.