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.
45mph is the answer if you do the math right
Answer:
DOUBLE CHECK BECUASE IM ONLY 68.030303039999999% SURE!!!
(ANSWER IS HERE) ( D) It lacked practical examples in supporting theory
Know it's not B becuase there was no scientific community back then.
Know it's not C becuase it actully had lots of evidence.
But I'm not sure about A
(a) 328.6 kg m/s
The linear impulse experienced by the passenger in the car is equal to the change in momentum of the passenger:

where
m = 62.0 kg is the mass of the passenger
is the change in velocity of the car (and the passenger), which is

So, the linear impulse experienced by the passenger is

(b) 404.7 N
The linear impulse experienced by the passenger is also equal to the product between the average force and the time interval:

where in this case
is the linear impulse
is the time during which the force is applied
Solving the equation for F, we find the magnitude of the average force experienced by the passenger:
