Answer:
a
The current that would be produced is 
b
Yes this arrangement can be used to cancel out earths magnetic field at points well above the Earth's surface.This is because this current loop acts as a magnetic dipole for point above the earth surface
c
No this arrangement can not be used to cancel out earths magnetic field at points on the Earth's surface .this because on the earth surface it shifts from its behavior as a magnetic dipole
Explanation:
From the question we are told that
The magnetic moment of earth is 
The radius of earth generally has a value of 
Magnetic moment is mathematically given as

A is the area of the of the earth(assumption that the earth is circular ) and this evaluated as

Now making
the subject in the above formula




Using the formula:
w = m x g ....... eq1
here w is weight of the object.
m is mass of the object, and
g is the acceleration of gravity.
mass, m = 14 lbm (given)
acceleration of gravity, g= 32.0 ft/
Now, substituting the values in equation (1):
w = 14lbm x 32.0 ft/
= 448 lbm ft/
since, 1 lbf = 32.174 lbm ft/
so, w = 448 x 
w = 13.924lbf
Hence, the mass of an object is 13.924 lbf.
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Answer:
If there is a net force acting on an object, the object will have an acceleration and the object's velocity will change. ... Newton's second law states that for a particular force, the acceleration of an object is proportional to the net force and inversely proportional to the mass of the object.
Explanation:
Answer:
250,000 Joules
Explanation:
The work required is the amount of energy added, so the answer is the difference in kinetic energy before and after the acceleration of the car. (The problem doesn't say it's flat where the car is, but well assume it since there is no information about an elevation change that would require us to concern ourselves with potential energy!)
W = E_final - E_initial
= 1/2 m v_final^2 - 1/2 m v_initial^2
= 1/2 m (v_final^2 - v_initial^2)
= 1/2 (1000 kg) ( (30 m/s)^2 - (20 m/s^2) )
= 1/2 (1000 kg) (500 m^2/s^2)
= 250,000 kg m^2/s^2
= 250,000 Joules