The velocity is 14 m/s
The parameters given on the question are
mass= 0.060 kg
kinetic energy= 5.9 joules
K.E= 1/2mv²
5.9= 1/2 × 0.060 × v²
5.9= 0.5 × 0.060v²
5.9= 003v²
v²= 5.9/0.03
v²= 196.66
v= √196.66
v= 14 m/s
Hence the velocity of the egg before it strikes the ground is 14 m/s
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Explanation:
It is given that,
Number of turns, N = 200
Area of cross section, A = 8.5 cm²
Magnetic field is directed out of the paper and is, B = 0.06 T
The magnetic field is out of the paper decreases to 0.02 T in 12 milliseconds. We need to find the direction of current induced. The induced emf is given by :

Since, 
I is the induced current

According to Lenz's law, the direction of induced current is such that it always opposes the change in current that causes it.
Here, the field is directed out of the plane of the paper, this gives the induced current in counterclockwise direction.
Answer:
1807.56 kJ
Explanation:
Parameters given:
Current, I = 8.9A
Time, t = 4.7hrs = 4.7 * 3600 = 16920 secs
Voltage, V = 12V
Electrical energy is given as:
E = I*V*t
Where I = Current
V = Voltage/Potential differenxe
t = time in seconds.
E = 8.9 * 12 * 16920
E = 1807056 J = 1807.056 kJ