Explanation:
It is given that,
Speed of the person, v = 3 mile/hr = 1.34 m/s
Speed of the truck, v' = 65 mile/hr = 29.05 m/s
(a) Since, 
For the person, 
For the truck, 
(b) The relativistic factor is given by :
For very small velocity, 
For the person :
For the person :
Hence, this is the required solution.
The change of speed is proportional to the Force. If you double the Force, you double the change of speed.
Answer:
Work done is 882000joule.
power is 29400watt.
Explanation:
given,
Mass(m)=500kg
Acceleration due to gravity(g)=9.8m/s²
Height(h)=6m
Time taken(t)=30s
Workdone=?
Power=?
now,
workdone=force*displaxement
= m*g*h
=500*9.8*6
=8,82,000joule
so, the work done by the man is 8,82,000joule.
then,
power=workdone/time taken
=8,82,000/30
=29,400watt
so, the required power to lift a load is 29,400watt.
<span>a)
Capacitance = k x ε° x area / separation
ε° = 8.854 10^-12 F/ m
k = 2.4max
average k = 0.78 / 1.27 * 2.4 +(1.27- 0.78) / 1.27 * 1 = 1.474 + 0.386 = 1.86
(61.4 % separation k = 2.4 --- 38.6 % k = 1 air --- average k = 0.614 * 2.34 + 0.386 * 1 = 1.86
area = 145 cm2 = 0.0145 m2
separation = 1.27 cm 0.0127 m
C = 1.86 * 8.854 10^-12 * 0.0145 / 0.0127 = 18.8 pF
b) Q = C * V --- 18.8 * 83 = 1560.4 pC = 1.5604 nC
c) E = V / d = 83 / 0.0127 = 6535.4 V/m </span>
