Answer:
Part a)

Part b)

Part c)

Part d)
from t = 0 to t = 4.9 s
so the reading of the scale will be same as that of weight of the block
Then its speed will reduce to zero in next 3.2 s
from t = 4.9 to t = 8.1 s
The reading of the scale will be less than the actual mass
Explanation:
Part a)
When elevator is ascending with constant speed then we will have



So it will read same as that of the mass

Part b)
When elevator is decending with constant speed then we will have



So it will read same as that of the mass

Part c)
When elevator is ascending with constant speed 39 m/s and acceleration 10 m/s/s then we will have



Reading is given as



Part d)
Here the speed of the elevator is constant initially
from t = 0 to t = 4.9 s
so the reading of the scale will be same as that of weight of the block
Then its speed will reduce to zero in next 3.2 s
from t = 4.9 to t = 8.1 s
The reading of the scale will be less than the actual mass
Answer:
Capacitive Reactance is 4 times of resistance
Solution:
As per the question:
R = 
where
R = resistance

f = fixed frequency
Now,
For a parallel plate capacitor, capacitance, C:

where
x = separation between the parallel plates
Thus
C ∝ 
Now, if the distance reduces to one-third:
Capacitance becomes 3 times of the initial capacitace, i.e., x' = 3x, then C' = 3C and hence Current, I becomes 3I.
Also,

Also,
Z ∝ I
Therefore,




Solving the above eqn:

Answer:
When an electric field exists in a conductor a current will flow.
This implies a voltage difference between two points on the conductor.
Electrostatics pertains to static charge distributions.
That means that an object such as a charged spherical conductor will be at the same potential (voltage) on both its outer and inner surfaces.
Given: Change of x is 35.4m, Velocity Final=7.10 m/s, Velocity Initial=0m/s
Find: Acceleration
Analysis:
Vf²=Vi²+2aΔx (Velocity final squared equals Velocity initial squared plus 2 times acceleration times change of x)
(7.10 m²/s)²=(0 m/s)²+2a(35.4 m)
50.41 m/s²=(70.8 m)a
a=0.712 m/s²