Answer:
Part a)

Part b)

Explanation:
Part A)
As we know that time period of the motion is given as

so we have



now at the point of maximum amplitude the force equation when Normal force is about to zero is given as

so we have



Part b)
Now if the amplitude of the SHM is 6.23 cm
and now at this amplitude if object will lose the contact then in that case again we have



so now we have


Answer:
C)the right cord pulls on the pulley with greater force than the left cord
Explanation:
As we can see the figure that B is connected to the right string while A is connected to the left string
Now force equation for B as it will move downwards is given as

similarly for block A which will move upwards we can write the equation as

now we know that pulley is also rotating so the tangential acceleration of the rope at the contact point with pulley must be same as that of acceleration of the blocks
so here pulley will rotate clockwise direction
So tension in the right string must be more than the left string
So correct answer will be
C)the right cord pulls on the pulley with greater force than the left cord
Answer:
8 signals received by twin A during the trip.
Explanation:
Given that,
Distance = 12 light year
Speed = 0.6 c
Time = 1 year
We need to calculate the time by A
Using formula of time

Put the value into the formula


Similarly,
The expression for distance cover by A



We need to calculate the time
Using formula of time



We need to calculate the signals received by twin A
Using formula for number of signals

Put the value into the formula


Hence, 8 signals received by twin A during the trip.
Whenever an object is falling, its potential energy
is decreasing and its kinetic energy is increasing.
Olivia's potential energy is decreasing and her kinetic energy
is increasing as she moves toward the right side of the picture,
all the way from W, through X, to the bottom of the arc.
Answer:
The maximum pressure that will be attained in the tank before the plug melts and releases gas should be less than 74.26 atm.
Explanation:
To calculate the final pressure of the system, we use the equation given by Gay-Lussac Law. This law states that pressure of the gas is directly proportional to the temperature of the gas at constant pressure.
Mathematically,

where,
are the initial pressure and temperature of the gas.
are the final pressure and temperature of the gas.
We are given:

Putting values in above equation, we get:

The maximum pressure that will be attained in the tank before the plug melts and releases gas should be less than 74.26 atm.