The only force on the system is the mass of the hoop F net = 2.8kg*9.81m/s^2 = 27.468 N The mass equal of the rolling sphere is found by: the sphere rotates around the contact point with the table.
So by applying the theorem of parallel axes, the moment of inertia of the sphere is computed by:I = 2/5*mR^2 for rotation about the center of mass + mR^2 for the distance of the axis of rotation from the center of mass of the sphere.
I = 7/5*mR^2 M = 7/5*m
Therefore, linear acceleration is computed by:F/m = 27.468 / (2.8 + 1/2*2 + 7/5*4) = 27.468/9.4 = 2.922 m/s^2
To solve this problem it is necessary to apply the concepts related to thermal stress. Said stress is defined as the amount of deformation caused by the change in temperature, based on the parameters of the coefficient of thermal expansion of the material, Young's module and the Area or area of the area.

Where
A = Cross-sectional Area
Y = Young's modulus
= Coefficient of linear expansion for steel
= Temperature Raise
Our values are given as,




Replacing we have,


Therefore the size of the force developing inside the steel rod when its temperature is raised by 37K is 38526.1N
Answer:
Part a)

Part b)

Explanation:
Part a)
Electric field due to large sheet is given as







now the electric field is given as


Part b)
Now since the electric field is required at same distance on other side
so the field will remain same on other side of the plate
