To solve the problem it is necessary to apply the concepts given in the kinematic equations of angular motion that include force, acceleration and work.
Torque in a body is defined as,

And in angular movement like

Where,
F= Force
d= Distance
I = Inertia
Acceleration Angular
PART A) For the given case we have the torque we have it in component mode, so the component in the X axis is the net for the calculation.

On the other hand we have the speed data expressed in RPM, as well


Acceleration can be calculated by



In the case of Inertia we know that it is equivalent to


Matching the two types of torque we have to,




PART B) The work performed would be calculated from the relationship between angular velocity and moment of inertia, that is,



The chromosome pair that attaches to fiber stretching is called Centrioles
Answer:
Part a)

Part b)

Part C)

Part d)
Due to large magnitude of friction between road and the car the momentum conservation may not be valid here as momentum conservation is valid only when external force on the system is zero.
Explanation:
Part a)
As we know that car A moves by distance 6.1 m after collision under the frictional force
so the deceleration due to friction is given as



now we will have




Part b)
Similarly for car B the distance of stop is given as 4.4 m
so we will have


Part C)
By momentum conservation we will have



Part d)
Due to large magnitude of friction between road and the car the momentum conservation may not be valid here as momentum conservation is valid only when external force on the system is zero.
Answer:
Number of Significant Figures: 2
The Significant Figures are 3 6
Explanation:
= 3.60 × 102
(scientific notation)
= 3.60e2
(scientific e notation)
= 360 × 100
(engineering notation)
(one)
= 360
(real number)
W = mg, Assuming g ≈ 9.8 m/s² on the earth surface.
735 N = m* 9.8
735/9.8 = m
75 = m
Mass , m = 75 kg. B.