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.
An object in motion stays in motion while an object at rest stays at rest.
If a ship will be sailing through warm and cold water, people think about making it less dense than the warmest water as they load the ship with cargo. I think you forgot to give the options along with the question. I hope that this is the answer that has actually come to your desired help.
Answer: equation for the reaction is given below
PCL2+PCL3=PCL5
Where pcl2=0.40atm,pcl3=0.27atm
Pcl5=0.0029atm
Using ∆G=-RTin(PCL5/PCl2*PCL3)
Where R=8.314J/K/mol and T=298K
∆G=-8.314*298in(0.0029/0.40*.27)
∆G=8962.6J/mol
Explanation:
Answer:
The minimum coefficient of friction is 0.544
Solution:
As per the question:
Radius of the curve, R = 48 m
Speed of the car, v = 16 m/s
To calculate the minimum coefficient of static friction:
The centrifugal force on the box is in the outward direction and is given by:

where
= coefficient of static friction
The net force on the box is zero, since, the box is stationary and is given by: