Answer:
v = 12.12 m/s
Explanation:
Given that,
Radius of the curvature, r = 30 m
To find,
The car's speed at the bottom of the dip.
Solution,
Let mg is the true weight of the passenger. When it is moving in the circular path, the centripetal force act on it. It is given by :

The normal reaction of the passenger is given by :

N = 1.5 mg
Let v is the car's speed at the bottom of the dip. It can be calculated as:



v = 12.12 m/s
So, the speed of the car at the bottom of the dip is 12.12 m/s. Hence, this is the required solution.
Answer:
0.12959085 J
Explanation:
k = Coulomb constant = 
q = Charge = 1.55 μC
d = Distance between charge = 0.5 m
Electric potential energy is given by

In this system with three charges which are equidistant from each other


The potential energy of the system is 0.12959085 J
Formula: Ca=Vt^2/r
centripetal acceleration(Ca)= ?
tangential speed(Vt)= 2.8m/s
Radius(r)=2m
Substitute: Ca=2.8^2/2
Ca=3.92m/s^2
The engine's efficiency is (35J)/125J) = 28% .
Do you have a different question to ask ?
Answer:
a. No, typical small pickup truck springs are not large enough to compress 0.61{\rm m}.
Explanation: