Incomplete question as the car's speed is missing.I have assumed car's speed as 6.0m/s.The complete question is here
An amusement park ride consists of a car moving in a vertical circle on the end of a rigid boom of negligible mass. The combined weight of the car and riders is 6.00 kN, and the radius of the circle is 15.0 m. At the top of the circle, (a) what is the force FB on the car from the boom (using the minus sign for downward direction) if the car's speed is v 6.0m/s
Answer:
Explanation:
Set up force equation
∑F=ma
∑F=W+FB
The minus sign for downward direction
Answer:
μ = tanθ = tan30 = 0.58
Explanation:
μ = force parallel/force perpendicular = mgsinθ/mgcosθ = tanθ
Answer:
Explanation:
Constant pressure molar heat capacity Cp = 29.125 J /K.mol
If Cv be constant volume molar heat capacity
Cp - Cv = R
Cv = Cp - R
= 29.125 - 8.314 J
= 20.811 J
change in internal energy = n x Cv x Δ T
n is number of moles , Cv is molar heat capacity at constant volume , Δ T is change in temperature
Putting the values
= 20 x 20.811 x 15
= 6243.3 J.
Answer:
0.026
Explanation:
The force of friction acts in the direction perpendicular to the norm force of the surface on which the object rests, induced by gravity. The magnitude of the friction force is
(Friction) = (mass) x (gravitational acceleration g) x (coefficient of friction)
from which the coefficient of friction can be determined:
(coefficient of friction) = (Friction) / ((mass)x(g)) = 3 N / (12 kg * 9.8 m/s^2) = 0.026
A bond with elements from B.
