Volume=mass/density
volume=455.6/19.3
volume=23.6 mL
For part a)
Since the conical surface is not exposed to the radiation coming from the walls only from the circular plate and assuming steady state, the temperature of the conical surface is also equal to the temperature of the circular plate. T2 = 600 K
For part b)
To maintain the temperature of the circular plate, the power required would be calculated using:
Q = Aσ(T₁⁴ - Tw⁴)
Q = π(500x10^-3)²/4 (5.67x10^-8)(600⁴ - 300⁴)
Q = 5410.65 W
To prevent the crate from slipping, the maximum force that the belt can exert on the crate must be equal to the static friction force.
Ff = 0.5 * 16 * 9.8 = 78.4 N
a = 4.9 m/s^2
If acceleration of the belt exceeds the value determined in the previous question, what is the acceleration of the crate?
In this situation, the kinetic friction force is causing the crate to decelerate. So the net force on the crate is 78.4 N minus the kinetic friction force.
Ff = 0.28 * 16 * 9.8 = 43.904 N
Net force = 78.4 – 43.904 = 34.496 N
To determine the acceleration, divide by the mass of the crate.
a = 34.496 ÷ 16 = 2.156 m/s^2
The finer‐grained<span> s</span>edimentary rocks<span> are called shale, siltstone, and mudstone</span>
