Answer:
The work done on the system can be accounted for by;
Both 
and 
Explanation:
The speed of the crate = Constant
Therefore, the acceleration of the crate = 0 m/s²
The net force applied to the crate, 
= 0
Therefore, the force of with which the crate is pulled = The force resisting the upward motion of the crate
However, we have;
The force resisting the upward motion of the crate = The weight of the crate + The frictional resistance of the ramp due to the surface contact between the ramp and the crate
The work done on the system = The energy to balance the resisting force = The weight of the crate × The height the crate is raised + The heat generated as internal energy to the system
The weight of the crate × The height the crate is raised = Gravitational Potential Energy =
The heat generated as internal energy to the system =
Therefore;
The work done on the system = + .
Answer:
a) P = 1240 lb/ft^2
b) P = 1040 lb/ft^2
c) P = 1270 lb/ft^2
Explanation:
Given:
- P_a = 2216.2 lb/ft^2
- β = 0.00357 R/ft
- g = 32.174 ft/s^2
- T_a = 518.7 R
- R = 1716 ft-lb / slug-R
- γ = 0.07647 lb/ft^3
- h = 14,110 ft
Find:
(a) Determine the pressure at this elevation using the standard atmosphere equation.
(b) Determine the pressure assuming the air has a constant specific weight of 0.07647 lb/ft3.
(c) Determine the pressure if the air is assumed to have a constant temperature of 59 oF.
Solution:
- The standard atmospheric equation is expressed as:
P = P_a* ( 1 - βh/T_a)^(g / R*β)
(g / R*β) = 32.174 / 1716*0.0035 = 5.252
P = 2116.2*(1 - 0.0035*14,110/518.7)^5.252
P = 1240 lb/ft^2
- The air density method which is expressed as:
P = P_a - γ*h
P = 2116.2 - 0.07647*14,110
P = 1040 lb/ft^2
- Using constant temperature ideal gas approximation:
P = P_a* e^ ( -g*h / R*T_a )
P = 2116.2* e^ ( -32.174*14110 / 1716*518.7 )
P = 1270 lb/ft^2
Answer:
The answer is "
"
Explanation:
please find the complete question in the attached file.
Given value:
calculating the radiation absorbed per kg:
Answer:
rpm= 916.7436 rev/min
Explanation:
First determine the perimeter of the wheel, to know the horizontal distance it travels in a revolution:
perimeter= π×diameter= π × 22 inches × 0.0254(m/inche)= 1.7555m
Time we divide the speed of the car, which is the distance traveled horizontally over time unit, by the perimeter of the wheel that is the horizontal distance traveled in a revolution, this dates us the revolutions over the time unit:
revolutions per time= velocity/perimeter
velocity= (60 mi/hr) × (1609.34m/mi) = 96560m/h
revolutions per time= (96560.6m/h) / (1.7555m)= 55004.614 rev/hr
rpm= (55004.614 rev/hr) × (hr/60min)= 916.7436 rev/min
