Work = (force) x (distance)
The worker does (40N) x (4m) = 160 joules of work.
Friction eats up (27N) x (4m) = 108 joules of that energy,
generating 108 joules of heat.
The remaining (160J - 108J) = 52 joules of energy moves the box.
Answer:
x = 0.75801 = 75.801%
T_2 = 72..78 degree F
Explanation:
From superheated R 134 a properties table
At 200 lb/in^2 and 200 degree F
steady flow energy equation is givena s
At 90 lb/in2 Tsat = 72.78 degree F
hfg = 77.345 Btu/lbm
h = hf + x hfg
solving for x we get
x = 0.75801 = 75.801%
The car’s momentum after 4.21s is 24617.4 kgm/s
<h3>
Newton's Second Law of Motion.</h3>
Newton's second law state that, the rate of change of momentum, is directly proportional to the applied force.
Given that a 1200 kg car passes traffic light at a velocity of 10.2 m/s to the north and accelerates at a rate of 2.45 m/s^2. To calculate the car’s momentum after 4.21 s, Let us first list all the parameters involved.
- Acceleration a = 2.45 m/s²
From Newton's second law,
F = (mv - mu) / t
ma = (mv - mu) / t
Substitute all the parameters into the formula above.
1200 × 2.45 = ( mv - 1200 × 10.2 ) / 4.21
2940 = ( mv - 12240 ) / 4.21
Cross multiply
12377.4 = mv - 12240
Make mv the subject of the formula
mv = 12377.4 + 12240
mv = 24617.4 kgm/s
Therefore, the car’s momentum after 4.21s is 24617.4 kgm/s
Learn more about Momentum here: brainly.com/question/25121535
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The circular path that the car is following reveals the presence
of a net force acting on the car.
If there were no net force, pointing toward the center of the circular
path, the car would travel in a straight line.
