Answer:
C-0.55 m/s to the right
Explanation:
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Answer:
0.55 to the right
Explanation:
Consider a satellite in a circular orbit around the earth. Why is it important to give a satellite a horizontal speed when placing it in orbit? What will happen if the horizontal speed is too small? What will happen if the speed is too large?
Answer:
In this case, the horizontal velocity of the rocket starts from the acceleration, so if its velocity drops to zero,
Explanation:
When a satellite is in orbit the most important force is the docking of gravity with the Earth
F = m a
where the acceleration is centripetal and F is the force of universal attraction
centripetal acceleration is
a = v² / r
F = m v² / r
In this case, the horizontal velocity of the rocket starts from the acceleration, so if its velocity drops to zero, the force also drops to serious and the satellite steels to Earth.
The speed of the satellite is provides the speed, by local for smaller speeds in satellite, it descends in its orbits and when the speed is amate you have the energy to stop an orb to go to a higher orbit.
Two teams are playing tug-of-war. Team A, on the left, is pulling on the rope with an effort of 5000 N. If the rope is moving at a constant velocity, how hard and in which direction is team B pulling?
A. 2500 N to the left
B. 5000 N to the right
C. 2500 N to the right
D. 5000 N to the left
Explanation:
If Team A is on the left, B is on the right
if the force is constant, it means that the effort applied is equal.
So Team B is pulling 5000N to the right.
Oil at 150 C flows slowly through a long, thin-walled pipe of 30-mm inner diameter. The pipe is suspended in a room for which the air temperature is 20 C and the convection coefficient at the outer tube surface is 11 W/m2 K. Estimate the heat loss per unit length of tube.
Answer:
1.01 W/m
Explanation:
diameter of the pipe d = 30 mm = 0.03 m
radius of the pipe r = d/2 = 0.015 m
external air temperature Ta = 20 °C
temperature of pipe wall Tw = 150 °C
convection coefficient at outer tube surface h = 11 W/m^2-K
From the above, we assumed that the pipe wall and the oil are in thermal equilibrium.
area of the pipe per unit length A = [tex]\pi r ^{2}[/tex] = [tex]7.069*10^{-4}[/tex] m^2/m
convectional heat loss Q = Ah(Tw - Ta)
Q = 7.069 x 10^-4 x 11 x (150 - 20)
Q = 7.069 x 10^-4 x 11 x 130 = 1.01 W/m
The heat loss per unit length of tube should be considered as the 1.01 W/m.
Calculation of the heat loss:Since
diameter of the pipe d = 30 mm = 0.03 m
radius of the pipe r = d/2 = 0.015 m
external air temperature Ta = 20 °C
temperature of pipe wall Tw = 150 °C
convection coefficient at outer tube surface h = 11 W/m^2-K
Now
area of the pipe per unit length A should be
= πr^2
= 7.069*10^-4 m^2/m
Now
convectional heat loss Q = Ah(Tw - Ta)
Q = 7.069 x 10^-4 x 11 x (150 - 20)
Q = 7.069 x 10^-4 x 11 x 130
= 1.01 W/m
hence, The heat loss per unit length of tube should be considered as the 1.01 W/m.
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