A 62 kg skydiver moving at terminal speed falls 50 m in 1 s. What power is the skydiver expending on the air?
1 answer:
You might be interested in
Refer to the diagram shown below.
The hoist is in static equilibrium supported by tensions in the two ropes.
For horizontal force balance, obtain
T₃ cos 50 = T₂ cos 38
0.6428T₃ = 0.788T₂
T₃ = 1.2259T₂ (1)
For vertical force balance, obtain
T₂ sin 38 + T₃ sin 50 = 350
0.6157T₂ + 0.766T₃ = 350 (2)
Substitute (1) into (2).
0.6157T₂ + 0.766(1.2259T₂) = 350
1.5547T₂ = 350
T₂ = 225.124 N
T₃ = 1.2259(225.124) = 275.979
Answer:
T₂ = 225.12 N
T₃ = 275.98 N
The hoist is in static equilibrium supported by tensions in the two ropes.
For horizontal force balance, obtain
T₃ cos 50 = T₂ cos 38
0.6428T₃ = 0.788T₂
T₃ = 1.2259T₂ (1)
For vertical force balance, obtain
T₂ sin 38 + T₃ sin 50 = 350
0.6157T₂ + 0.766T₃ = 350 (2)
Substitute (1) into (2).
0.6157T₂ + 0.766(1.2259T₂) = 350
1.5547T₂ = 350
T₂ = 225.124 N
T₃ = 1.2259(225.124) = 275.979
Answer:
T₂ = 225.12 N
T₃ = 275.98 N
<h2>Answer: Gravitational attraction will be the same</h2>
According to the law of universal gravitation, which is a classical physical law that describes the gravitational interaction between different bodies with mass:
(1)
Where:
is the module of the force exerted between both bodies
is the universal gravitation constant.
and
are the masses of both bodies.
is the distance between both bodies
Now, if we double both masses and the distance also doubles, this means:
and
will be now
and
will be now
Let's rewrite the equation (1) with this new values:
(2)
Solving and simplifying:
(3)
As we can see, equation (3) is the same as equation (1).
So, if the masses both double and the distance also doubles the <u>Gravitational attraction between both masses will remain the same.</u>