Input heat, Qin = 4 x 10⁵ J
Output heat, Qout = 3.5 x 10⁵ J
From the first Law of thermodynamics, obtain useful work performed as
W = Qin - Qout
= 0.5 x 10⁵ J
By definition, the efficiency is
η = W/Qin
= 100*(0.5 x 10⁵/4 x 10⁵)
= 12.5%
Answer: The efficiency is 12.5%
Answer:
F = 352 N
Explanation:
we know that:
F*t = ΔP
so:
F*t = M
-M
where F is the force excerted by the wall, t is the time, M the mass of the ball,
the final velocity of the ball and
the initial velocity.
Replacing values, we get:
F(0.05s) = (0.8 kg)(11m/s)-(0.8 kg)(-11m/s)
solving for F:
F = 352 N
Answer:
r = 4.24x10⁴ km.
Explanation:
To find the radius of such an orbit we need to use Kepler's third law:

<em>where T₁: is the orbital period of the geosynchronous Earth satellite = 1 d, T₂: is the orbital period of the moon = 0.07481 y, r₁: is the radius of such an orbit and r₂: is the orbital radius of the moon = 3.84x10⁵ km. </em>
From equation (1), r₁ is:
Therefore, the radius of such an orbit is 4.24x10⁴ km.
I hope it helps you!