Answer:
a. v = 1.679 × 10⁹ m/s b. The answer of v = 1.679 × 10⁹ m/s is not reasonable
Explanation:
Here is the complete question
You read in a science magazine that on the Moon, the speed of a shell leaving the barrel of a modern tank is enough to put the shell in a circular orbit above the surface of the Moon (there is no atmosphere to slow the shell).
A.....What should be the speed for this to happen? Assume that the radius of the Moon is rM = 1.74×106m, and the mass of the Moon is mM = 7.35×1022kg.(Express your answer to two significant figures and include the appropriate units.)
B.....Is this number reasonable?
Solution
From the question, the centripetal force in orbit for the shell = gravitational force of moon on shell.
So, mv²/r = GMm/r²
So v = √(GM/r) where v = velocity of shell, G = universal gravitational constant = 6.67 × 10⁻¹¹ Nm²/kg², M = mass of moon = 7.35 × 10²² kg and r = radius of moon = 1.74 × 10⁶ m.
v = √(GM/r) = v = √(6.67 × 10⁻¹¹ Nm²/kg² × 7.35 × 10²² kg/1.74 × 10⁶ m)
v = √(49.0245/1.74 × 10¹⁷) m/s
v = √28.175 × 10¹⁷ m/s
v = √2.8175 × 10¹⁸ m/s
v = 1.679 × 10⁹ m/s
b. The answer of v = 1.679 × 10⁹ m/s is not reasonable because, it is over 1000000 km/s which is greater than the speed of light which is 300000 km/s
1 revolution = 2π radians
revs = (1/2π) · (rads) / (2π)
revs/sec = (1/2π) · (rads/sec)
RPM = (1/2π) · (rads/min) = (30/π) · (rads/sec)
Answer:37 J
Explanation:
Given
Step :1
Heat added Q=44 J
Work done=-20 J
Step :2
Heat added Q=-61 J
work done
as the process is cyclic
work done in compression is 37 J
Answer: 2933 kg
Explanation:
Given
From the law of conservation of linear momentum, we know that
m(i)v(i) = m(f)v(f), where there is no external force acting on the system
m(i) = initial mass of the freight car system
m(f) = maximum mass of the freight car system
v(i) = initial linear velocity of the system
v(f) = final linear velocity of the system
2000 * 4.4 = m(f) * 3
3m(f) = 8800
m(f) = 8800 / 3
m(f) = 2933.3 kg
Therefore the maximum mass of grain that it can accept is 2933 kg