50POINTS! What is the escape velocity for lunar module? Lunar module mass 15,200 kg radius of moon 1.74x106m, mass of moon 7.34x
2 answers:
Answer:
2.73 km/s
Explanation:
The escape velocity of an object in the gravitational field of the moon is (on the surface of the planet)
where
is the gravitational constant
is the mass of the Moon
is the radius of the Moon
As we can see, the escape velocity does not depend on the mass of the lunar module.
Substituting the numbers into the formula, we find
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Answer: The energy absorbed by the reaction from the water is 996 Joules.
Explanation:
Energy absorbed by the reaction or energy lost by the water to the reaction,Q.
Mass of the the reaction ,m = 60 g
Specific heat of water = c = 4.15 J\g ^oC
Change is temperature=
Negative sigh indicates that energy was given by the water to the reaction.
The energy absorbed by the reaction from the water is 996 Joules.
Answer:
The force of car 3 on car 2 ≈ 1810.82 N
Explanation:
The equation for the change in momentum of the two cars are;
Conservation of linear momentum
150( 2.2 - v) = 265(1.5-v)
150 × 2.2 - 265×1.5 = (150+265)v
150 × 2.2 - 265×1.5 = -67.5 = 415×v
∴ v = -67.5/415 = -0.1627 m/s West = 0.1627 m/s East
The impulse of the net force is the amount of momentum change experienced given by the equation;
Impulse force = -
Where;
= The final velocity
= The initial velocity
For the the 265 kg mass, we have;
= 0.1627 m/s
= 1.5 m/s
Which gives the impulse a s F×Δt = 265×0.1627 - 265×1.5 = -354.38 kg·m/s
The change in kinetic energy of the collision = 1/2×265×(0.1627^2 - 1.5^2) =-294.62 J
Whereby the distance moved in one second is 0.1627 m, we have;
Work done = Force × Distance = Force × 0.1627 = 294.62
Force = 294.62/0.1627 = 1810.82 N.