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a. <span>FM GmMmr2
</span>= 6.67 x 10-11N.m2kg27 .35 x 1022 kg 70 kg 3.78 x 108 m2
<span>= 2.40 x 10-3 N
b. </span><span>FE GmEmr2
= 6.67 x 10-11 N.m2kg 25 .97 x 1034 kg (70kg) 6.38 x 106 m2
=685 N
FMFE 2.40 x 10-3N685 N= 0.0004%</span>
Answer:
the stopping distance is greater than the free length of the track, the vehicle leaves the track before it can brake
Explanation:
This problem can be solved using the kinematics relations, let's start by finding the final velocity of the acceleration period
v² = v₀² + 2 a₁ x
indicate that the initial velocity is zero
v² = 2 a₁ x
let's calculate
v =
v = 143.666 m / s
now for the second interval let's find the distance it takes to stop
v₂² = v² - 2 a₂ x₂
in this part the final velocity is zero (v₂ = 0)
0 = v² - 2 a₂ x₂
x₂ = v² / 2a₂
let's calculate
x₂ =
x₂ = 573 m
as the stopping distance is greater than the free length of the track, the vehicle leaves the track before it can brake
Answer:
9
Explanation:
2.13 rad/s * 26.9 sec
2.13 * 26.9
57.297
3282.88 deg / 360 deg = 9.12
It makes 9 complete revolutoins
Answer: 
Explanation:
A direct proportionality means a linear relationship between two variables and rate of change means an application of derivatives. Hence, the mathematical model is:

Answer:
<em>d. The sail should be reflective because in this case the momentum transferred to the sail per unit area per unit time is larger than for absorbing sail, therefore the radiation pressure is larger for the reflective sail.</em>
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Explanation:
Let us take the momentum of a photon unit as u
we know that the rate of change of momentum is proportional to the force exerted.
For a absorbing surface, the photon is absorbed, therefore the final momentum is zero. From this we can say that
F = (u - 0)/t = u/t
for a unit time, the force is proportional to the momentum of the wave due to its energy density. Therefore,
F = u
For a reflecting surface, the momentum of the wave strikes the sail and changes direction. Since we know that the speed of light does not change, then the force is proportional to
F = (u - (-u))/t = 2u/t
just as the we did above, it becomes
F = 2u.
From this we can see that the force for a reflective sail is twice of that for an absorbing sail, and we know that the pressure is proportional to the force for a given area. From these, we conclude that <em>the sail should be reflective because in this case the momentum transferred to the sail per unit area per unit time is larger than for absorbing sail, therefore the radiation pressure is larger for the reflective sail.</em>
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