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Reptile [31]
4 years ago
9

The most energy-efficient way to send a spacecraft to the Moon is to boost its speed while it is in circular orbit about the Ear

th such that its new orbit is an ellipse. The boost point is the perigee of the ellipse, and the point of arrival at the Moon is the apogee. Calculate the percentage increase in speed required to achieve such an orbit. Assume that the spacecraft is initially in a low-flying circular orbit about Earth. The distance between Earth and the Moon is approximately 60 Re, where Re is the radius of Earth
Physics
1 answer:
jeyben [28]4 years ago
5 0

Answer:

98.33 %

Explanation:

On an elliptical orbit, angular momentum will be conserved .

Angular momentum = I ω = mvR

So mv₁R₁ = mv₂R₂

= v₁R₁ = v₂R₂

where v₁ is velocity and R₁ radius in low orbit (perigee)and v₂ and R₂ is velocity and radius in high orbit ( apogee ).

Here R₁ = Radius of the earth , R₂ is distance between moon and earth.

R₁ / R₂ = 1/60

v₁ /v₂ = R₂ / R₁  = 60

v₂ / v₁ = 1 / 60

1 - (v₂ / v₁ ) = 1 -( 1 / 60)

(v₁ -v₂)/v₁ = ( 60-1 )/60

(v₁ -v₂)/v₁ x 100 = 5900/60 = 98.33 %

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a level physics moments question please explain how the answer to (a) is 7.5N and the answer to (b) is 5.7N​
Mkey [24]

Explanation:

a) Sum of moments about the pivot:

∑τ = Iα

(10 kg × 10 m/s²) (0.3 m) + F (-0.4 m) = 0

F = 75 N

b) Sum of moments about the pivot:

∑τ = Iα

(20 N) (0.3 m) + (20 N) (-0.2 m) + F (-0.7 m) = 0

6 Nm − 4 Nm + F (-0.7 m) = 0

F = 2.9 N

5 0
4 years ago
A tortoise can move with a speed of 10.0cm/s, while a rabbit can move 10 times faster. In a race, both of them started at the sa
Xelga [282]

Answer:

C. 199.9 s

Explanation:

3 minutes = 3×60 = 180 seconds.

the turtle moves in that time 180×10 = 1800 cm.

in other words the rabbit gave it that much head-start (it does not matter if that was at the begin of in the middle of the race).

the rabbit moves with 10×10cm/s = 100cm/s.

the rabbit needs therefore 1800/100 = 18 seconds for the

1800 cm.

at that time the turtle has added another 18×10 = 180 cm.

for which the rabbit needs 180/100 = 1.8 seconds.

during that time the turtle has added 1.8×10 = 18 cm.

and so on.

in formal mathematics this looks like this :

1800 + 10x = 100x

after x seconds of the rabbit running both will have run the same distance, and it is a tie.

1800 = 90x

x = 20 seconds

so, at that point, the rabbit was actively running for 20 seconds and raced 20×100 = 2000 cm

and the turtle was actively running for 180 + 20 = 200 seconds, and also covered 200×10 = 2000 cm.

but our question tells us that the turtle won by 10 cm.

so, the race was over a little bit before these 200 seconds (for a tie).

this means, the rabbit could not run the last 10 cm for the tie (because the race was over and the turtle had won).

the rabbit would have needed 10/100 seconds for these 10 cm.

as speed = distance/time

we need to divide distance by speed

distance/1 / distance/time

to get time.

so,

10cm/1 / 100cm/s = 10s/100 = 1/10 s

so, we need to deduct this 1/10 s from the 200 seconds of the turtle (and also from the 20 seconds for the rabbit).

the race lasted of course the whole time the turtle was running (while the rabbit was resting, officially still participating in the race with speed 0 for 3 minutes).

and so, the race was 199.9 s long.

8 0
2 years ago
You push a heavy crate down a ramp at a constant velocity. Only four forces act on the crate. Which force does the greatest magn
Brilliant_brown [7]

The friction force does the greatest magnitude of work on the crate

Consider all four forces. The normal force does no work at all, since there is no motion in the direction of that force, perpendicular to the ramp. The force of gravity is smaller than the force of friction, since you still need to push the crate to get constant velocity. The force of you pushing is also smaller than the force of friction, since you are moving down a ramp, and are therefore assisted by gravity. Therefore the force doing greatest magnitude of work is the force of friction. Note that, even though the frictional work is negative, it still has the greatest magnitude

Learn more about friction force here:

brainly.com/question/4618599

#SPJ4

6 0
2 years ago
A rocket has initail mass M begins to move from space , with an exhaust constant speed . Find the mass of the rocket while has t
irina1246 [14]

Answer:

m=\frac{m_{0}}{e}

Explanation:

Equation of the rocket is,

m\frac{dv}{dt} =F-v'\frac{dm}{dt}

Here, v' is the relative velocity of rocket.

In space F is zero.

So,

m\frac{dv}{dt} =-v'\frac{dm}{dt}\\dv=-v'\frac{dm}{m} \\v=-v'ln\frac{m}{m_{0} }

Now the momentum can be obtained by multiply by m on both sides.

P=-v'mln\frac{m}{m_{0} }

Now for maxima, \frac{dP}{dm}=0

-v'ln\frac{m}{m_{0} }-v'm\frac{m_{0}}{m }m_{0=0

Now,

ln(\frac{m}{m_{0} } )=-1\\\frac{m}{m_{0} }=\frac{1}{e} \\m=\frac{m_{0}}{e}

Therefore, the mass of the rocket while having maximum momentum is \frac{m_{0}}{e}

3 0
4 years ago
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Fittoniya [83]
The top of the trajectory is the point where it changes from rising to falling. At that exact instant, its vertical speed is zero.
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