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4vir4ik [10]
3 years ago
6

1. A 46000N helicopter feels a net force of 9200N. What litt force is exerted by the air on the propellers?

Physics
1 answer:
Lina20 [59]3 years ago
4 0

Answer:

Force exerted by the air on the propellers = 46000 - 9200

= 36800 N

Hope this helps!

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A 1000 kg satellite and a 2000 kg satellite follow exactly the same orbit around the earth. What is the ratio F1/F2 of the gravi
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the <em>ratio F1/F2 = 1/2</em>

the <em>ratio a1/a2 = 1</em>

Explanation:

The force that both satellites experience is:

F1 = G M_e m1 / r²       and

F2 = G M_e m2 / r²

where

  • m1 is the mass of satellite 1
  • m2 is the mass of satellite 2
  • r is the orbital radius
  • M_e is the mass of Earth

Therefore,

F1/F2 = [G M_e m1 / r²] / [G M_e m2 / r²]

F1/F2 = [G M_e m1 / r²] × [r² / G M_e m2]

F1/F2 = m1/m2

F1/F2 = 1000/2000

<em>F1/F2 = 1/2</em>

The other force that the two satellites experience is the centripetal force. Therefore,

F1c = m1 v² / r    and

F2c = m2 v² / r

where

  • m1 is the mass of satellite 1
  • m2 is the mass of satellite 2
  • v is the orbital velocity
  • r is the orbital velocity

Thus,

a1 = v² / r ⇒ v² = r a1    and

a2 = v² / r ⇒ v² = r a2

Therefore,

F1c = m1 a1 r / r = m1 a1

F2c = m2 a2 r / r = m2 a2

In order for the satellites to stay in orbit, the gravitational force must equal the centripetal force. Thus,

F1 = F1c

G M_e m1 / r² = m1 a1

a1 = G M_e / r²

also

a2 = G M_e / r²

Thus,

a1/a2 = [G M_e / r²] / [G M_e / r²]

<em>a1/a2 = 1</em>

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