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Harrizon [31]
3 years ago
14

The rate at which work is done is known as which of the following?

Physics
1 answer:
sveta [45]3 years ago
3 0

Answer:

A , power

Explanation:

Hope this is useful. Have a lovely rest of your day! God bless you.

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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
Tamiku [17]

Answer:

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>

4 0
3 years ago
at 1 p.m. a car traveling at a constant velocity of 78 km per hour towards the West it's 34 km to the west of our school how far
Fittoniya [83]

83 km/h * 2.5 hours (3:30 - 1:00) = 207.5 km  

207.5 km - 15 km = 192.5 km

3 0
3 years ago
Which of the following is not a region of the brain?
lorasvet [3.4K]
Definitely the last one, brain stem
7 0
3 years ago
Which two layers of Earth are mainly composed of iron and nickel? inner core outer core lower mantle upper mantle crust
stich3 [128]
Inner core and outer core
3 0
4 years ago
Read 2 more answers
A 1 200-kg automobile moving at 25 m/s has the brakes applied with a deceleration of 8.0 m/s2. How far does the car travel befor
Alja [10]

Answer:

Δx = 39.1 m

Explanation:

  • Assuming that deceleration keeps constant during the braking process, we can use one of the kinematics equations, as follows:

        v_{f} ^{2} - v_{o} ^{2} = 2* a * \Delta x (1)

        where  vf is the final velocity (0 in our case), v₀ is the initial velocity

        (25 m/s), a is the acceleration (-8.0 m/s²), and Δx is the distance

        traveled since the brakes are applied.

  • Solving (1) for Δx, we have:

        \Delta x = \frac{-v_{o} ^{2} }{2*a} = \frac{-(25m/s)^{2}}{2*(-8.0m/s2} = 39.1 m (2)        

7 0
3 years ago
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