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navik [9.2K]
3 years ago
11

Occupants in a single space shuttle in orbit feel weightless. describe a scheme whereby occupants in a pair of shuttles (or even

one shuttle and another massive object in orbit) would be able to use a long cable to continuously experience a comfortably normal earthlike gravity.
Physics
1 answer:
agasfer [191]3 years ago
8 0
The   scheme  whereby occupants  in  a  pair  of  shuttles is  as  follows
use  a  strong  cable  with  large  weight  on  the  end
Then  use  the  orbital  naneuvering   system(OMS)  to   set  the   whole  work  as  spinning   about   their  common  center of  gravity.
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PLS HELP<br><br> compare longitudinal and transverse waves
babymother [125]
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6 0
2 years ago
Read 2 more answers
One mole of an ideal gas expands isothermally from 0.01 m3 to 0.05 m3. The final pressure of gas is 1.2 atm. [15 Marks] determin
erik [133]

Answer:

Initial pressure = 6 atm. Work = 0.144 J

Explanation:

You need to know the equation P1*V1=P2*V2, where P1 is the initial pressure, V1 is the initial volume, and P2 and V2 are the final pressure and volume respectively. So you can rearrange the terms and find that (1.2*0.05)/(0.01) = initial pressure = 6 atm. The work done by the system can be obtained calculating the are under the curve, so it is 0.144J

4 0
3 years ago
Which event causes tides
Reptile [31]

Answer:

High tides and low tides are caused by the Moon.

Explanation:

The Moon's gravitational pull generates something called the tidal force.

3 0
3 years ago
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A 3,000 kg truck moving at +10 m/s hits a 1,000 kg parked car which moves off at +15 m/s What is the velocity of
Rina8888 [55]

Answer:

v₃ = 5 [m/s]

Explanation:

To solve this problem we must use the definition of linear momentum, which tells us that momentum is equal to the product of mass by Velocity.

P = m*v

where:

P = linear momentum [kg*m/s]

m = mass [kg]

v = velocity [m/s]

We must also clarify that the momentum is preserved i.e. it is equal before the collision and after the collision

Pbeforecollision = Paftercollision

(m₁*v₁) + (m₂*v₂) = (m₁*v₃) + (m₂*v₄)

where:

m₁ = mass of the truck = 3000 [kg]

v₁ = velocity of the truck = 10 [m/s]

m₂ = mass of the car = 1000 [kg]

v₂ = velocity of the car before the collision = 0 (the car is parked)

v₃ = velocity of the truck after the collision [m/s]

v₄ = velocity of the car after the collision = 15 [m/s]

(3000*10) + (1000*0) = (3000*v₃) + (1000*15)

30000 = 3000*v₃ + 15000

3000*v₃ = 30000 - 15000

3000*v₃ = 15000

v₃ = 5 [m/s]

7 0
2 years ago
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