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Crazy boy [7]
3 years ago
5

If body a and b are both within system can forces between them affect the acceleration of the system?

Physics
2 answers:
Mekhanik [1.2K]3 years ago
6 0
Accelerating a system would change the momentum of the system. 

Momentum is conserved until there is an external force on a system.

Internal forces do not change the momentum of a system because forces come in pairs and they are equal and opposite to each other(Newton's 3rd law), so all internal forces cancel each other out.

Internal forces can't accelerate the center of mass of a system.


Yuri [45]3 years ago
5 0
It affects the muscular and digestive
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9420 km/hr is the correct answer

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Is The force of friction always opposite to the motion? ​
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8 0
4 years ago
Describe a scenario where a car's speed could stay the same, but the acceleration changes.
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An object's acceleration is the rate its velocity (speed and direction) changes. Therefore, an object can accelerate even if its speed is constant - if its direction changes.

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3 years ago
Older railroad tracks in the U.S. are made of 12 m-long pieces of steel. When the tracks are laid, gaps are left between the sec
Shkiper50 [21]

Answer:

0.005 m

Explanation:

length of steel (L°) = 12 m

initial temperature (T) = 16 degrees

expected temperature (T') = 50 degrees

We can find how large the gaps should be if the track is not to buckle when the temperature is as high as 50 degrees from the formula below

ΔL = ∝L°ΔT where

  • ΔL = expansion / gap
  • ∝ = linear expansion coefficient of steel = 12x10^{-6} C^{-1}
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  • ΔT = change in temperature

ΔL = 12x10^{-6} x 12 x (50-16) = 0.005 m

3 0
4 years ago
It takes a minimum distance of 41.14 m to stop a car moving at 11.0 m/s by applying the brakes (without locking the wheels). Ass
padilas [110]

Answer:

Find the time it took for the car to stop at 11.0m/s

V = deltax/t

t = 41.14/11.0 = 3.74s

Now find at what rate it was decelerating, so find the acceleration during that interval of time.

vf = vi + at

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a = -2.94m/s^2

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Now find how much time it would take for the car to stop at 28.0m/s but with the same acceleration, the car is the same so its acceleration to stop the car will remain the same.

vf = vi + at

0 = 28.0 - 2.94t

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Once the time is obtained, you can find the final position, xf, by plugging the time acceleration and velocity values.

xf = 0 + (28m/s)(9.52s) + 1/2(-2.94)(9.52s)^2

xf = 266.6m - 133.23m = 133m

8 0
4 years ago
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