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Bas_tet [7]
3 years ago
5

If a 990 kg car is on the road and the Ff is 360 n what is the normal force

Physics
2 answers:
Veseljchak [2.6K]3 years ago
8 0
Given that a car is in the road, there is only movement in the x-direction. There is no movement in the y-direction.

Looking at the y-direction for the normal force:

F = N - mg
0 = N - mg, (no movement in y-dir.)
N = mg
N = (990)(9.8)
N = 9702 newtons

The normal force exerted on the car by the road is 9702 newtons.
Arturiano [62]3 years ago
8 0

Answer:

9700 N upward

Explanation:

Got right on my test.

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Read 2 more answers
Car a with mass 1,783 kg collides with stationary 1600 kg car b. they become locked together after the collision and move with s
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The initial speed of car A is 15.18 m/s.

Momentum is defined as mass in motion. If there are two objects (the two objects in motion or only one object in motion and the other in stationary) that collide and no other forces work in the system, the law of momentum conservation applies in the system.

p=p'

pa+pb = pa'+pb'

(ma×va) + (mb×vb) = (ma×va') + (mb×vb')

  • ma = mass of object A (kg) = 1,783 kg
  • mb = mass of object B (kg) = 1,600 kg
  • va = speed of object A before collides (m/s)
  • va' = speed of object A after collides (m/s) = 8 m/s
  • vb = speed of object B before collides (m/s) = 0 m/s
  • vb' = speed of object B after collides (m/s) = 8 m/s
  • p = momentum before collision (Ns)
  • p' = momentum after collision (Ns)

(ma×va) + (mb×vb) = (ma×va') + (mb×vb')

(1,783×va) + (1,600×0) = (1,783×8) + (1,600×8)

(1,783×va) + 0 = 14,264+12,800

(1,783×va) = 27,064

va \:=\: \frac{27,064}{1.783}

va = 15.18 m/s

Learn more about The law of momentum conservation here: brainly.com/question/7538238

#SPJ4

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Have a longer revolution time since they definitely do not get warmer, They do not have fewer moons (Jupiter has about 100 and earth has 1) they are not smaller in diameter (Earth v Jupiter)
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