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defon
3 years ago
14

a 2100-kg car drives with a speed of 18 m/s onb a flat road around a curve that has a radius of curvature of 83m. The coefficien

t of static friction between the car and the road is 0.78. What is the magnitude of the force of static friction acting on the car
Physics
1 answer:
Law Incorporation [45]3 years ago
7 0

Answer:

<u>The magnitude of the friction force is 8197.60 N</u>

Explanation:

Using the definition of the centripetal force we have:

\Sigma F=ma_{c}=m\frac{v^{2}}{R}

Where:

  • m is the mass of the car
  • v is the speed
  • R is the radius of the curvature

Now, the force acting in the motion is just the friction force, so we have:

F_{f}=m\frac{v^{2}}{R}

F_{f}=2100\frac{18^{2}}{83}

F_{f}=8197.60 \: N

<u>Therefore the magnitude of the friction force is 8197.60 N</u>

I hope it helps you!

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Answer:

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A virtual image is formed 17.0 cm from a concave mirror having a radius of curvature of 39.0 cm. (a) Find the position of the ob
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Explanation:

Image distance, v = -17 cm (-ve for virtual image)

Radius of curvature of concave mirror, R = 39 cm

Focal length, f = -19.5 cm (-ve for a concave mirror)

(a) Using mirror's formula as :

\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}

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\dfrac{1}{u}=\dfrac{1}{-19.5}-\dfrac{1}{(-17)}  

u = 132.6 cm    

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m=\dfrac{-17}{132.6}

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3 years ago
A 1400-kg car traveling east at 25m/s collides with a 1800-kg car traveling at a speed of 20m/s in a direction that makes angle
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M1*V1 + M2*V2 = M1*V + M2*V.
1400*25 + 1800*20[180+40]=1400*V+1800*V.
Divide both sides by 100:
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Answer:

w = 706.32 [N]

Explanation:

The force due to gravitational acceleration can be calculated by means of the product of mass by gravitational acceleration.

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Then we have:

w = 72*9.81\\w = 706.32 [N]

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