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Inessa [10]
2 years ago
12

A 1000-kg sports car of mass accelerates from rest to 20 m/s in 6.6 s. What is the frictional force exerted by the road on the c

ar? Group of answer choices 1500 N 1750 N 2750 N 3000 N
Physics
1 answer:
lukranit [14]2 years ago
4 0

The frictional force exerted by the road on the car is 3000N

Given the mass of the car is 1000 kg , the velocity of the car is 20m/s

and the time is 6.6s

We need to find the frictional force exerted by the road on the car

We know that Force = mass * acceleration

Now here Mass is given but acceleration is not given

So, we will find acceleration by using the formula v = u+at

Where u = 0

v = 20m/s

a = ?

t = 6.6

Substitute the values in the formula We get

v = u+at

20 = 0+(a)(6.6)

20/6.6 = a

∴ a = 3.03 m/s^2

Rounding to nearest tenth is 3m/s^2

Hence the acceleration is 3m/s^2

Now substituting the value of acceleration in F = ma

Where m = mass

a = acceleration

F = 1000*3

∴ F = 3000N

Hence the frictional force exerted by the road on the car is 3000 N

Learn more about Frictional force here

brainly.com/question/23161460

#SPJ4

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LUCKY_DIMON [66]

Answer:

θ_p = 53.0º

Explanation:

For reflection polarization occurs when a beam is reflected at the interface between two means, the polarization in total when the angle between the reflected and the transmitted beam is 90º

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We replace

     n1 sin θ_p = n2 sin (90 - θ_p)

Let's use the trigonometry relationship

    Sin (90- θ_p) = sin 90 cos θ_p - cos 90 sin θ_p = cos θ_p

In the law of reflection  incident angle equals reflected angle,  

    ni sin θ_p = ns cos θ_p

    n₂ / n₁ = sin θ_p / cos θ_p

    n₂ / n₁ = tan θ_p

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The refractive index of air is 1 (n1 = 1) the refractive index of seawater varies between 1.33 and 1.40 depending on the amount of salts dissolved in the water

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n₂ = 1.40

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      Tep = 54.5º

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learn more:

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