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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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How much force is needed to accelerate a 62 kg skier at 3 m/sec2?
Fittoniya [83]

Answer:

<h2>The answer is 186 N</h2>

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

<h3>force = mass × acceleration</h3>

From the question

mass = 62 kg

acceleration = 3 m/s²

We have the final answer as

force = 62 × 3

We have the final answer as

<h3>186 N</h3>

Hope this helps you

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3 years ago
If a fly gets his wings cut of, is it still a fly?
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Answer:

Yes I think... But the joke is it is a walk then.

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<span>3.834 m/s. In this problem we need to have a centripetal force that is at least as great as the gravitational attraction the object has. The equation for centripetal force is F = mv^2/r and the equation for gravitational attraction is F = ma Since m is the same in both cases, we can cancel it out and then set the equations equal to each other, so a = v^2/r Substitute the known values (radius is diameter/2) and solve for v 9.8 m/s^2 <= v^2/1.5 m 14.7 m^2/s^2 <= v^2 3.834057903 m/s <= v So the minimum velocity needed is 3.834 m/s.</span>
4 0
4 years ago
The curved section of a speedway is a circular arc having a radius of 190 m. this curve is properly banked for racecars moving a
Anestetic [448]

The banking angle of the curved part of the speedway is determined as 32⁰.

<h3>Banking angle of the curved road</h3>

The banking angle of the curved part of the speedway is calculated as follows;

V(max) = √(rg tanθ)

where;

  • r is radius of the path
  • g is acceleration due to gravity

V² = rg tanθ

tanθ = V²/rg

tanθ = (34²)/(190 x 9.8)

tanθ = 0.62

θ = arc tan(0.62)

θ = 31.8

θ ≈ 32⁰

Learn more about banking angle here: brainly.com/question/8169892

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2 years ago
Jack tries to place a magnet on the door of his refrigerator. He observes that the magnets don't stick. He guesses that the door
qwelly [4]
He is made a inference to the observation that the magnet doesn't stick<span />
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