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PolarNik [594]
2 years ago
15

An Olympic runner completes the 200-meter sprint in 23 seconds. What is the runner’s average speed? (Round your answer to the

nearest tenth of a meter per second. ) 0. 9 m/s 1. 2 m/s 8. 7 m/s 10. 1 m/s.
Physics
1 answer:
Amanda [17]2 years ago
3 0

The average speed of the runner is 8.7m/s.

Therefore, Option C) 8.7m/s is the correct answer.

Given the data in the question;

Distance covered; d = 200m

Time taken; t = 23s

Average speed; s = \ ?

Speed is the rate at which an object covers a certain distance. It is expressed as:

s = \frac{d}{t}

Where s is speed, d is distance and t is time taken.

We substitute our given distance into the equation

s = \frac{200m}{23s} \\\\s = 8.6956m/s\\\\s = 8.7m/s

The average speed of the runner is 8.7m/s

Therefore, Option C) 8.7m/s is the correct answer.

Learn more: brainly.com/question/21503615

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an object has a length of 5.0 cm, aheight of 3.0 cm and a width of 15.0 cm. it has a mass of 24 grams. what is its density?
kondaur [170]
5x3=N
Nx15=F
Fx24=Your Awnser
5 0
2 years ago
A 3 kg toy car with a speed of 7 m/s collides head-on with a 2 kg car traveling in the opposite direction with a speed of 5 m/s.
Mademuasel [1]

Answers:

kinetic energy lost = 86.4J

Explanation:

let Kf be the kinetic energy after the collision and Ki be the kinetic energy before the collision. let the 3kg car be 1 and 2kg car be 2.

Kf = K1(f) + K2(f)

Ki = K1(i) + k2(i)

loss in kinetic energy = Kf - Ki

                                 = 1/2(3)(2.20)^2 + 1/2(2)(2.20)^2 - 1/2(3)(7)^2 - 1/2(2)(-5)^2

                                 = 12.1 - 98.5

                                 = -86.4 J

therefore, the kinetic energy lost in the collision is 86.4 J.

6 0
2 years ago
Which of the following is NOT a major factor in understanding climate?
ivann1987 [24]
I think is altitude because tbh it don’t even mean nothing
6 0
2 years ago
What disagreement did Sigmund Freud have with both Josef Breuer and Jean Martin Charcot?
GREYUIT [131]

Answer:

The correct answer is 'A'

Explanation:

I guessed and was correct.

4 0
3 years ago
An inclined plane of effective length 4.5m is used to raise a load of 500N through a height of 1.5m .If the list is raise by for
IrinaK [193]

Answer:

1850 N

Explanation:

The formula for friction force between the load and plane is given as ;

F= μ*N

N = mg cos θ

To find θ, which is the angle the inclined plane makes with the ground at the height of 1.5 m

Sin θ = 1.5/4.5

Sin θ = 0.3333

Sin⁻{0.3333} = 19.50°

θ = 19.50°

Finding N , where m= 500 N , and g= 9.81

N = mg cos θ

N= 500 * 9.81 * cos 19.50°

N= 4624 N

Coefficient of kinetic friction is calculated as;

μ=F/W

μ = 200/500 = 0.4

The magnitude of kinetic friction is given as;

Fk= μ * N

Fk = 0.4 * 4624

Fk= 1850 N

8 0
2 years ago
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