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Luda [366]
2 years ago
11

A 50.0-kg person steps on a scale in an elevator. The scale reads 5000 N. What is the magnitude of the acceleration of the eleva

tor
Physics
1 answer:
zaharov [31]2 years ago
5 0

The magnitude of the acceleration of the elevator is 90m/s^2

Due to Newton's Law ∑ Forces in direction of motion is equal to mass

multiplied by the acceleration

We have here two forces 5000 N in direction of motion and the weight of the person in opposite direction of motion

The weight of the person is his mass multiplied by the acceleration of gravity

So ,

W = mg , where m is the mass and g is the acceleration of gravity

m = 50 kg and g = 9.8 m/s²

Substitute these values in the rule above

W = 50 × 9.8 = 490 N

The scale reads 5000 N

F = 5000 N , W = 490 N , m = 50kg

F - W = ma

5000 - 490 = 50 a

4510 = 50 a

Divide both sides by 50

a = 90.2 m/s²

Hence the acceleration is 90m/s^2

Learn more about Acceleration here

brainly.com/question/14344386

#SPJ4

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Question:

A 63.0 kg sprinter starts a race with an acceleration of 4.20m/s square. What is the net external force on him? If the sprinter from the previous problem accelerates at that rate for 20m, and then maintains that velocity for the remainder for the 100-m dash, what will be his time for the race?

Answer:

Time for the race will be t = 9.26 s

Explanation:

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As the sprinter starts the race so initial velocity = v₁ = 0

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Distance = s₂ = 100 m

We first need to find the final velocity (v₂) of sprinter at the end of the first 20 meters.

Using 3rd equation of motion

(v₂)² - (v₁)² = 2as₁ = 2(4.2)(20)

v₂ = 12.96 ms⁻¹

Time for 20 m distance = t₁ = (v₂ - v ₁)/a

t₁ = 12.96/4.2 = 3.09 s

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t₂ = 80/12.96 = 6.17 s

Total time = T = t₁ + t₂ = 3.09 + 6.17 = 9.26 s

T = 9.26 s

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