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3 years ago

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A 16ft seesaw is pivoted in the center. At what distance from the center would a 200lb person sit to balance a 120lb person on t

he opposite end?

1 answer:

pickupchik [31]3 years ago

7 0

Answer:

9.6 ft

Explanation:

Distance is inversely proportional to weight

distance = k / (weight), where

k is a constant

or you could say,

distance * weight = k

In this scenario,

120 * 16 = 200 * distance

On rearranging, making, distance the subject of formula, we have

Distance = 120 * 16 / 200

Distance = 1920 / 200

Distance = 9.6 ft

So the 200 pounds person should sit 9.6 feet away from the centre to balance the see saw

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By testing it out. Try it.

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A future use of space stations may be to provide hospitals for severely burned persons. It is very painful for a badly burned pe

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Answer:

1.5min

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To solve the problem it is necessary to take into account the concepts related to Period and Centripetal Acceleration.

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With our values we know that

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Therefore solving to find V, we have:

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3 years ago

Newly discovered planet has twice the mass and three times the radius of the earth. What is the free-fall acceleration at its su

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Answer:

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3 years ago

Read 2 more answers

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$493 \; \text{W}\cdot \text{m}^{-2}$ .

<h3>Explanation</h3>

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$\dfrac{P}{A} = \sigma \cdot T^{4}$

where,

$P$ the total power emitted,
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$\sigma = 5.67 \times 10^{-8} \;\text{W}\cdot \text{m}^{-2} \cdot \text{K}^{-4}$ .

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Keep as many significant figures in $T$ as possible. The error will be large when $T$ is raised to the power of four. Also, the real value will be much smaller than $493\; \text{W}\cdot \text{m}^{-2}$ since the emittance of a human body is much smaller than assumed.

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