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3 years ago

How much work does an athlete do if she raises a 5N kettle bell 2 m off the ground?

Physics

1 answer:

Degger [83]3 years ago

6 0

Answer:

<h2>The answer is 10 N</h2>

Explanation:

To find the work done by an object given the distance covered by the object and it's force we use the formula

<h3>work done = force × distance</h3>

From the question

force = 5 N

distance = 2 m

We have

work done = 5 × 2

We have the final answer as

Hope this helps you

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Which of the following represents an upright image? <br> A. -do<br> B. +m<br> C. -m<br> D. +do

Tom [10]

Answer:

B. +m

Explanation:

The magnification of an image is defined as the ratio between the size of the image and of the object:

$m = \frac{y'}{y}$

where we have

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There are two possible situations:

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3 years ago

A spring has a 12 cm length. When a 200-g mass is hung from the spring, it extends to 27 cm. The hanging mass were pulled downwa

BartSMP [9]

Answer:

The equation of the time-dependent function of the position is $x(t)=5\cos(8.08t)$

(b) is correct option.

Explanation:

Given that,

Length = 12 cm

Mass = 200 g

Extend distance = 27 cm

Distance = 5 cm

Phase angle =0°

We need to calculate the spring constant

Using formula of restoring force

$F=kx$

$mg=kx$

$k=\dfrac{mg}{x}$

$k=\dfrac{200\times10^{-3}\times9.8}{(27-12)\times10^{2}}$

$k=13.06\ N/m$

We need to calculate the time period

Using formula of time period

$T=2\pi\sqrt{\dfrac{m}{k}}$

Put the value into the formula

$T=2\pi\sqrt{\dfrac{0.2}{13.6}}$

$T=0.777\ sec$

At t = 0, the maximum displacement was 5 cm

So, The equation of the time-dependent function of the position

$x(t)=A\cos(\omega t)$

Put the value into the formula

$x(t)=5\cos(2\pi\times f\times t)$

$x(t)=5\cos(2\pi\times\dfrac{1}{T}\times t)$

$x(t)=5\cos(2\pi\times\dfrac{1}{0.777}\times t)$

$x(t)=5\cos(8.08t)$

Hence, The equation of the time-dependent function of the position is $x(t)=5\cos(8.08t)$

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