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3 years ago

16) A man ran a 5 mile race. The race looped around a city park and back

Physics

1 answer:

finlep [7]3 years ago

3 0

Answer:

The man's total displacement is equal to 0.

Explanation:

Given that,

A man ran a 5 mile race. The race looped around a city park and back to the starting line.

We need to find the total displacement of the man.

We know that,

Displacement = shortest path covered

Also,

Displacement = final position - initial position

As it reaches back to its starting line, it means, the displacement is equal to 0.

Hence, the man's total displacement is equal to 0.

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Explanation:

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3 years ago

A pendulum makes 60 vibrations in 15 secs what’s its frequency

o-na [289]

Answer:

4 hertz

Explanation:

The defination of freqyency = the total no of cycle made by a wave in one second .

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cycle or vibrations=60

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now,

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hence, the its frequency = 4hertz

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3 years ago

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Firdavs [7]

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3 years ago

What is the difference between continuous and a discontinuous spectrum?

aleksklad [387]

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5 0

3 years ago

If a person weighs 818 N on earth and 5320 N on the surface of a nearby planet, what is the acceleration due to gravity on that

alexandr402 [8]

Answer:

$g'=63.74\ m/s^2$

Explanation:

It is given that,

Weight of the person on Earth, W = 818 N

Weight of a person is given by the following formula as :

$W=mg$

g is the acceleration due to gravity on earth

$m=\dfrac{W}{g}$

$m=\dfrac{818\ N}{9.8\ m/s^2}$

m = 83.46 kg

The mass of an object is same everywhere. It does not depend on the location.

Let W' is the weight of the person on the surface of a nearby planet, W' = 5320 N

g' is the acceleration due to gravity on that planet. So,

$g'=\dfrac{W'}{m}$

$g'=\dfrac{5320\ N}{83.46\ kg}$

$g'=63.74\ m/s^2$

So, the acceleration due to gravity on that planet is $63.74\ m/s^2$ . Hence, this is the required solution.

6 0

3 years ago