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3 years ago

A 87 arrow is fired from a bow whose string exerts an average force of 105 on the arrow over a distance of 75 .

Physics

1 answer:

timofeeve [1]3 years ago

6 0

The solution would be like this for this specific problem:

V^2 = 2AS = 2FS/M

V = sqrt(2FS/M) = sqrt(2*105*.75/.087) = 44.52817783 = 42.5 mps

So the speed of the arrow as it leaves the bow is 42.5 mps.

I am hoping that this answer has satisfied your query and it will be able to help you in your endeavor, and if you would like, feel free to ask another question.

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bazaltina [42]

The equilibrant force of the two given forces is 14.14 N.

<h3 /><h3 /><h3>What is equilibrant force?</h3>

This is a single force that balances other given forces.

The given parameters:

First force, F₁ = 10 N
Second force, F₂ = 10 N
Angle between the forces, θ = 90⁰

The equilibrant force of the two given forces is calculated as follows;

$F = \sqrt{F_1 ^2 + F_2 ^2} \\\\F = \sqrt{(10)^2 + (10)^2} \\\\F = 14.14 \ N$

Thus, the equilibrant force of the two given forces is 14.14 N.

Learn more about equilibrant force here: brainly.com/question/8045102

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2 years ago

How long will it take to go 150000m traveling at 50km/hr

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Answer:

3000 hurs

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3 years ago

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Stells [14]

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v = 7.67 m/s

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3 years ago

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