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2 years ago

How far will 20 N of force stretch a spring with a spring constant of 140 N/m?

Physics

1 answer:

densk [106]2 years ago

7 0

Answer:

0.143 m

Explanation:

The relationship between force applied on a string and stretching of the spring is given by Hooke's law:

$F=kx$

where

F is the force exerted on the spring

k is the spring constant of the spring

x is the stretching of the spring from its equilibrium position

In this problem, we have:

F = 20 N is the force applied on the spring

k = 140 N/m is the spring constant

Solving for x, we find how far the spring will stretch:

$x=\frac{F}{k}=\frac{20}{140}=0.143 m$

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Elena L [17]

Answer:

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Explanation:

Given,

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$d = \frac{PE}{mg}$

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h = 228.5 m

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Height of the floor above ground is equal to 599.5 m.

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2 years ago

A 6.00-mH solenoid is connected in series with a 5.0-μF capacitor and an AC source. The solenoid has internal resistance 3.0 Ω w

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Answer:

5773.50269 Hz

23 A

Explanation:

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$R$ = Resistance = 3 Ω

$\epsilon$ = Maximum emf = 69 V

Resonant angular frequency is given by

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The resonant angular frequency is 5773.50269 Hz

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The current amplitude at the resonant angular frequency is 23 A

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Vladimir79 [104]

Answer:

$m = 2.23 \times 10^{-32} kg$

Explanation:

Given data:

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we know that

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solving for

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$a = \frac{v^2}{\frac{vT}{2\pi}}$

$a = \frac{2\pi v}{T}$

we know that

f =ma

$G\times \frac{\pi m}{(vT)^2} = a = \frac{2\pi m v}{T}$

solving for m

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2 years ago

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