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Anna35 [415]
3 years ago
13

A 25-coil spring with a spring constant of 350 N/m is cut into five equal springs with five coils each. What is the spring const

ant of each of the 5-coil springs
Physics
1 answer:
DanielleElmas [232]3 years ago
7 0
<h2>Answer:</h2>

1750N/m

<h2>Explanation:</h2>

According to Hooke's law, the spring constant (k) of an elastic material is the ratio of the force (F) applied to the material to the extension (x) of the material caused by this force. i.e

k = F / x         --------------(i)

From the question, the elastic material (spring) of 25 coils has a spring constant of 350N/m. This means that for every 350N force applied to the spring, the spring extends by 1m.

Now if the spring is cut into five equal parts each with five coils, imagine they are attached together such that the same force of 350N is applied to still cause a total extension of 1m. Each spring contributes 1/5 of this extension.

Therefore, the extension caused by each spring is 1/5 of 1m = 0.2m

Since the same force of 350N is applied, substitute F = 350N and x =  0.2m into equation (i) as follows;

k = 350 / 0.2

k = 1750N/m

Therefore, the spring constant of each of the 5-coil spring is 1750N/m

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True

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A concave mirror produces real and inverted image when object placed at different locations beyond focus of the mirror.

A concave mirror produces virtual and erect and magnified image when an object placed between focus and the pole of the mirror.

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what is the amplitude of the transverse wave modeled in the figure below if the height of a crest is 3 cm above is the resting p
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Explanation:

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3 years ago
I. Choose the correct answer:
Ann [662]

Answer:

1. c

2. b

3. a

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3 0
3 years ago
The electric field of a sinusoidal electromagnetic wave obeysthe equation
Firdavs [7]

Answer with Explanation:

We are given that

E=-(375V/m)sin(5.97\times 10^{15}(rad/s)t+(1.99\times 1067(rad/m)x)

a.General equation of electric field wave

E=E_0sin(\omega t+kx)

Where E_0=Amplitude of electric field wave

By comparing

\omega=5.97\times 10^{15}rad/s

k=1.99\times 10^7rad/m

a.Amplitude of electric field wave=E_0=375V/m

b.Amplitude of magnetic field wave,B_0=\frac{E_0}{c}

Where c=3\times 10^8 m/s

Amplitude of magnetic field wave=B_0=\frac{375}{3\times 10^8}=125\times 10^{-8} T

c.Frequency of wave,f=\frac{\omega}{2\pi}=\frac{5.97\times 10^{15}}{2\pi}=0.95\times 10^{15}Hz

d.Wavelength,\lambda=\frac{2\pi}{k}

\lambda=\frac{2\pi}{1.99\times 10^7}=3.16\times 10^{-7} m

e.Period of wave,T=\frac{1}{f}=\frac{1}{0.95\times 10^{15}}=1.05\times 10^{-15} s

f.Speed of wave,v=f\lambda=0.95\times 10^{15}\times 3.16\times 10^{-7}=3.00\times 10^8 m/s

5 0
4 years ago
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