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IRISSAK [1]
3 years ago
15

A mass attached to a spring is displaced from its equilibrium position by 5cm and released. The system then oscillates in simple

harmonic motion with a period of 1s. If that same mass–spring system is displaced from equilibrium by 10cm instead, what will its period be in this case?(A) 0.5s(B) 2s(C) 1s(D) 1.4s
Physics
1 answer:
aliina [53]3 years ago
5 0

Answer:

T= 1 s

Explanation:

Given that

When x=  cm ,T= 1

we know that time period of spring mas system given as

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

T= Time period

m= mass

k=spring constant

So from above equation we can say that time period of system does not depends on the value of x.

So when x= 10 cm ,still time period will be 1 s.

T= 1 s

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Answer:

B) Gets smaller

Explanation:

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Z = R + X   where R is the resistive component and X the reactance component, which is due either to a presence of an  inductor or a capacitor. In any case the total impedance depends on R the resistive, and the phase angle φ is:  

tan⁻¹ φ =  X/R

Have a look to a pure capactive circuit (we are talking about AC current) in this case current leads voltage by 90⁰. If we add a resistor in the circuit  the current still will lead a voltage but in this condition the phase angle will be smaller,

If  R increase, X/R  decrease and tan⁻¹ φ also decrease

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3 years ago
A 120-V rms voltage at 1000 Hz is applied to an inductor, a 2.00-μF capacitor and a 100-Ω resistor, all in series. If the rms va
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Answer:

The inductance of the inductor is 35.8 mH

Explanation:

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Using formula of current

I = \dfrac{V}{Z}

Z=\sqrt{R^2+(L\omega-\dfrac{1}{C\omega})^2}

Put the value of Z into the formula

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Put the value into the formula

0.680=\dfrac{120}{\sqrt{(100)^2+(L\times2\pi\times1000-\dfrac{1}{2\times10^{-6}\times2\pi\times1000})^2}}

L=35.8\ mH

Hence, The inductance of the inductor is 35.8 mH

4 0
3 years ago
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The electric field between the plates of the cathode ray tube of an older television set can be as high as 2.5x104 N/C. Determin
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Answer:

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F = qE

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