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2 years ago

What is the equivalent resistance of a circuit that contains three 10.0 Ω

Physics

2 answers:

yanalaym [24]2 years ago

8 0

Answer:

O D.30.0 Ω

Explanation:

this is the correct answer!

Cerrena [4.2K]2 years ago

8 0

Question :

What is the equivalent resistance of a circuit that contains three 10.0 Ω resistors connected in series to a 6.0 V battery?

Options given :

A. 0.600 Ω
Β. 60.0 Ω
C. 16.0 Ω
D. 30.0 Ω

Solution :

As we know that,

When a circuit is connected in series resistance is calculated by :

Rs = R1 + R2 + R3 .... Rn

Here,

R is resistance

As a circuit contains three resistors. So R1 , R2 and R3 would be :

R1 = 10 Ω
R2 = 10 Ω
R3 = 10 Ω

Now,

>> Rs = 10 + 10 + 10

>> Rs = 30 Ω

Therefore,

Option D is correct.

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c) 2.57s

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If you divide the first equation and the third equation, you can calculate w:

$\frac{x}{a}=\frac{Acos(\omega t)}{-\omega^2 Acos(\omega t)}\\\\\omega=\sqrt{-\frac{a}{x}}=\sqrt{-\frac{-107m/s^2}{0.134m}}=28.25\frac{rad}{s}$

with this value you can compute the frequency:

$f=\frac{\omega}{2\pi}=\frac{28.25rad/s}{2\pi}=4.49Hz$

the mass of the block is given by the formula:

$f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}\\\\m=\frac{k}{4\pi^2f^2}=\frac{427N/m}{(4\pi^2)(4.49Hz)^2}=0.536kg$

c) to find the amplitude of the motion you need to know the time t. This can computed by dividing the equation for v with the equation for x and taking the arctan:

$\frac{v}{x}=-\omega tan(\omega t)\\\\t=\frac{1}{\omega}arctan(-\frac{v}{x\omega })=\frac{1}{28.25rad/s}arctan(-\frac{-12.1m/s}{(0.134m)(28.25rad/s)})=2.57s$

Finally, the amplitude is:

$x=Acos(\omega t)\\\\A=\frac{0.134m}{cos(28.25rad/s*2.57s )}=0.45m$

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