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3 years ago

A 5 kg mass is oscillating on a spring with a time period of 2.8 seconds. What is the spring constant k of the spring?

Physics

1 answer:

katrin [286]3 years ago

5 0

Answer:

Explanation:

<u>Simple Harmonic Oscillator</u>

It consists of a weight attached to one end of a spring being allowed to move forth and back.

If m is the mass of the weight and k is the constant of the spring, the period of the oscillation is given by:

$\displaystyle T=2\pi {\sqrt {\frac {m}{k}}}$

If the period is known, we can find the value of the constant by solving for k:

$\displaystyle k=m\left(\frac{2\pi}{T}\right)^2$

Substituting the given values m=5 Kg and T=2.8 seconds:

$\displaystyle k=5 \left(\frac{2\pi}{2.8}\right)^2$

k = 25.18 N/m

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3 years ago

What is the voltage, V2, in units of Volts, across resistor R2 in the circuit shown below where VS = 4V, R1 = 14 Ohms and R2 = 3

svetlana [45]

<h2>Correct answer:</h2>

$\boxed{v_{out}=2,85V}$

<h2>Explanation:</h2>

We can use voltage divider to solve this problem that is defined as the passive linear circuit producing an output voltage $v_{out}$ that is a fraction of its input voltage $v_{in}$ . So we can use the formula:

$v_{out}=\frac{R_{2}}{R_{1}+R_{2}}v_{in}, \ where \ v_{in}=v_{s}=4V \\ \\ \therefore v_{out}=\frac{35}{14+35}(4) \\ \\ \therefore v_{out}=2,85V$

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2 years ago

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Deffense [45]

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