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disa [49]
3 years ago
14

A horizontal spring on a frictionless surface has a spring constant of 10 N/m with a mass of 2kg attached to the end of the spri

ng. If the spring is stretched 2m passed its point of equilibrium and released, how many times does the mass pass through equilibrium per second
Physics
1 answer:
Alex3 years ago
4 0

Answer:

0.356 times the mass pass through equilibrium per second.

Explanation:

Given that,

Spring constant = 10 N/m

Mass = 2 kg

Stretched spring = 2m

We need to calculate the frequency

Using formula of frequency

f=\dfrac{1}{2pi}\sqrt{\dfrac{k}{m}}

Where, m = mass

k = spring constant

Put the value into the formula

f=\dfrac{1}{2\pi}\sqrt{\dfrac{10}{2}}

f=0.356\ Hz

We know that,

Hertz = cycle per second

Hence, 0.356 times the mass pass through equilibrium per second.

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<span>5.98 x 10^-2 ohms. Resistance is defined as: R = rl/A where R = resistance in ohms r = resistivity (given as 1.59x10^-8) l = length of wire. A = Cross sectional area of wire. So plugging into the formula, the known values, including the area of a circle being pi*r^2, gives: R = 1.59x10^-8 * 3.00 / (pi * (5.04 x 10^-4)^2) R = (4.77 x 10^-8) / (pi * 2.54016 x 10 ^-7) R = (4.77 x 10^-8) / (7.98015 x 10^-7) R = 5.98 x 10^-2 ohms So that wire has a resistance of 5.98 x 10^-2 ohms.</span>
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A non-conducting sphere of radius R = 3.0 cm carries a charge Q = 2.0 mC distributed uniformly throughout its volume. At what di
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given data

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