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3 years ago

A stretch spring has an elastic potential energy of 35 J when it is stretched 0.54m. What is the spring constant of the spring?

Physics

1 answer:

AlekseyPX3 years ago

5 0

Answer:

240.1 N/m

Explanation:

Applying the formula for the potential energy stored in a stretched spring,

E = 1/2ke².................... Equation 1

Where E = potential energy, k = spring constant, e = extension.

make k the subject of the equation,

k = 2E/e².................. Equation 2

Given: E = 35 J, e = 0.54 m

Substitute into equation 2

k = 2(35)/0.54²

k = 70/0.2916

k = 240.1 N/m.

Hence the spring constant of the spring is 240.1 N/m

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Answer:

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3 years ago

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Marina CMI [18]

For a cylinder that has both ends open resonant frequency is given by the following formula:
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4 years ago

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3 years ago

Through what potential difference must electrons be accelerated so that they will exhibit wave nature in passing through a pinho

Mila [183]

Answer:

0.15 mV

Explanation:

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$\lambda=0.10 \mu m = 1.0\cdot 10^{-7} m$

The de Broglie wavelength of an electron is

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$e\Delta V = \frac{1}{2}mv^2$

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So, we can find the potential difference needed:

$\Delta V=\frac{mv^2}{2e}=\frac{(9.11\cdot 10^{-31})(7278)^2}{2(1.6\cdot 10^{-19})}=1.5\cdot 10^{-4}V = 0.15 mV$

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