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3 years ago

9

A spring that is compressed 14.5 cm from its equilibrium position stores 2.99 J of potential energy. Determine the spring consta

nt.

1 answer:

strojnjashka [21]3 years ago

7 0

Answer:

284.4233 N/m

Explanation:

k = Spring constant

x = Compression of spring = 14.5 cm

U = Potential energy = 2.99 J

The potential energy of a spring is given by

$U=\dfrac{1}{2}kx^2$

Rearranging to get the value of k

$\\\Rightarrow k=\dfrac{2U}{x^2}\\\Rightarrow k=\dfrac{2\times 2.99}{0.145^2}\\\Rightarrow k=284.4233\ N/m$

The spring constant is 284.4233 N/m

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2 years ago

Compute the dot product of the vectors u and v, and find the angle between the vectors. Bold v equals 7 Bold i minus Bold j and

OLga [1]

Answer:

$\theta = 106.3 degree$

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As we know that

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also we can find the magnitude of two vectors as

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similarly we have

$|w| = \sqrt{(7^2) + (-1)^2}$

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3 years ago

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FrozenT [24]

Answer:

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4 0

3 years ago