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uranmaximum [27]

4 years ago

8

A spring, when compressed 0.20 m from the equilibrium position, stores 24 J. What is the value of the spring constant

1 answer:

Delvig [45]4 years ago

3 0

Given that:

spring compressed ( x) = 0.2 m

Work (W) = 24 J

determine , spring constant (x)=?

We know hat ,

Work (W) = (1/2). k.x²

24 = 1/2 ( k × 0.2²)

48 = k × 0.2²

<em> k = 1200 N/m</em>

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Answer:

The pressure is $p_1 = 4051.4 \ Pa$

Explanation:

From the question we are told that

The gauge pressure at the mouth is $p_1$

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Generally according to Bernoulli's equation

$p_1 = [\frac{1}{2} \rho v_2^2 + h_2 \rho g +p_2] -[\frac{1}{2} \rho * v_1^2 + h_1 \rho g ]$

Now $\rho = 1000 \ kg m^{-3}$ which is the density of water

$p_2$ is the gauge pressure of the atmosphere which is zero

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$[(0.5 * 1000 * 0.31^2) + (0.08*1000 * 9.8)]$

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