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1 year ago

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a 1200 kg car rolling on a horizontal surfact has a speed of 25 m/s when it strikes a horizontal coiled spring and is brought ot

rest in a distance of 2.5m. what is the spring constant of this spring

1 answer:

gizmo_the_mogwai [7]1 year ago

5 0

The required spring constant:

The spring constant of the spring is $12\times 10^4 \text{ N/m}$ .

Calculation:

The mass of the car is m=1200 kg, the speed of the car is v=25 m/s, and after colliding the spring is brought to rest at a distance of x=2.5m. Let the spring constant of the spring is, k.

From the conservation of energy,

Total initial kinetic energy= Total final potential energy of the spring

Therefore,

$\frac{1}{2}mv^2=\frac{1}{2}kx^2$

Now, substituting the values of the mass of the car, speed of the car, and displacement, we get:

$\begin{aligned}1200\times(25)^2&=k\times (2.5)^2\\\Rightarrow k&=\frac{1200\times(25)^2}{(2.5)^2}\\&=12\times 10^4 \text{ N/m}\end{aligned}$

To know more about spring constant, refer to:

brainly.com/question/14159361

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A sample of an ideal gas is in a tank of constant volume. The sample absorbs heat energy so that its temperature changes from 33

Anuta_ua [19.1K]

Answer:

$\frac{v_{2}}{v_{1}}=2$ .

Explanation:

The average kinetic energy per molecule of a ideal gas is given by:

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Now, we know that $\bar{K} = (1/2)m\bar{v}^{2}$

Before the absorption we have:

$(1/2)m\bar{v_{1}}^{2}=\frac{3k_{B}T_{1}}{2}$ (1)

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If we want the ratio of v2/v1, let's divide the equation (2) by the equation (1)

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$\frac{v_{2}}{v_{1}}=\sqrt{\frac{T_{2}}{T_{1}}}$

$\frac{v_{2}}{v_{1}}=\sqrt{\frac{1340}{335}}$

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I hope it helps you!

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2 years ago

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In that system, it is possible to determine the velocity needed to maintain the orbit. The formula is: v = $\sqrt{\frac{GM}{r} }$ , where:

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M = 42.553. $10^{36}$ kg

The mass of the massive object at the center of the ring is 42.553. $10^{36}$ kg.

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