Calculate the velocity of a moving train with a mass 10,000kg and 150,000J of energy.

1 answer:

5 0

Kinetic energy possessed by moving train is $E_k = 150000J$ .

And mass $m = 10000kg$ .

The formula for kinetic energy is $E_k = \frac{m\times v^2}{2}$ .

Solve for velocity:

$E_k = \frac{m\times v^2}{2}\implies 2E_k = m\times v^2$

$v^2=\frac{2E_k}{m}\implies \underline{v = \sqrt{\frac{2E_k}{m}}}$

Now put in the data:

$v = \sqrt{\frac{2\times150000J}{10000kg}}\approx5.48\frac{m}{s}$

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Answer:

The 95% confidence interval is (-0.2451, 06912)

Step-by-step explanation:

From the question, we have;

The number of small cars in the sample of small cars, n₁ = 12

The number of small cars that were totaled, x = 8

The number of large cars in the sample of small cars, n₂ = 15

The number of large cars that were totaled, y = 5

Therefore, the proportion of small cars that were totaled, pX = x/n₁

∴ pX = 8/12 = 2/3

The proportion of large cars that were totaled, pY = y/n₁

∴ pY = 5/15 = 1/3

The 95% confidence interval for the difference pX - pY is given as follows;

$pX-pY\pm z^{*}\sqrt{\dfrac{pX\left (1-pX \right )}{n_{1}}+\dfrac{pY\left (1-pY \right )}{n_{2}}}$

$\dfrac{2}{3} -\dfrac{1}{3} \pm 1.96 \times \sqrt{\dfrac{\dfrac{2}{3} \times \left (1-\dfrac{2}{3} \right )}{12}+\dfrac{\dfrac{1}{3} \times \left (1-\dfrac{1}{3} \right )}{15}}$