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Monica [59]
3 years ago
9

A train is travelling at speed of 20m/s. Brakes are applied so as to produce a uniform acceleration of -0.5m/s2. find how far th

e train will go before it is brought to rest. ​
Physics
1 answer:
aivan3 [116]3 years ago
7 0

Use v^2 - u^2 = 2as

Here, u = 20 m/s , a = -0.5 m/s^2 and v = 0

- 400 = -s ======> s = 400 m

Hope it helps, please mark me as brainliest and give me a good rating. Thank You!

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A mass of gas under constant Pressure occupies a volume of 0.5 M3 At a temperature of 20゚C using the formula for Cubic expansion
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By what angle should the second polarized sheet be rotated relative to the first to reduce the transmitted intensity to one-half
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Answer:

\mathbf{u(t) =\dfrac{m_2g }{k}(1 - cos \omega_n t) + \dfrac{\sqrt{2gh}}{\omega_n}\times \dfrac{m_1}{m_1+m_2}sin \omega _n t}

Explanation:

From the information given:

The equation of the motion can be represented as:

(m_1 +m_2) \hat u + ku = m_2 g--- (1)

where:

m_1 = mass of the body 1

m_2 = mass of the body 2

\hat u = acceleration

k = spring constant

u = displacement

g = acceleration due to gravity

Recall that the formula for natural frequency \omega _n = \sqrt{\dfrac{k}{m_1+m_2}}

And the equation for the general solution can be represented  as:

u(t) = A cos \omega_nt + B sin \omega _n + \dfrac{m_2g}{k} --- (2)

To determine the initial velocity, we have:

\hat u_2^2 = 2gh

\hat u_2 = \sqrt{2gh}

where h = height

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\hat u (t) = - \omega_n A sin \omega_n t+ \omega_n B cos \omega _n t + 0

now if t = 0

Then

\hat u (0) = - \omega_n A sin \omega_n (0)+ \omega_n B cos \omega _n (0) + 0

= \omega _n B

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By replacing the value of \hat u_2 with \sqrt{2gh

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u(0)= \dfrac{ m_2 \times \sqrt{2gh}}{(m_1+m_2)}

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0 = A (1)+0+ \dfrac{m_2g}{k}

A =- \dfrac{m_2g}{k}

Also, if  \hat u (0) = \omega_nB

Then :

\dfrac{m_2}{m_1+m_2}\sqrt{2gh} = \omega_n B

making B the subject; we have:

B = \dfrac{m_2}{m_1 + m_2}\dfrac{\sqrt{2gh}}{\omega_n}

Finally, replacing the value of A and B back to the general solution at equation (2); we have the equation of the subsequent motion u(t) which is:

\mathbf{u(t) =\dfrac{m_2g }{k}(1 - cos \omega_n t) + \dfrac{\sqrt{2gh}}{\omega_n}\times \dfrac{m_1}{m_1+m_2}sin \omega _n t}

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