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Contact [7]
2 years ago
14

Speed of a train is 5m/s, find its value in km/hr.​

Physics
2 answers:
antiseptic1488 [7]2 years ago
4 0

Answer: 18km/hr

0.005km/s

1 sec --> 0.005

3600sec --> 0.005*3600 is 18

So, 18km/hr

Free_Kalibri [48]2 years ago
3 0

▪▪▪▪▪▪▪▪▪▪▪▪▪  {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪

Here's the solution ~

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:5 \: m \: s {}^{ - 1}

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:\dfrac{5}{1000} km \:  {s}^{ - 1}

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:\dfrac{5}{1000} \times 3600 \:  km \:  {h}^{ - 1}

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \: \dfrac{5}{5}  \times 18 \: km \: h {}^{ - 1}

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \: 18 \: km \: h {}^{ - 1}

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Solution :

Given :

M = 0.35 kg

$m=\frac{M}{2}=0.175 \ kg$

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But $K.E._{top} = 0$ and $P.E._{bottom} = 0$

Therefore, potential energy at the top = kinetic energy at the bottom

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      = 2.62 m/s

It is the velocity of M just before collision of 'm' at the bottom.

We know that in elastic collision velocity after collision is given by :

$v_1=\frac{m_1-m_2}{m_1+m_2}v_1+ \frac{2m_2v_2}{m_1+m_2}$

here, $m_1=M, m_2 = m, v_1 = 2.62 m/s, v_2 = 0$

∴ $v_1=\frac{0.35-0.175}{0.5250}+\frac{2 \times 0.175 \times 0}{0.525}

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3 0
3 years ago
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gladu [14]

Where they slide over each other.

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3 years ago
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Alika [10]

sorry - late reply...just stumbled across tis...hope u can still use it :)


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<span>
</span>

<span>where di = distance to image = +12cm (+ for real image)</span>


and do = distance to object = +8cm


Substitute and solve for f, the focal length

<span><span>
1/12 + 1/8 = 1/f
</span><span>
1/f = (8 + 12) / 12 * 8 = 20/96
</span><span>
so f = 96/20 = 4.8 cm</span>
</span>
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Answer:

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