Question

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Answer 1

$\large\underline{\sf{Solution-4}}$

Mass of bullet, m = 10g = 0.01 kg

Initial velocity, u = 300 m/ sec

Distance covered, s = 3 cm = 0.03 m

Since, bullet comes to rest.

So, Final velocity, v = 0

We know,

$\rm :\longmapsto\: {v}^{2} - {u}^{2} = 2as$

$\rm :\longmapsto\: {0}^{2} - {300}^{2} = 2a(0.03)$

$\rm :\longmapsto\: - 90000 = 0.06a$

$\bf\implies \:a = - \dfrac{90000}{0.06} = 1500000 \: m {sec}^{ - 2} (numerically)$

Now, we know

$\red{\rm :\longmapsto\:Force = m \times a \: }$

$\red{\rm :\longmapsto\:Force = 0.01 \times 1500000 \: }$

$\red{\rm \implies\:Force = 15000 \: N\: }$

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$\large\underline{\sf{Solution-5}}$

Given that, Forces of

$\sf\:5 \sqrt{2}N \: and \: 6 \sqrt{2}N \: act \: on \: a \: particle \: at \: an \: angle \: of \: 60 \degree$

Let assume that

$\red{\rm :\longmapsto\:F_1 = 5 \sqrt{2} \: N}$

$\red{\rm :\longmapsto\:F_2 = 6 \sqrt{2} \: N}$

So, resultant force, F is given by

$\boxed{\tt{ F = \sqrt{ {F_1}^{2} + {F_2}^{2} + 2F_1F_2 \: cos \theta}}}$

$\rm :\longmapsto\: F = \sqrt{ {(5 \sqrt{2} )}^{2} + {(6 \sqrt{2} )}^{2} + 2(5 \sqrt{2})(6 \sqrt{2}) \: cos 60 \degree}$

$\rm :\longmapsto\:F = \sqrt{50 + 72 + 2 \times 60 \times \dfrac{1}{2} }$

$\rm :\longmapsto\:F = \sqrt{122 + 60 }$

$\rm :\longmapsto\:F = \sqrt{182}$

$\bf\implies \:F = 13.49 N$

Now, Given that

Mass = 1000 kg

We know,

Acceleration is given by

$\red{\rm :\longmapsto\:a = \dfrac{F}{m} }$

$\red{\rm :\longmapsto\:a = \dfrac{13.49 }{1000} }$

$\red{\rm :\longmapsto\:a = 0.01349 \: m \: {sec}^{ - 2} }$