Question

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Answer my questions which are given in the attachment

❤Conditions=> Fullexplanation❤

Class => 11

Subject => Physics

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Jai bajarang bali

@Dr. S.M.

Answer 1

$here \: is \: the \: answer \: of \: your \: question$
1. In the above question we have given that,

u = 72 km/hr.
Now u in m/s = 72 x 5/18 = 20m/s

v = 0 m/s

a = ?

As static friction is present between the tyres and road.

So, g = - 0.5 x 10 = - 5.0 m/so

As we know that,

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

s = v² - u/2a

s = 0 - (20)²/-10

s = - 400/-10

$s \: = \: 40 \: m$


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2. Consider a body of mass (m) sliding down a rough inclined plane of angle theta as shown figure (in attachment).

In equilibrium,

$f \: = \: mg \: sin \: \alpha \: .....(1)$

Let R = r

$r \: = \: mg \: cos \: \alpha \: .....(2)$
Divide (1) by (2)


$f \: = \: mg \: sin \: \alpha \div r \: = \: mg \: sin \: \alpha$

On dividing we get,

$f \: \div \: r \: = \: tan \: \alpha$

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Hope it helps you.