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Answer 1

$\sf\huge{\underline{\mathbb{Question}}}$

A ball of mass 150 g moving with a velocity of 15 $ms^{-1}$ is brought to rest by a player in 0.1s.Calculate the force acting on the body?!

$\sf\huge{\underline{Answer}}$

Given: mass = 150g = $150\times 10^{-3}$kg

Velocity= 15$ms^{-1}$

t=0.1s

hence we have,

Acceleration =$\frac{velocity}{time}$

= $\frac{15}{0.1}$

a(acceleration)= 150$ms^{-2}$

Force = $mass\times acceleration$

Force = $150\times 10^{-3}\times 150$

Force =22.5 newton!!..

2.$\sf\huge{\underline{\mathbb{Question}}}$

The gravitational force between two objects is 49 N. How much the distance between these objects be decreased that the force between them becomes double!!??..

$\sf\huge{\underline{solution}}$

F(g)=>Gravitational force = 49N

let the two masses be m1 and m2, and the distance between r1..

then we have.

F(g)= $\frac{G\times m1\times m2}{r1^{2}}$

49 = $\frac{G\times m1\times m2}{r1^{2}}$==(1)

second case【2】

For force should be double i. e, $49\times 2$= 98N

let the reduced distance between the two masses be r2..

then the force is given by.!!

98 = $\frac{G\times m1\times m2}{r2^{2}}$==(2)

Dividing equation (2) by (1)

$\frac{98}{49}$ = $\frac{G\times m1\times m2}{r2^{2}}$÷$\frac{G\times m1\times m2}{r1^{2}}$

2 = $\frac{r1^{2}}{r2^{2}}$

2= ${[tex]\frac{r1}{r2}$}^2[/tex]

√2= $\frac{r1}{r2}$

$√2\times r2$ = r2

r1 = $1.41\times r2$

r2=$\frac{1\times r1}{1.41}$

r2 = 0.7092 r1

then decrease in distance should be 0.7092 r1

[tex]0.7092\times r1[/ is the distance which should be reduced between the two masses!!