Question

Physics questions and solution​

Answer 1

Answer:

Your answer is 28 a)scalar quantity

b)vector quantity

Hope it helps you friend

Answer 2

$\mathbb{\large{\underline{\underline{EXPLANATION:-}}}}$

$\textbf{\underline{\underline{Que 22}}}</p><p>$

$\sf{Given}\begin{cases}\sf{t=10s}\\ \sf{a=-2.5ms^{2}}\\ \sf{u=??}\\ \sf{v=0ms^{-1}}\end{cases}$

$\textbf{\underline{Using equation of motion,,}}</p><p>$

$\mathtt{\boxed{\blue{v=u+at}}}$

$\sf{\implies{v=u+(-2.5)(10)}}$

$\sf{\implies{v=u-25}}$

$\sf{\implies{u=25ms^{-1}}}$

Therefore initial velocity,,

$\sf{\underline{\boxed{u=25ms^{-1}}}}$

$\rule{200}2$

$\textbf{\underline{\underline{Que 23}}}</p><p>$

$\sf{Given}\begin{cases}\sf{t=2s}\\ \sf{a=10ms^{2}}\\ \sf{u=0ms^{-1}}\\ \sf{s=??}\end{cases}$

$\textbf{\underline{Using equation of motion,,}}</p><p>$

$\mathtt{\boxed{\blue{s=ut+\frac{1}{2}at^{2}}}}$

$\sf{\implies{s=0t+\frac{1}{2}(10)(2)^{2}}}$

$\sf{\implies{s=\frac{1}{2}\times{40}}}$

$\sf{\implies{s=+20m}}$

Hence the distance covered by the body,,

$\sf{\underline{\boxed{s=20m}}}$

$\rule{200}2$

$\textbf{\underline{\underline{Que 24}}}</p><p>$

$\sf{Given}\begin{cases}\sf{t=10s}\\ \sf{a=0.2ms^{2}}\\ \sf{u=5ms^{-1}}\\ \sf{s=??}\\ \sf{v=??}\end{cases}$

$\textbf{\underline{Using equation of motion,,}}</p><p>$

$\mathtt{\boxed{\blue{s=ut+\frac{1}{2}at^{2}}}}$

$\sf{\implies{s=5\times{10}+\frac{1}{2}(0.2)(10)^{2}}}$

$\sf{\implies{s=50+\frac{1}{2}\times{0.2}\times{100}}}$

$\sf{\implies{s=50+\frac{1}{2}\times{20}}}$

$\sf{\implies{s=50+10}}$

$\sf{\implies{s=60m}}$

Therefore, Distance travelled by the motorcycle,,

$\sf{\underline{\boxed{s=60m}}}$

$\textbf{\underline{To find final velocity,,}}</p><p>$

$\mathtt{\boxed{\blue{v=u+at}}}$

$\sf{\implies{v=5+(0.2)(10)}}$

$\sf{\implies{v=5+2}}$

$\sf{\implies{v=7ms^{-1}}}$

Therefore,, final velocity of the motorcycle,,

$\sf{\underline{\boxed{v=7ms^{-1}}}}$

$\rule{200}2$

$\textbf{\underline{\underline{Que 25}}}</p><p>$

$\sf{Given}\begin{cases}\sf{t=2.5s}\\ \sf{a=??}\\ \sf{u=18ms^{-1}}\\ \sf{v=0ms^{-1}}\end{cases}$

$\textbf{\underline{Using equation of motion,,}}</p><p>$

$\mathtt{\boxed{\blue{0=18+a(2.5)}}}$

$\sf{\implies{\frac{-18}{2.5}=a}}$

$\sf{\implies{a=\cancel\dfrac{-180}{25}}}$

$\sf{\implies{a=-7.2ms^{-2}}}$

Hence,, retardation,,

$\sf{\underline{\boxed{a=-7.2ms^{-2}}}}$

$\rule{200}2$

$\textbf{\underline{\underline{Que 26}}}</p><p>$

$\sf{Given}\begin{cases}\sf{t=3min=180s}\\ \sf{a=0.2ms^{-2}}\\ \sf{u=0ms^{-1}}\\ \sf{v=??}\\ \sf{s=??}\end{cases}$

$\textbf{\underline{Using equation of motion,,}}</p><p>$

$\mathtt{\boxed{\blue{v=u+at}}}$

$\sf{\implies{v=0+(2.5)(180)}}$

$\sf{\implies{v=25\times{18}}}$

$\sf{\implies{v=450ms^{-1}}}$

Therefore,, speed acquired,,

$\sf{\underline{\boxed{v=450ms^{-1}}}}$

$\rule{200}2$

$\textbf{\underline{To find distance ,,}}</p><p>$

$\textbf{\underline{Using equation of motion,,}}</p><p>$

$\mathtt{\boxed{\blue{s=ut+\frac{1}{2}at^{2}}}}$

$\sf{\implies{s=0+\frac{1}{2}(0.2)(180)^{2}}}$

$\sf{\implies{s=\frac{1}{2}(0.2)(32400)}}$

$\sf{\implies{s=3240m}}$

Therefore Distance travelled by the train,,

$\sf{\underline{\boxed{s=3240m}}}$

$\rule{200}2$

$\textbf{\underline{\underline{Que 29}}}</p><p>$

$\sf{Given}\begin{cases}\sf{t=30s}\\ \sf{a=0.5ms^{-2}}\\ \sf{u=20ms^{-1}}\\ \sf{s=??}\end{cases}$

$\textbf{\underline{Using equation of motion,,}}</p><p>$

$\mathtt{\boxed{\blue{s=ut+\frac{1}{2}at^{2}}}}$

$\sf{\implies{s=20\times{30}+\frac{1}{2}(0.5)(30)^{2}}}$

$\sf{\implies{s=600+\frac{1}{2}(0.5)(900)}}$

$\sf{\implies{s=600+225}}$

$\sf{\implies{s=825m}}$

Therefore Distance travelled by the train,,

$\sf{\underline{\boxed{s=825m}}}$

$\rule{200}2$

$\textbf{\underline{\underline{Que 30}}}</p><p>$

$\sf{Given}\begin{cases} \sf{a=??}\\ \sf{u=15ms^{-1}}\\ \sf{s=18m}\\ \sf{v=0ms^{-1}}\end{cases}$

$\textbf{\underline{Using equation of motion,,}}</p><p>$

$\mathtt{\boxed{\blue{v^{-2}-u^{2}=2as}}}$

$\sf{\implies{0^{-2}-15^{2}=2a(18)}}$

$\sf{\implies{-225=36a}}$

$\sf{\implies{\cancel\dfrac{-225}{36}=a}}$

$\sf{\implies{a=-6.25ms^{-2}}}$

Therefore,, deceleration,,

$\sf{\underline{\boxed{a=-6.25ms^{-2}}}}$

$\rule{200}2$

$\textbf{\underline{\underline{Que 32}}}</p><p>$

$\textbf{\underline{(a) The train's velocity}}</p><p>$

Ans- Since,the v-t graph is a straight horizontal line, therefore the train has uniform velocity.

$\textbf{\underline{(b) The train's acceleration}}</p><p>$

Ans- As we know that if velocity is uniform then acc is zero..Thus here Acc=0.

❣️❣️Hope it helps.. Please mark this answer as brainliest... thanks❣️❣️