CBSE 11TH PHYSICS - Online Test

Initial velocity u = 60 m/s

As it stops then final velocity v = 0 m/s

Time taken t = 2.0 s

We know,

v-u = at

=> 0-60= 2a

=> a = −602=−30 m/s2

-ve sign shows velocity is decreasing.

Initial velocity u = 10m/s

As at the maximum height ball ll stop so final velocity v = 0 m/s

Only acceleration working on it is acceleration due to gravity g = -9.8m/s​​​​​​2

Let height = h

So we know

v2−u2=2as

=>(0)2−(10)2=2×(−9.8)h

=> h=−100−19.6$h=$ 5.10 m

Also let time taken to reach maximum height = t

Then 

We know 

v = u +at

=> 0 = 10 +(-9.8)t

=> t = −10−9.8=1.02s

W=nR(Tf−Ti)V∝T10273=3TfTf=81.9KTi=273K

W=1×8.31×(81.9−273)=−1.6×103J

When the lift fall freely under gravity, a = g

Therefore Apparent weight, (We experience weight due to reaction)

R=m(g−a)=m(g−g)=0

This is the condition for weightlessness.

Energy liberated per second 

E=mc2=4.19×109×3×108×3×108=37.71×1025J

power output of sun is equal to energy output per second 

P=Wt=37.71×10251=3.77×1026W