Physics Questions and Answers

m=1200kg,  $V_1$= $10ms^{-1}$   $V_2$ = $15ms^{-1}$,  w= ?

work=►K.E = $K.E_2$ =  $K.E_2$ - $K.E_1$

⇒work= $\frac{1}{2}{mv^2_2}-\frac{1}{2}{mv^2_1}$

⇒work= $\frac{1}{2}m({v^2_2}-{v^2_1}$)

⇒work= $\frac{1}{2}× 1200× (15^2-10^2)$

⇒work = 600 ×  (225 -100)

⇒work= 600 × 125

⇒work= 7.5×$10^4 j$

Nuclear fusion is the process in which lighter nuclei are combined to form heavier ones. Matter lost in this process is converted into energy. The energy produced inside the stars is due to the process of nuclear fusion.

L=1.5m,  A=$620m^2$,  ø=25-15=10°c

K=0.6071 , $Wm^{-1}k^{-1}$ ,  $\frac{q}{t}$ =?

$\frac{q}{t}$ =KA,     $\frac{►t}{L}$ 

where k is the thermal conductivity constant and $\frac{q}{t}$  is the rate of heat transfer

=$\frac{q}{t}$=0.6071×620×$\frac{10}{1.5}$

$\frac{q}{t}$ = 2509.35w ≈ 2.5kw

As light ray enters a drop of water the light is refracted at the surface and at the end of the drop, it is totally internally reflected in which the reflected light returns to the front surface, where it again undergoes refraction as it moves from water to air. The result of this is a dispersed light of colours of different wavelengths.

Object A will undergo unstable equilibrium as it can move farther from its original position

Object B will undergo stable equilibrium as it can return to its original position

Object C will undergo neutral equilibrium as it can remain in a new position