Physics Questions and Answers

Quantity of heat required to raise the temperature of the ice cube from \(-10^oC\) to \(0^oC\)
⇒H=mc►ø = 0.02 × 2100 × 0 -(-10)

⇒ H = 0.02 × 2100 × (0 + 10) = 0.02 × 2100 × 10

⇒ H = 420 J

quantity of heat required to melt ice at \(0^oC\)

⇒ H = mL = 0.02 × 3.34 × \(10^5\)

⇒H=6680 j

Quantity of heat required to raise the temperature of the melted ice cube (water) from \(0^oC\) to \(-10^oC\)

⇒H=mc►ø = 0.02 × 4200 × (10-0)

⇒H = 0.02 × 4200 × 10

⇒H = 840j

;420 + 6680 + 840 = 7940 j

 

A= \(0.8m^2\)    d= 20mm =\(\frac{20}{1000}\) = 0.02m

v =120v;  \(ε_oA\)= \(8.85 × 10^-12 fm^-1\)

C = \(\frac{ε_oA}{d}\)

C =\(\frac{8.85 × 10^-12×0.8}{0.02}\)

C= \(3.54 × 10^-10 F\)

Q= CV

⇒  \(3.54 × 10^-10\) × 120 = \(4.25×10^-8c\)

     Q=\(42.5 × 10^-9c\)  = 42.5nC

      

\(r_1\) = 4.5cm , \( P_1\) =is the total pressure on the bubble at a depth of 12m from the surface.

\(P_1\) = 12 + 10.34 =22.34m

 \(V_1\) = \(\frac{4}{3}\)π× \(r^3_1\)

= \(\frac{4}{3}× π×{4.5^3cm^3}\)

\(P_2\) = 10.34m

\(V_2\)  = \(\frac{4}{3} {π}{r^3_2}\)

from boyles law:

\(P_1V_1\)  = \(P_2V_2\) 

⇒ 22.34× \(\frac{4}{3}× π×{4.5^3}\) = 10.34 × \(\frac{4}{3}×π×{r^3_2}\)

⇒ 22.34 × \(4.5^3\) = 10.34 × \(r^3_2\)

⇒ \(r^3_2 = \sqrt[3]{196.88}\)

⇒ \(r_2\) = 5.82cm

\(r_1\) = 4.5cm , \( P_1\) =is the total pressure on the bubble at a depth of 12m from the surface.

\(P_1\) = 12 + 10.34 =22.34m

 \(V_1\) = \(\frac{4}{3}\)π× \(r^3_1\)

= \(\frac{4}{3}× π×{4.5^3cm^3}\)

\(P_2\) = 10.34m

\(V_2\)  = \(\frac{4}{3} {π}{r^3_2}\)

from boyles law:

\(P_1V_1\)  = \(P_2V_2\) 

⇒ 22.34× \(\frac{4}{3}× π×{4.5^3}\) = 10.34 × \(\frac{4}{3}×π×{r^3_2}\)

⇒ 22.34 × \(4.5^3\) = 10.34 × \(r^3_2\)

⇒ \(r^3_2 = \sqrt[3]{196.88}\)

⇒ \(r_2\) = 5.82cm

ε= 80%, L=200× 10 = 2000N, \(d_L\)=10m
 

\(d_E\)= 110-10 = 100m, E=?

ε = \(\frac{work done on the load}{work done by the effort}\)  × 100%

⇒  work done on load = 2000 ×10 =20,000 j


⇒ work done by effort = E × 100  = 100E

⇒ 80 =   \(\frac{20,000}{100E}\) × 100%

⇒ 80 =   \(\frac{20,000}{E}\) 

⇒ E = \(\frac{20,000}{80}\)

              ⇒  E = 250N

m=2kg, r=1.2m, f=120rev/min ; T=?

f=   \(\frac{120rev}{1min}\) × \(\frac{1min}{60s}\) = 2rev\s

w=2πf= 2×π×2= 4πrad\s

T= \(\frac{mv^2}{r}\),  but v = wr 

⇒ T = \(\frac{(m)(wr)^2}{r}\)

⇒ T = \(mw^2r\)

= T = 2× (4π)^2 × 1.2

   T = 379N ( to 3 s.f)

r=0.2mm = \(2 × {10^{-4}}\) e= 0.05% of L,=0.005L, m=1.5kgg=\(10 ms^{-2}\) ,  y=?,    

(1000 mm = 1m)

Y = \(\frac{FL}{Ae}\)

w = mg = 1.5 × 10 = 15

A = \(\pi r^2\) = \(\frac{22}{7}×(2×10^{-4})^2\) = \(1.26×10^{-7}(m^2)\)

⇒ Y = \(\frac{15L}{1.26×10^{-7}×0.005L}\)

⇒ Y = \(\frac{15}{1.26×10^{-7}×0.005}\)

⇒ Y = \(\frac{15}{6.29×10^{-10}}\)

Y = \(2.4×10^{10}(Nm^{-2})\)

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