Physics Questions and Answers

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

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 = \(2.4×10^{10}(Nm^{-2})\)