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 = $2.4×10^{10}(Nm^{-2})$