Physics Questions and Answers

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