JAMB Physics Past Questions & Answers - Page 12

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.