Physics Questions and Answers

As light ray enters a drop of water the light is refracted at the surface and at the end of the drop, it is totally internally reflected in which the reflected light returns to the front surface, where it again undergoes refraction as it moves from water to air. The result of this is a dispersed light of colours of different wavelengths.

Object A will undergo unstable equilibrium as it can move farther from its original position

Object B will undergo stable equilibrium as it can return to its original position

Object C will undergo neutral equilibrium as it can remain in a new position

An intrinsic semiconductor, also called an undoped semiconductor, is a pure semiconductor. In intrinsic semiconductors the number of excited electrons and the number of holes are equal: n = p

\(T_ {^1/_2}\) = 12 days ;V=?

\(T_ {^1/_2}\) = \(\frac{0.693}{γ}\)

⇒γ ⇒\(\frac{0.693}{t_{^1/_2}}\)

⇒γ =\(\frac{0.693}{12}\)

⇒γ= 0.05775 \(day^{-1}\)

We first find the lorry's initial velocity
 

a = \(4ms^-2\), s= 250m,  t= 10,  u=?

⇒ ut +  \(\frac{1}{2}×4×(10^2)


⇒250= 10u + 200


⇒ 250-200 = 10u


⇒ 50 = 10u


⇒U =\frac{50}{10}\) = \(5ms^-1\)

For the next 10s, we set t=20s

= s = (5 × 20) + \(\frac{1}{2}×4×{20^2}\)

= s  = 100 + 800

⇒ 900 m

;Distance covered:900-250 =650m
 

ε = 24.0 V,  r=1Ω,  R=5.0Ω

terminal p·d = ?

ε = IR+Ir = I(R+r) 

⇒ IR+Ir = I(R+r)
 
⇒ I = \(\frac{24}{5+1}\)

⇒ I =\(\frac{24}{6}\)

⇒ I = 4 A

terminal p·d = 4 × 5  

terminal p·d  =  20.0 V

h=32·8m,      p=  \(1×10^3kgm^-3\)

g=9·8 \(ms^-2\) ,   \(P_a\)=101·3KPa

\(P_T\)= ?

P=hpg

⇒\(32·8×1×10^3 × 9·8\)=321440 \(P_a\)

⇒321·44KPa

⇒ \(P_T\) =321.44+101.3

;\(P_T\)=422·7KPa

Relative Humidity=  \(\frac {partial  pressure  of  H2O}{saturated  vapor  pressure  of  H2O}\)×100%

S.v.p. of water at 40°c = 55.3

⇒ Relative Humidity=  \(\frac{38.8}{55.3}\)×100%

Relative Humidity= 70%

U= 75 \(ms^-1\) =22º,   g= 10 \(ms^-2\) ,   R=?
 

   R=     \(\frac{U^2sin2\theta}{g}\)

⇒R=     \(\frac{75^2×sin2(22)}{10}\)

⇒R=      \(\frac{5625×Sin44}{10}\)

⇒R=      \(\frac{5625×0.6947}{10}\)

⇒R=      \(\frac{3907.69}{10}\)

 ⇒    391m

W =  mg  = 220 x 9.8 = 2156 N

⇒Sin 38º = \(\frac{2156}{T_1}\)

⇒ \(T_1\) =    \(\frac{2156}{Sin 38}\)

⇒ \(T_1\) =     3502 N

Cos 38º =  \(\frac{T_2}{T_1}\)

⇒ \(T_2\)  =   3502 x Cos 38º

⇒ \(T_2\) =   2760 N

;    \(T_1\) =   3502 N,  \(T_2\) = 2760 N.