2.5kw
250kw
300kw
3.0kw
Correct answer is A
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
(ii) and (iv) only
(i), (iii) and (iv) only
(ii), (iii) and (iv) only
(i), (ii) and (iv) only
Correct answer is C
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.
Name the type of equilibrium for each position of the ball
A - unstable, B - neutral, C - stable
A - stable, B - neutral, C - unstable
A - stable, B - unstable, C - neutral
A - unstable, B - stable, C - neutral
Correct answer is D
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
The number of holes in an intrinsic semiconductor
is not equal to the number of free electrons
is greater than the number of free electrons
is equal to the number of free electrons
is less than the number of free electrons
Correct answer is C
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
The half life of a radioactive material is 12 days. Calculate the decay constant.
0.8663 \(day^{-1}\)
0.04331 \(day^{-1}\)
0.17325 \(day^{-1}\)
0.05775 \(day^{-1}\)
Correct answer is D
\(T_ {^1/_2}\) = 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 \(day^{-1}\)
650
900
800
250
Correct answer is A
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
422.7
220.14
464.53
321.74
Correct answer is A
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
70
62
80
42
Correct answer is A
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%
195 m
271 m
391 m
136 m
Correct answer is C
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
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