The answer would be D hope it helps and sorry if it is wrong. :)
A. Using the third equation of motion:
v2 = u2 + 2as
from the question;
the jet was initially at rest
hence u = 0
a = 1.75m/s2
s = 1500m
v2 = 02 + 2(1.75)(1500)
v2 = 5250
v = √5250
v = 72.46m/s
hence it moves with a velocity of 72.46m/s.
b. s = ut + 1/2at2
1500 = 0(t) + 1/2(1.75)t2
1500 × 2 = 2× 1/2(1.75)t2
3000 = 1.75t2
1714.29 = t2
41.4 = t
hence the time taken for the plane to down the runway is 41.4s.
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Answer:
57 %
Explanation:
input power = 16.4 kW = 16.4 x 10^3 W = 16400 W
Water pumped per second = 67 L/s
Mass of water pumped per second, m = Volume of water pumped epr second x density of water
m = 67 x 10^-3 x 1000 = 67 kg/s
height raised, h = 14 m
Output Power = m x g x h / t = 67 x 10 x 14 = 9380 W
efficiency = output power / input power = 9380 / 16400 = 0.57
% efficiency = 57 %
thus, the efficiency of the pump is 57 %.
Answer:
Explanation:
Flux through the coil = nBA , n is no of turns , B is magnetic flux and A a is area of the coli
= 200 x 5.6 x 10⁻⁵ x 11.8 x 10⁻⁴
= 13216 x 10⁻⁹ weber .
b ) When the coil becomes parallel to magnetic field , flux through it will become zero.
c ) e m f induced = change in flux / time
= 13216 x 10⁻⁹ / 4.9 x 10⁻²
= 2697.14 x 10⁻⁷ V
= 269.7 x10⁻⁶
269.7 μV.
1. one-Half
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3. Any object that revolves around another object
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