Answer:
<em>2 m/s</em>
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Explanation:
The electromagnetic flow-metre work on the principle of electromagnetic induction. The induced voltage is given as
where 
is the induced voltage = 2.88 mV = 2.88 x 10^-3 V
is the distance between the electrodes in this field which is equivalent to the diameter of the tube = 1.2 cm = 1.2 x 10^-2 m
is the velocity of the fluid through the field = ?
is the magnetic field = 0.120 T
substituting, we have
2.88 x 10^-3 = 0.120 x 1.2 x 10^-2 x
2.88 x 10^-3 = 1.44 x 10^-3 x
= 2.88/1.44 = <em>2 m/s</em>
The electric output of the plant is 48.19 MW
First we need to calculate the water power, it is given by the formula
WP=ρQgh
Here, ρ=1000 kg/m3 is density of water,Q is the flow rate, g is the gravity, and h is the water head
Therefore, WP=1000*65*9.81*90=57388500 W=57.38 MW
Now the overall efficiency of the hydroelectric power plant is given as
η=
Plugging the values in the above equation
0.84=EP/57.38
EP=48.19 MW
Therefore, the electric output of the plant is 48.19 MW.
Answer:
θ = sin⁻¹
Explanation:
From one of the equations of motion, v² = u² + 2as.......... equation 1
Since the object thrown was moving against gravity, then the acceleration, a would change to -g and the initial velocity u would change to V₀ sin θ because the object is travelling at angle of θ to the horizontal. By inputting all these parameter into equation 1, we would arrive at:
v² = (u sin θ)² - 2gd
(u sin θ)² = 2gd
d = (u sin θ)²/2g
sin² θ = 2gd
sin θ =
θ = sin⁻¹
Balanced forces<span> act on the same object and </span>Action-Reaction forces<span> act on different objects.</span>
