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.
To have a weight of 2.21N., the ball's mass is (2.21/9.8) = .226kg.
<span>a) d = 1/2 (vt), = 1/2 (18 x .17), = 1.53m. </span>
<span>b) Acceleration of the ball = (v/t), = 18/.17, = 105.88m/sec^2. </span>
<span>f = (ma), = .226 x 105.88, = 23.92N. </span>
To solve this exercise it is necessary to apply the equations related to the magnetic moment, that is, the amount of force that an image can exert on the electric currents and the torque that a magnetic field exerts on them.
The diple moment associated with an iron bar is given by,

Where,
Dipole momento associated with an Atom
N = Number of atoms
y previously given in the problem and its value is 2.8*10^{-23}J/T


The number of the atoms N, can be calculated as,

Where
Density
Molar Mass
A = Area
L = Length
Avogadro number


Then applying the equation about the dipole moment associated with an iron bar we have,



PART B) With the dipole moment we can now calculate the Torque in the system, which is



<em>Note: The angle generated is perpendicular, so it takes 90 ° for the calculation made.</em>
Answer:
inelastic, since the girl moves in the same direction as the thrown ball
Explanation:
yess this ok
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Answer:
Explanation:
Inital KE = (1/2) m v^2 = (1/2) * 1500 * 50^2 = 1,875,000 J
Final KE = (1/2) * 1500 * 100^2 = 7,500,000 J
But ,
4 * 1875000 = 7500000
so the KE has increased by 4 times.