Answer:
The maximum electric power output is ![P_{max} =1.339*10^{9} \ W](https://tex.z-dn.net/?f=P_%7Bmax%7D%20%3D1.339%2A10%5E%7B9%7D%20%5C%20W)
Explanation:
From the question we are told that
The capacity of the hydroelectric plant is ![\frac{V}{t} = 690 \ m^3 /s](https://tex.z-dn.net/?f=%5Cfrac%7BV%7D%7Bt%7D%20%20%20%3D%20%20690%20%5C%20m%5E3%20%2Fs)
The level at which water is been released is ![h = 220 \ m](https://tex.z-dn.net/?f=h%20%20%3D%20%20220%20%5C%20m)
The efficiency is
0.90
The electric power output is mathematically represented as
Where
is the potential energy at level h which is mathematically evaluated as
![PE_l = mgh](https://tex.z-dn.net/?f=PE_l%20%20%3D%20%20mgh)
and
is the potential energy at ground level which is mathematically evaluated as
![PE_o = mg(0)](https://tex.z-dn.net/?f=PE_o%20%20%3D%20%20mg%280%29)
![PE_o = 0](https://tex.z-dn.net/?f=PE_o%20%20%3D%20%200)
So
here ![m = V * \rho](https://tex.z-dn.net/?f=m%20%20%3D%20%20%20V%20%2A%20%20%5Crho)
where V is volume and
is density of water whose value is ![\rho = 1000 kg/m^3](https://tex.z-dn.net/?f=%5Crho%20%3D%201000%20kg%2Fm%5E3)
So
![P = \frac{V}{t} * gh \rho](https://tex.z-dn.net/?f=P%20%20%3D%20%5Cfrac%7BV%7D%7Bt%7D%20%2A%20gh%20%5Crho)
substituting values
![P =690 * 9.8 * 220 * 1000](https://tex.z-dn.net/?f=P%20%20%3D690%20%2A%209.8%20%2A%20220%20%2A%201000)
![P =1.488*10^{9} \ W](https://tex.z-dn.net/?f=P%20%20%3D1.488%2A10%5E%7B9%7D%20%5C%20W)
The maximum possible electric power output is
![P_{max} = P * \eta](https://tex.z-dn.net/?f=P_%7Bmax%7D%20%3D%20P%20%2A%20%5Ceta)
substituting values
![P_{max} =1.488*10^{9} * 0.90](https://tex.z-dn.net/?f=P_%7Bmax%7D%20%3D1.488%2A10%5E%7B9%7D%20%2A%200.90)
![P_{max} =1.339*10^{9} \ W](https://tex.z-dn.net/?f=P_%7Bmax%7D%20%3D1.339%2A10%5E%7B9%7D%20%5C%20W)
Answer:
a = 2.72 [m/s2]
Explanation:
To solve this problem we must use the following kinematics equation:
![v_{f} =v_{o} + a*t](https://tex.z-dn.net/?f=v_%7Bf%7D%20%3Dv_%7Bo%7D%20%2B%20a%2At)
where:
Vf = final velocity = 1200 [km/h]
Vo = initial velocity = 25 [km/h]
t = time = 2 [min] = 2/60 = 0.0333 [h]
1200 = 25 + (a*0.0333)
a = 35250.35 [km/h2]
if we convert these units to units of meters per second squared
![35250.35[\frac{km}{h^{2} }]*(\frac{1}{3600^{2} })*[\frac{h^{2} }{s^{2} } ]*(\frac{1000}{1} )*[\frac{m}{km} ] = 2.72 [\frac{m}{s^{2} } ]](https://tex.z-dn.net/?f=35250.35%5B%5Cfrac%7Bkm%7D%7Bh%5E%7B2%7D%20%7D%5D%2A%28%5Cfrac%7B1%7D%7B3600%5E%7B2%7D%20%7D%29%2A%5B%5Cfrac%7Bh%5E%7B2%7D%20%7D%7Bs%5E%7B2%7D%20%7D%20%5D%2A%28%5Cfrac%7B1000%7D%7B1%7D%20%29%2A%5B%5Cfrac%7Bm%7D%7Bkm%7D%20%5D%20%3D%202.72%20%5B%5Cfrac%7Bm%7D%7Bs%5E%7B2%7D%20%7D%20%5D)
This is your perfect answer
The base unit for time is the second (the other SI units are: metre for length, kilogram for mass, ampere for electric current, kelvin for temperature, candela for luminous intensity, and mole for the amount of substance). The second can be abbreviated as s or sec.
Answer:
f = 12 cm
Explanation:
<u>Center of Curvature</u>:
The center of that hollow sphere, whose part is the spherical mirror, is known as the ‘Center of Curvature’ of mirror.
<u>The Radius of Curvature</u>:
The radius of that hollow sphere, whose part is the spherical mirror, is known as the ‘Radius of Curvature’ of mirror. It is the distance from pole to the center of curvature.
<u>Focal Length</u>:
The distance between principal focus and pole is called ‘Focal Length’. It is denoted by ‘F’.
The focal length of the spherical (concave) mirror is approximately equal to half of the radius of curvature:
![f = \frac{R}{2}](https://tex.z-dn.net/?f=f%20%3D%20%5Cfrac%7BR%7D%7B2%7D)
where,
f = focal length = ?
R = Radius of curvature = 24 cm
Therefore,
![f = \frac{24\ cm}{2}](https://tex.z-dn.net/?f=f%20%3D%20%5Cfrac%7B24%5C%20cm%7D%7B2%7D)
<u>f = 12 cm</u>
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