The hydro power plant transforms one form of energy into another. However, the total amount
of energy of the water stays the same until it enters the turbine.
Explain how this statement is supported by the three column charts above.
*attached is the column charts
Answers
The hydro power plant consists of a (artificial) dam that builds gravitational potential energy, P.E. from natural flowing water sources, by locating the dam along the water path. The stored potential energy, P. is converted into kinetic energy, K.E. as the water falls from the dam, down to the turbines, located at a much lower level according to the following principle of conservation of energy equation;
Total Mechanical Energy, M.E. = The potential energy of the water, P.E. + The kinetic energy of the water, K.E. = Constant
M.E. = P.E. + K.E. = Constant
Where;
P.E. = m·g·h
K.E. = (1/2)·m·v²
m = Mass
g = The acceleration due to gravity
h = The height of the dam
v = The velocity
The charts can be explained as follows;
Given that the potential energy P.E. = m·g·h, we have that the potential energy is directly proportional to the height of the dam, and therefore, at mid height, the potential energy would be half the maximum value, and we have;
At mid height, P.E. = (1/2)·[tex]\mathbf{P.E._{max}}[/tex]
At the top of the dam, the (vertical) velocity of the water = 0, therefore, the kinetic energy = 0
Therefore, at the top of the dam, we get;
M.E. = [tex]P.E._{max}[/tex] + 0 =
M.E. = [tex]\mathbf{P.E._{max}}[/tex]
Similarly, at the bottom of the dam, the height, h = 0, therefore, being proportional to the height, P.E. = 0, and the velocity is maximum, and at the bottom, we have;
M.E. = 0 + [tex]K.E._{max}[/tex]
The first chart, water is halfway down the dam
At the halfway down therefore, we have;
P.E. = (1/2)·[tex]\mathbf{P.E._{max}}[/tex]
M.E. = [tex]P.E._{max}[/tex] = (1/2)·
∴ K.E. = [tex]P.E._{max}[/tex] - (1/2)·
Therefore the first chart, water is halfway down the dam;
Halfway, K.E. = (1/2)·[tex]\mathbf{P.E._{max}}[/tex] = P.E.
∴ K.E. = P.E. as shown on the chart
The second chart, water has reached the turbine
Water reaches the turbine at the bottom, and as explained above, we get;
M.E. = 0 + [tex]K.E._{max}[/tex]
∴ M.E.≈ [tex]K.E._{max}[/tex]
Therefore, when water has reached the turbine at the bottom of the dam, the kinetic energy is approximately proportional to the total mechanical energy as shown in the chart
The third chart, water is at the top of the dam
Here as shown above, we have;
The total mechanical energy, M.E. ≈ [tex]\mathbf{P.E._{max}}[/tex] as shown on the chart
Learn more about potential and kinetic energy here;
https://brainly.com/question/18683052
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The plane of a5cm*8cm rectangular loop of wire is parallel to a 0.19T magnetic field the loop carries a current of 6.2 A. What torque acts on aloop? What is the magnetic moment of the loop?
Answers
Answer:
Torque; τ = 4.712 × 10^(-3) J
Magnetic moment; M = 0.0248 J/T
Explanation:
Torque is gotten from the formula;
τ = BIA
Where;
B is magnetic field
I is current
A is area
We are given;
B = 0.19T
I = 6.2A
Rectangle dimensions = 5cm by 8cm = 0.05m by 0.08m
Thus;
Area; A = 0.05m × 0.08m = 0.004 m²
Thus;
τ = 0.19 × 6.2 × 0.004
τ = 4.712 × 10^(-3) J
Formula for the magnetic moment is given by;
M = IA
M = 6.2 × 0.004
M = 0.0248 J/T
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Answer:
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[tex]{\fcolorbox{white}{lightgreen}{\bf{\textcircled{$\checkmark$}}{Verified\:answer}}}[/tex]
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Answer:
low-energy alpha particles passing through a field of high mass-number-particles
Explanation:
A boy shoves his stuffed toy zebra down a frictionless chute. It starts at a height of 1.45 m above the bottom of the chute with an initial speed of 1.23 m/s . The toy animal emerges horizontally from the bottom of the chute and continues sliding along a horizontal surface with a coefficient of kinetic friction of 0.231 . How far from the bottom of the chute does the toy zebra come to rest? Assume g=9.81 m/s2 .
Answers
Answer:
The answer is "4.97 m".
Explanation:
[tex]u = 1.23\ \frac{m}{s}\\\\[/tex]
[tex]H= 1.45 \ m\\\\[/tex]
[tex]\mu = 0.231\\\\[/tex]
The law of conservation tells us that heat energy at the top with kinetic energy at the top equals kinetic energy at the base.
[tex]mgh+\frac{1}{2}mu^2=\frac{1}{2}mv^2\\\\2gh +u^2 =v^2\\\\v=\sqrt{u^2+2gh}[/tex]
[tex]v=\sqrt{(1.23\ \frac{m}{s})^2+2(9.81 \frac{m}{s^2}) +(1.45\ m)[/tex]
[tex]=\sqrt{1.5129+19.62 +1.45}\\\\=\sqrt{22.5829}\\\\=4.75\ \frac{m}{s}[/tex]
Friction force is given by the formula
[tex]f=-\mu mg \\\\ma= -\mu mg\\\\a=-\mu g\\\\[/tex]
[tex]= -(0.231) \ (9.81\ \frac{m}{s^2})\\\\=-2.26611 \ \frac{m}{s^2}[/tex]
Now by using an equation of motion as
[tex]v^2-u^2= 2as[/tex]
From the above the distance traveled is
[tex]S=\frac{v^2-u^2}{2a}[/tex]
[tex]S=\frac{(0)^2-(4.75\ \frac{m}{s})^2}{2(-2.26611\ \frac{m}{s^2})}\\\\[/tex]
[tex]=\frac{-(4.75\ \frac{m}{s})^2}{2(-2.26611\ \frac{m}{s^2})}\\\\=\frac{-22.5625}{-4.53222}\\\\=4.97[/tex]
In other words, the distance from the bottom of the chute to the point where the toy zebra comes to rest is [tex]s = 4.97\ m[/tex]
