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Zielflug [23.3K]
3 years ago
6

All of the following statements are true of hydroelectric power EXCEPT: A. Hydroelectric power stations rely on dams to help har

ness the energy of moving water. B. The water within hydroelectric power stations will transfer energy from gravitational potential energy to kinetic energy. C. Hydroelectric power stations can only produce enough energy for a small town as they do not produce large quantities of energy. D. Hydroelectric power stations are very reliable source of energy.
Physics
2 answers:
Artist 52 [7]3 years ago
5 0
The correct answer should be C. Hydroelectric power stations can only produce enough energy for a small town as they do not produce large quantities

Hydroelectric power stations can power even large cities that have millions of people.
kicyunya [14]3 years ago
3 0
In my opinion, the correct answer among the choices listed above is option C. <span>Hydroelectric power stations cannot only produce enough energy for a small town as they do produce large quantities of energy for a large area. Hope this answers the question.</span>
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A projectile is shot directly away from Earth's surface. Neglect the rotation of the Earth. What multiple of Earth's radius RE g
natali 33 [55]

(a) 5.65 times the Earth's radius

The escape velocity for a projectile on Earth is

v_e=\sqrt{\frac{2GM}{R}}

where

G is the gravitational constant

M is the Earth's mass

R is the Earth's radius

If the projectile has an initial speed of 0.421 escape speed,

v=0.421 v_e

So its initial kinetic energy will be

K=\frac{1}{2}m(0.421 v)^2=0.089 m(\sqrt{\frac{2GM}{R}})^2=0.177 \frac{GMm}{R}

where m is the mass of the projectile

At the point of maximum altitude, all this energy is converted into gravitational potential energy:

K=U\\0.177 \frac{GMm}{R}=\frac{GMm}{r}

where r is the distance from the Earth's centre reached by the projectile. We can write r as a multiple of R, the Earth's radius:0.177 \frac{GMm}{R}=\frac{GMm}{nR}

And solving the equation we find

n=\frac{1}{0.177}=5.65

So, the projectile reaches a radial distance of 5.65 times the Earth's radius.

b) 2.36 times the Earth's radius

The kinetic energy needed to escape is:

K=\frac{1}{2}mv_e^2 = \frac{1}{2}m(\sqrt{\frac{2GM}{R}})^2=\frac{GMm}{R}

This time, the projectile has 0.421 times this energy:

K=0.421 \frac{GMm}{R}

Again, at the point of maximum altitude, all this energy will be converted into potential energy:

0.421 \frac{GMm}{R}=\frac{GMm}{nR}

and by solving for n we find

n=\frac{1}{0.421}=2.36

So, the projectile reaches a radial distance of 2.36 times the Earth's radius.

c) E=U=\frac{GMm}{R}

The least initial mechanical energy needed for the projectile to escape Earth is equal to the gravitational potential energy of the projectile at the Earth's surface:

E=U=\frac{GMm}{R}

Indeed, the kinetic energy of the projectile must be equal to this value. In fact, if we use the formula of the escape velocity inside the formula of the kinetic energy, we find

K_e=\frac{1}{2}mv_e^2 = \frac{1}{2}m(\sqrt{\frac{2GM}{R}})^2=\frac{GMm}{R}

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2 years ago
The relationship among mass force and acceleration is explained by
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Newton's second law of motion. F = m a .
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