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Vesnalui [34]
2 years ago
9

What is the source of geothermal energy? User: What is the source of geothermal energy?

Physics
2 answers:
bazaltina [42]2 years ago
8 0

The answer is; C

In particular points in the earth’s surface, underground water is naturally heated to steam that can be harness for geothermal energy. The steam that ejects from the ground with high kinetic energy can be used to turn turbines that generate electricity. The underground water is usually heated by the hot rocks beneath that are subjected to the immense heat of magma or the enormous pressure of overlying crust.  


Alinara [238K]2 years ago
7 0

<em>ANSWER</em>

<u><em>C. natural underground reservoirs of steam and hot water.</em></u>

<u><em>HOPE THIS HELPED COULD YOU MARK ME BRAINLIEST?</em></u>

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Can you tell from the coefficient of restitution whether a collision has added kinetic energy to a system, taken some away, or l
Maurinko [17]
For an inelastic collision where coefficient of restitution,e, is equal to 0, the momentum is conserved but not the kinetic energy. So, there is addition or elimination of kinetic energy.

On the otherhand, when e = 1, like for an elastic collision, kinetic energy and momentum is conserved. Thus, the system's kinetic energy is unchanged.
6 0
3 years ago
For a given initial projectile speed Vo, calculate what launch angle A gives the longest range R. Show your work, don't just quo
pickupchik [31]
The optimal angle of 45° for maximum horizontal range is only valid when initial height is the same as final height. 

<span>In that particular situation, you can prove it like this: </span>

<span>initial velocity is Vo </span>
<span>launch angle is α </span>

<span>initial vertical velocity is </span>
<span>Vv = Vo×sin(α) </span>

<span>horizontal velocity is </span>
<span>Vh = Vo×cos(α) </span>

<span>total time in the air is the the time it needs to fall back to a height of 0 m, so </span>
<span>d = v×t + a×t²/2 </span>
<span>where </span>
<span>d = distance = 0 m </span>
<span>v = initial vertical velocity = Vv = Vo×sin(α) </span>
<span>t = time = ? </span>
<span>a = acceleration by gravity = g (= -9.8 m/s²) </span>
<span>so </span>
<span>0 = Vo×sin(α)×t + g×t²/2 </span>
<span>0 = (Vo×sin(α) + g×t/2)×t </span>
<span>t = 0 (obviously, the projectile is at height 0 m at time = 0s) </span>
<span>or </span>
<span>Vo×sin(α) + g×t/2 = 0 </span>
<span>t = -2×Vo×sin(α)/g </span>

<span>Now look at the horizontal range. </span>
<span>r = v × t </span>
<span>where </span>
<span>r = horizontal range = ? </span>
<span>v = horizontal velocity = Vh = Vo×cos(α) </span>
<span>t = time = -2×Vo×sin(α)/g </span>
<span>so </span>
<span>r = (Vo×cos(α)) × (-2×Vo×sin(α)/g) </span>
<span>r = -(Vo)²×sin(2α)/g </span>

<span>To find the extreme values of r (minimum or maximum) with variable α, you must find the first derivative of r with respect to α, and set it equal to 0. </span>

<span>dr/dα = d[-(Vo)²×sin(2α)/g] / dα </span>
<span>dr/dα = -(Vo)²/g × d[sin(2α)] / dα </span>
<span>dr/dα = -(Vo)²/g × cos(2α) × d(2α) / dα </span>
<span>dr/dα = -2 × (Vo)² × cos(2α) / g </span>

<span>Vo and g are constants ≠ 0, so the only way for dr/dα to become 0 is when </span>
<span>cos(2α) = 0 </span>
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<span>α = 45° </span>
4 0
2 years ago
To practice problem-solving strategy 22.1: gauss's law. an infinite cylindrical rod has a uniform volume charge density ρ (where
BabaBlast [244]

Let say the point is inside the cylinder

then as per Gauss' law we have

\int E.dA = \frac{q}{\epcilon_0}

here q = charge inside the gaussian surface.

Now if our point is inside the cylinder then we can say that gaussian surface has charge less than total charge.

we will calculate the charge first which is given as

q = \int \rho dV

q = \rho * \pi r^2 *L

now using the equation of Gauss law we will have

\int E.dA = \frac{\rho * \pi r^2* L}{\epcilon_0}

E. 2\pi r L = \frac{\rho * \pi r^2* L}{\epcilon_0}

now we will have

E = \frac{\rho r}{2 \epcilon_0}

Now if we have a situation that the point lies outside the cylinder

we will calculate the charge first which is given as it is now the total charge of the cylinder

q = \int \rho dV

q = \rho * \pi r_0^2 *L

now using the equation of Gauss law we will have

\int E.dA = \frac{\rho * \pi r_0^2* L}{\epcilon_0}

E. 2\pi r L = \frac{\rho * \pi r_0^2* L}{\epcilon_0}

now we will have

E = \frac{\rho r_0^2}{2 \epcilon_0 r}


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3 years ago
Which answer choice?? I have trouble with this physics concept :/
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8 0
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Apparent brightness depends on which two properties?
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Answer:

I believe it is luminosity and distance

Explanation:

So B

6 0
2 years ago
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