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1 year ago

How can geo thermal energy be used ?

Physics

2 answers:

jek_recluse [69]1 year ago

5 0

Answer:

Geothermal energy can heat, cool, and generate electricity: Geothermal energy can be used in different ways depending on the resource and technology chosen—heating and cooling buildings through geothermal heat pumps, generating electricity through geothermal power plants, and heating structures through direct-use

Explanation:

Tju [1.3M]1 year ago

3 0

geothermal energy, form of energy conversion in which heat energy from within Earth is captured and harnessed for cooking, bathing, space heating, electrical power generation, and other uses. Heat from Earth's interior generates surface phenomena such as lava flows, geysers, fumaroles, hot springs, and mud pots.

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What is TRUE of carbon monoxide?

Aleks04 [339]

¡Hello!

Carbon monoxide is an odorless, colorless and tasteless gas produced by the incomplete combustion of carbon in fossil fuels such as wood, propane, charcoal, oil, gas, coal or other fuel.

6 0

2 years ago

A student wants to determine the impulse delivered to the lab cart when it runs into the wall. The student measures the mass of

forsale [732]

Impulse = Force * times and also Impulse = change in momentum.

Given that the mass does not change, change if momentum = mass * (final velocity - initial velocity)

Given that you know mass and initial velocity (which is the velicity before the cart hits the wall) you need the final velocity (which is the velocity after the cart hits the wall).

Answer: the velocity of the cart after it hits the wall.

6 0

2 years ago

The interior space of large box is kept at 30 C. The walls of the box are 3 m high and have a ‘sandwich’ construction consisting

White raven [17]

Answer:

$\frac{\dot Q}{A} =20.129\ W.m^{-2}$

$T_1=27.58\ ^{\circ}C$ & $T_2=2.41875\ ^{\circ}C$

Explanation:

Given:

interior temperature of box, $T_i=30^{\circ}C$

height of the walls of box, $h=3\ m$

thickness of each layer of bi-layered plywood, $x_p=1.25\ cm=0.0125\ m$

thermal conductivity of plywood, $k_p=0.104\ W.m^{-1}.K^{-1}$

thickness of sandwiched Styrofoam, $x_s=5\ cm=0.05\ m$

thermal conductivity of Styrofoam, $k_s=0.04\ W.m^{-1}.K^{-1}$

exterior temperature, $T_o=0^{\circ}C$

From the Fourier's law of conduction:

$\dot Q=\frac{dT}{(\frac{x}{kA}) }$

$\dot Q=\frac{dT}{R_{th} }$ ....................................(1)

Now calculating the equivalent thermal resistance for conductivity using electrical analogy:

$R_{th}=R_p+R_s+R_p$

$R_{th}=\frac{x_p}{k_p.A}+\frac{x_s}{k_s.A}+\frac{x_p}{k_p.A}$

$R_{th}=\frac{1}{A} (\frac{x_p}{k_p}+\frac{x_s}{k_s}+\frac{x_p}{k_p})$

$R_{th}=\frac{1}{A} (\frac{0.0125}{0.104}+\frac{0.05}{0.04}+\frac{0.0125}{0.104})$

$R_{th}=\frac{1.4904}{A}$ .....................(2)

Putting the value from (2) into (1):

$\dot Q=\frac{30-0}{\frac{1.4904}{A} }$

$\dot Q=\frac{30\ A}{1.4904}$

$\frac{\dot Q}{A} =20.129\ W.m^{-2}$ is the heat per unit area of the wall.

The heat flux remains constant because the area is constant.

For plywood-Styrofoam interface from inside:

$\frac{\dot Q}{A} =k_p.\frac{T_i-T_1}{x_p}$

$20.129=0.104\times \frac{30-T_1}{0.0125}$

$T_1=27.58\ ^{\circ}C$

&For Styrofoam-plywood interface from inside:

$\frac{\dot Q}{A} =k_s.\frac{T_1-T_2}{x_s}$

$20.129=0.04\times \frac{27.58-T_2}{0.05}$

$T_2=2.41875\ ^{\circ}C$

4 0

2 years ago

When objects are heated they tend to expand because

densk [106]

Hello!
===
When objects are heated, their molecules tend to vibrate fast. As they vibrate, the space between each atom increases. This keeps on happening, and the object expands until it has cooled down.
===
Hope this helps! :)

5 0

3 years ago

An elaborate pulley consists of four identical balls at the ends of spokes extending out from a rotating drum. A box is connecte

Klio2033 [76]

Answer:

its speed will be less than V

Explanation:

When the ball falls a distance d, its final kinetic energy plus rotational kinetic energy of the drum equals its initial potential energy.

K = U

With its speed V at the end of d, we have

1/2mV² + 1/2Iω² = mgd where I = rotational inertia of drum and balls, ω = angular speed of drum and balls and m = mass of box

1/2mV² + 1/2Iω² = mgd

1/2mV² = mgd - 1/2Iω²

V² = [2(mgd - 1/2Iω²)/m]

V = √[2(mgd - 1/2Iω²)/m]

When the four balls are moved inward closer to the drum, their rotational inertia increases and also its angular speed which thus causes an increase in rotational kinetic energy. But, since the box still falls the same distance of d, its final kinetic energy plus rotational kinetic energy of the drum plus balls still equals its initial potential energy

K = U

I' = new rotational inertia of drum and balls, ω' = new angular speed of drum and balls

With its new speed is now V' at the end of d,

1/2mV'² + 1/2I'ω'² = mgd

1/2mV'² = mgd - 1/2I'ω'²

V² = [2(mgd - 1/2I'ω'²)/m]

V' = √[2(mgd - 1/2I'ω'²)/m]

Since I' and ω' increase, the rotational kinetic energy of the drum and balls (1/2I'ω'²) increases. Thus, the difference (mgd - 1/2I'ω'²) < (mgd - 1/2Iω²) which implies that the kinetic energy of the box decreases. Hence, since its kinetic energy decreases, its speed V' also decreases.

So, V' < V

So, its speed will be less than V.

3 0

2 years ago

How can geo thermal energy be used ?​

How can geo thermal energy be used ?