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1 year ago

Explain how geothermal energy and tidal energy can be used effectively?

2 answers:

4 0

heating and cooling buildings through geothermal heat pumps, generating electricity through geothermal power plants, and heating structures through direct-use

tidal streams, barrages, and tidal lagoons.

Vadim26 [7]1 year ago

3 0

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 applications.

Tidal can be harnessed in three different ways; tidal streams, barrages, and lagoons.

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What is the MOST LIKELY effect of deforestation that results from urbanization of an area?

KengaRu [80]

A. Erosion is the answer

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3 years ago

Read 2 more answers

A ball is dropped off a building and falls past a window that is 2.2m long. If it takes .28s for the ball to cross the window wh

Sidana [21]

0.616 is the distance from the top
of the building to the top of the window.

8 0

1 year ago

A cold beverage can be kept cold even a warm day if it is slipped into a porous ceramic container that has been soaked in water.

Arisa [49]

Answer:

The rate at which the container is losing water is 0.0006418 g/s.

Explanation:

Under the assumption that the can is a closed system, the conservation law applied to the system would be: $E_{in}-E_{out}=E_{change}$ , where $E_{in}$ is all energy entering the system, $E_{out}$ is the total energy leaving the system and, $E_{change}$ is the change of energy of the system.
As the purpose is to kept the beverage can at constant temperature, the change of energy ( $E_{change}$ ) would be 0.
The energy that goes into the system, is the heat transfer by radiation from the environment to the top and side surfaces of the can. This kind of transfer is described by: $Q=\varepsilon*\sigma*A_S*(T_{\infty}^4-T_S^4)$ where $\varepsilon$ is the emissivity of the surface, $\sigma=5.67*10^{-8}\frac{W}{m^2K}$ known as the Stefan–Boltzmann constant, $A_S$ is the total area of the exposed surface, $T_S$ is the temperature of the surface in Kelvin, $T_{\infty}$ is the environment temperature in Kelvin.
For the can the surface area would be ta sum of the top and the sides. The area of the top would be $A_{top}=\pi* r^2=\pi(0.0252m)^2=0.001995m^2$ , the area of the sides would be $A_{sides}=2*\pi*r*L=2*\pi*(0.0252m)*(0.09m)=0.01425m^2$ . Then the total area would be $A_{total}=A_{top}+A_{sides}=0.01624m^2$
Then the radiation heat transferred to the can would be $Q=\varepsilon*\sigma*A_S*(T_{\infty}^4-T_S^4)=1*5.67*10^{-8}\frac{W}{m^2K}*0.01624m^2*((32+273K)^4-(17+273K)^4)=1.456W$ .
The can would lost heat evaporating water, in this case would be $Q_{out}=\frac{dm}{dt}*h_{fg}$ , where $\frac{dm}{dt}$ is the rate of mass of water evaporated and, $h_{fg}$ is the heat of vaporization of the water ( $2257\frac{J}{g}$ ).
Then in the conservation balance: $Q_{in}-Q_{out}=Q_{change}$ , it would be $1.45W-\frac{dm}{dt}*2257\frac{j}{g}=0$ .
Recall that $1W=1\frac{J}{s}$ , then solving for $\frac{dm}{dt}$ : $\frac{dm}{dt}=\frac{1.45\frac{J}{s} }{2257\frac{J}{g} }=0.0006452\frac{g}{s}$

5 0

2 years ago

A rocket of mass m is to be launched fromplanet X, which has a mass M and a radius R. What is the minimum speed that the rocket

Sunny_sXe [5.5K]

Answer:

The minimum speed = $\sqrt{\frac{2GM}{R} }$

Explanation:

The minimum speed that the rocket must have for it to escape into space is called its escape velocity. If the speed is not attained, the gravitational pull of the planet would pull down the rocket back to its surface. Thus the launch would not be successful.

The minimum speed can be determined by;

Escape velocity = $\sqrt{\frac{2GM}{R} }$

where: G is the universal gravitational constant, M is the mass of the planet X, and R is its radius.

If the appropriate values of the variables are substituted into the expression, the value of the minimum speed required can be determined.

7 0

2 years ago

If displacement covered by a particle is zero then distance cover by it

sleet_krkn [62]

Answer:

When displacement is zero, the particle may be at rest, therefore, distance travelled = 0.

Again, when displacement is zero, the final position matches with the initial position after some time, but the distance travelled will not be zero.

7 0

2 years ago