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2 years ago

What type of energy is stored in solar panels

Physics

1 answer:

Andru [333]2 years ago

4 0

Answer:Thermal Storage

Concentrating solar plants can capture heat from the sun and store the energy in water, molten salts, or other fluids. This stored energy is later used to generate electricity, enabling the use of solar energy even after sunset.

Explanation: Solar panels use the photons produced by sunlight to generate direct current (DC) electricity. When the photons hit the panel they are absorbed by the panel's semiconducting silicon material. During this process electrons separate from the atoms and move around the solar cell.

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Energy resources that exist in limited amounts and, fonts used, cannot be replaced except over the course of millions of years a

kobusy [5.1K]

Energy resources that cannot be replaced are called "non-renewable" energy resources.

5 0

3 years ago

Which of theses substances would be a poor conductor of electricity

lys-0071 [83]

Answer:

B. Water and sugar.

Explanation:

In the given options water and sugar would be the poor conductor of electricity. Other given options such as water and salt, water and Hcl and water and NaOH are better conductor of electricity because Hcl ,NaOH, salt (Nacl) can break into their ionic form whereas water and sugar will not.

8 0

3 years ago

Suppose a nonconducting sphere, radius r2, has a spherical cavity of radius r1 centered at the sphere's center. Assuming the cha

leva [86]

Answer:

Explanation:

a ) Between r = 0 and r = r₁

Electric field will be zero . It is so because no charge lies in between r = 0 and r = r₁ .

b ) From r = r₁ to r = r₂

At distance r , charge contained in the sphere of radius r

volume charge density x 4/3 π r³

q = Q x r³ / R³

Applying Gauss's law

4πr² E = q / ε₀

4πr² E = Q x r³ / ε₀R³

E= Q x r / (4πε₀R³)

E ∝ r .

c )

Outside of r = r₂

charge contained in the sphere of radius r = Q

Applying Gauss's law

4πr² E = q / ε₀

4πr² E = Q / ε₀

E = Q / 4πε₀r²

E ∝ 1 / r² .

6 0

3 years ago

A cat runs at 2 m/s for 3 s and then slows to a stop with an acceleration of 0.80 m/s2. What is

sp2606 [1]

Answer:

see that the correct one is B

Explanation:

To solve this exercise let us use the kinematic relations

v² = v₀² - 2 a x

as they indicate that the car stops, therefore the final speed is yield v = 0

x = v₀² / 2a

let's calculate

x = 2²/(2 0.8)

x = 2.5 m / s²

When reviewing the answers we see that the correct one is B

7 0

2 years ago

Calculate the orbital period of a dwarf planet found to have a semimajor axis of a = 4.0x 10^12 meters in seconds and years.

padilas [110]

Explanation:

We have,

Semimajor axis is $4\times 10^{12}\ m$

It is required to find the orbital period of a dwarf planet. Let T is time period. The relation between the time period and the semi major axis is given by Kepler's third law. Its mathematical form is given by :

$T^2=\dfrac{4\pi ^2}{GM}a^3$

G is universal gravitational constant

M is solar mass

Plugging all the values,

$T^2=\dfrac{4\pi ^2}{6.67\times 10^{-11}\times 1.98\times 10^{30}}\times (4\times 10^{12})^3\\\\T=\sqrt{\dfrac{4\pi^{2}}{6.67\times10^{-11}\times1.98\times10^{30}}\times(4\times10^{12})^{3}}\\\\T=4.37\times 10^9\ s$

Since,

$1\ s=3.17\times 10^{-8}\ \text{years}\\\\4.37\times 10^9\ s=4.37\cdot10^{9}\cdot3.17\cdot10^{-8}\\\\4.37\times 10^9\ s=138.52\ \text{years}$

So, the orbital period of a dwarf planet is 138.52 years.

3 0

3 years ago