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mote1985 [20]
3 years ago
8

Why is it not possible to make use of solar cells to meet all of our energy needs?why​

Physics
2 answers:
Nady [450]3 years ago
5 0

Answer:

In the solar cells, the energy is obtained only during the day, when the Sun shines. ... So, it increases the cost of using solar panels as the source of energy. So, the solar cell is not used to meet all our energy needs

Explanation:

n the solar cells, the solar panel convert solar energy into electricity, which is stored in storage battery. The storage battery give the direct current but all the appliances are working by the alternating current, so first of all direct current is converted into alternating current by any suitable appliances before it can be used to run various devices. So it increases the cost of using solar panels as the source of energy. 

aleksley [76]3 years ago
4 0

Answer:

In the solar cells, the energy is obtained only during the day, when the Sun shines. .So, it increases the cost of using solar panels as the source of energy. So, the solar cell is not used to meet all our energy needs

hope this helps

have a good day :)

Explanation:

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When a wave passes from one medium to another, its _________ remains constant.
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3 years ago
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Answer:

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6 0
3 years ago
Every few hundred years most of the planets line up on the same side of the Sun.(Figure 1)Calculate the total force on the Earth
mylen [45]

Answer: 3.7 \times 10^{-4} N

Explanation:

The gravitational pull between two object is given by:

F = G\frac{Mm}{r^2}

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We have to calculate the net force on Earth due to Venus, Jupiter and Saturn when they are in one line. It means when they are the closest distance.

F_{net] = G\frac{M_eM_v}{r_v^2}+G\frac{M_eM_j}{r_j^2}+G\frac{M_eM_s}{r_s^2}

Mass of Earth, Me = 5.98 × 10²⁴ kg

Mass of Venus, Mv = 0.815 Me

Mass of Jupiter, Mj = 318 Me

Mass of Saturn, Ms = 95.1 Me

closest distance between Earth and Venus, rv = 38 × 10⁶ km = 0.25 AU

closest distance between Jupiter and Earth, rj = 588 × 10⁶ km = 3.93 AU

closest distance between Earth and Saturn, rs = 1.2 × 10⁹ km = 8.0 AU

where 1 AU = 1.5 × 10¹¹ m

Inserting the values:

F_{net} = G\frac{M_e\times 0.815 M_e}{(0.25AU)^2}+G\frac{M_e\times 318 M_e}{(3.93AU)^2}+G\frac{M_e\times 95.1 M_e}{(8.0AU)^2}\\ \Rightarrow F_{net} = \frac{(GM_e^2)}{(1AU)^2}(\frac{0.815}{0.25^2}+\frac{318}{3.93^2}+\frac{95.1}{8.0^2})=\frac{6.67\times 10^{-11} \times (5.98\times 10^{24})^2}{(1.5\times 10^{11})^2}(35.1) = 3.7 \times 10^{-4} N

4 0
3 years ago
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ra1l [238]

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4 0
3 years ago
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