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asambeis [7]
3 years ago
8

Problems with solar energy include _____.

Physics
2 answers:
german3 years ago
4 0
First choice:  the inability of current technology to capture
large amounts of the Sun's energy

Well, it's true that large amounts of it get away ... our 'efficiency' at capturing it is still rather low.  But the amount of free energy we're able to capture is still huge and significant, so this isn't really a major problem.

Second choice:  the inability of current technology to store
captured solar energy

No.  We're pretty good at building batteries to store small amounts, or raising water to store large amounts.  Storage could be better and cheaper than it is, but we can store huge amounts of captured solar energy right now, so this isn't a major problem either.

Third choice:  inconsistencies in the availability of the resource

I think this is it.  If we come to depend on solar energy, then we're
expectedly out of luck at night, and we may unexpectedly be out
of luck during long periods of overcast skies.

Fourth choice:  lack of demand for solar energy

If there is a lack of demand, it's purely a result of willful manipulation
of the market by those whose interests are hurt by solar energy.
Yuri [45]3 years ago
4 0

The answer is the first three.

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Excess radiation can cause cancer. true or false
RSB [31]
It would be true. Excess radiation causes cancer.
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3 years ago
What do we call the small changes that
Lelu [443]

Answer:

The "butterfly Effect"

Explanation:

The "butterfly effect" will probably have big changes in the future.

6 0
2 years ago
What is the magnification formula?
erica [24]

Answer:

the height of the image ÷ by the height of the object.

Explanation:

7 0
3 years ago
A certain copper wire has a resistance of 13.0 Ω . At some point along its length the wire was cut so that the resistance of one
alekssr [168]

Answer with Explanation:

Let r be the resistance of short piece of copper wire.

Resistance of copper wire=R=13\Omega

Resistance is directly proportional to length.

If a wire has greater resistance then,the wire will be greater in length.

Therefore,resistance of long piece of wire=7r

Total resistance of copper  wire=Sum of resistance of two piece of wires

r+7r=13

8r=13

r=\frac{13}{8}ohm

Resistance of long piece of wire=7\times\frac{13}{8}=\frac{91}{8}\Omega

Resistance of short piece of wire =\frac{13}{8}\Omega

Resistivity of wire and cross section area of wire remains same .

Let L be the total  length  of wire and L' be the length of short  piece of wire.

We know that

R=\frac{\rho L}{A}=\frac{\rho}{A}L=KL

\frac{R}{L}=K

Where K=\frac{\rho}{A}=Constant

Using the formula

\frac{13}{L}=\frac{\frac{13}{8}}{L'}

\frac{L'}{L}=\frac{13}{8}\times \frac{1}{13}=\frac{1}{8}

L'=\frac{L}{8}

Length of short piece of wire=L'=\frac{L}{8}

Length of long piece of  wire=L-L'=L-\frac{L}{8}=\frac{8L-L}{8}=\frac{7}{8}L

% of length of short piece of   wire=\frac{\frac{L}{8}}{L}\times 100=12.5%%

The resistance of the short piece=\frac{13}{8}\Omega

The resistance of the long piece=\frac{91}{8}\Omega

8 0
3 years ago
A 34.1-mL sample of benzene at 20.8°C was cooled to its melting point, 5.5 °C, and then frozen at 5.5 °C. Calculate the quantity
posledela

Answer:

4541.8 J

Explanation:

First we find the mass of benzene available

mass = density x volume

         = 0.867 x 34.1

         = 29.5647 g

Then we find the amount of heat transferred by two processes:

heat tranferred = heat lost during temp drop + heat lost during freezing

                         = mcΔT + mL

                         = 29.5647 x 1.74 x (20.8 - 5.5) + 29.5647 x 127

                         = 4541.7883434 J

                         = 4541.8 J

                         

3 0
3 years ago
Read 2 more answers
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