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3 years ago

Which of the following are disadvantages of solar energy?

Physics

2 answers:

gogolik [260]3 years ago

7 0

Answer:

no power unless the suns out

Explanation:

Ronch [10]3 years ago

3 0

There are a few disadvantages, so let's start with the most simple. Night time, the sun isn't up so solar panels can't gather energy. Also, when it's snowing or raining or it's cloudy. Another thing is as of right now it's costly to get solar panels installed. Furthermore, a solar panel has to be maintained as in brushed cause debris can get caught on it and therefore block the sunlight. If you have anymore questions just comment and I will help further. I hope this helps!

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The density is 1.5 g/cm^3. the mass is 1500. what is the volume

Naddik [55]

The density of an object can be calculated using the formula Density = Mass/Volume. In this case however we are searching for the volume and must rearrange the formula so that we are solving for the volume. If you multiply both sides by volume and then divide both sides by mass you end up with the equation Volume = Mass/Density.

Volume = 1500g/1.5g/cm^3
Volume = 1000 cm^3

8 0

3 years ago

How is a scientific theory different from a scientific law?

Anuta_ua [19.1K]

Answer:

Scientific theory is a well-supported explanation of an occurrence of a natural phenomenon. Scientific law is a description of a natural phenomenon that is always observed to occur under specific circumstances.

Explanation:

5 0

3 years ago

What frequency (in Hz) is received by a person watching an oncoming ambulance moving at 116 km/h and emitting a steady 950 Hz so

Arte-miy333 [17]

To solve this problem we will apply the concepts related to the Doppler Effect, defined as the change in apparent frequency of a wave produced by the relative movement of the source with respect to its observer. Mathematically it can be written as

$f_{obs} = f(\frac{v_w}{v_w-v_s})$

Here,

$f_s$ = Frequency of the source

$v_w$ = Speed of the sound

$v_s$ = Speed of source

Now the velocity we have that

$v_s = 116km/h (\frac{1000m}{1km})(\frac{1h}{3600s})$

$v_s = 32.22m/s$

Then replacing our values,

$f_{obs} = (950Hz) (\frac{345m/s}{345m/s-32.22m/s})$

$f_{obs} = 1047.86Hz$

Therefore the frequency of the observer is 1047.86Hz

8 0

3 years ago

It is found experimentally that the electric field in a certain region of Earth's atmosphere is directed vertically down. At an

Maru [420]

Answer:

$q=7.965*10^-^6C$

Explanation:

From the question we are told that

Altitude of $d_1 390m$

Magnitude $M_1=60.0 N/C$

Altitude of $d_2=240 m$

Magnitude is $M_2= 100 N/C$

Distance of cube $d_c=150 m$

Generally the flux $\phi$ is mathematical given as

$\phi=60(150)^2cos180+100(150)^2*cos0$

$\phi=-9*10^5$

Generally Quantity of charge q is mathematically given as

$q=\varepsilon _0 *\phi$

$q=8.85*10^-^1^2 *9*10^5$

$q=7.965*10^-^6C$

5 0

3 years ago

How does the work required to accelerate a particle from 10 m/s to 20 m/s compare to that required to accelerate it from 20 m/s

poizon [28]

To solve this problem we will apply the energy conservation theorem for which the work applied on a body must be equivalent to the kinetic energy of this (or vice versa) therefore

$W = \Delta KE$

$\Delta W = \frac{1}{2} (m)(v_f)^2 -\frac{1}{2} (m)(v_i)^2$

Here,

m = mass

$v_{f,i}$ = Velocity (Final and initial)

First case) When the particle goes from 10m/s to 20m/s

$\Delta W = \frac{1}{2} (m)(v_f)^2 -\frac{1}{2} (m)(v_i)^2$

$\Delta W = \frac{1}{2} (m)(20)^2 -\frac{1}{2} (m)(10)^2$

$W_1 = 150(m) J$

Second case) When the particle goes from 20m/s to 30m/s

$\Delta W = \frac{1}{2} (m)(v_f)^2 -\frac{1}{2} (m)(v_i)^2$

$\Delta W = \frac{1}{2} (m)(30)^2 -\frac{1}{2} (m)(20)^2$

$W_1 = 250(m) J$

As the mass of the particle is the same, we conclude that more energy is required in the second case than in the first, therefore the correct answer is A.

5 0

3 years ago