With current technologies and solar collectors on the ground, the best we can hope for is that solar cells will generate an average (day and night) power of about 230 watts/m2watts/m2 . What total area would we need to cover with solar cells to supply all the power needed for the United States

The question is missing

The total annual U.S. energy consumption is about 2 × 10^20 J

Explanation:

So, if the annual Energy is

2 × 10^20J

Let find the power usage of this energy

We know that

Power = Energy / Time

Now, the time is one year, so we have to convert one year to seconds

1year = 365days

365days = 365 × 24hours

365 × 24 hours = 365 × 24 × 3600 seconds

Then,

1year = 31,536,000seconds

Then,

P = E / T

P = 2 × 10^20 / 31,536,000

P = 6.342 × 10^12 Watts

The power intensity is given as

I = P/A

Then,

A = P / I

Where,

P is power, I is intensity and A is area

Given that, I = 230 W/m²

A = 6.342 × 10^12 / 230

A = 2.757 × 10^10 m²

d

explanation:

v=uf/(u-f)

= 5.10/(5-10) convex lens has positive focal length

=50/-5

=-10 cm (formed at other side)

speed of light in vacuum is 300,000 km/s

speed of light in other mediums is given by the expression v = c/n, where c is the speed of light in vacuum and n is the refractive index of medium.

for the first medium v = 200,000 km/s

  so, refractive index of medium = 300000/200000 = 1.5

    1.5 refractive index is usually for glass. so first medium is glass.

for the second medium v = 250,00 km/s

so, refractive index of medium = 300000/250000 = 1.2

    1.2 refractive index is usually for water. so second medium is water.

for the third medium v = 0 km/s

so, refractive index of medium = 300000/0 = ∞

    ∞ refractive index is usually for opaque mediums. so third medium is opaque medium.