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3 years ago

200 kW of solar radiation is shining on a 300 m^2 parking lot. What is the insulation on the parking lot?

1 answer:

7 0

That's "insolation" ... not "insulation".

'Insolation' is simply the intensity of solar radiation over some area.

If 200 kW of radiation is shining on 300 m² of area, then the insolation is

(200 kW) / (300 m²) = (666 and 2/3) watt/m² .

Note that this is the intensity of the incident radiation. It doesn't say anything
about how much soaks in or how much bounces off.

Wait !
I just looked back at the choices, and realized that I didn't answer the question
at all. I have no idea what "1 sun" means. Forgive me. I have stolen your
points, and I am filled with remorse.

Wait again !
I found it, through literally several seconds of online research.

1 sun = 1 kW/m².

So 2/3 of a kW per m² = 2/3 of 1 sun

That's between 0.5 sun and 1.0 sun.

I feel better now, and plus, I learned something.

You might be interested in

What would decrease the resistance of wires carrying an electric current

Murljashka [212]

I believe the answer is A. shorter wires

3 0

3 years ago

Read 2 more answers

An electron accelerated from rest through a voltage of 780 v enters a region of constant magnetic field. part a part complete if

maxonik [38]

The electron is accelerated through a potential difference of $\Delta V=780 V$ , so the kinetic energy gained by the electron is equal to its variation of electrical potential energy:
$\frac{1}{2}mv^2 = e \Delta V$
where
m is the electron mass
v is the final speed of the electron
e is the electron charge
$\Delta V$ is the potential difference

Re-arranging this equation, we can find the speed of the electron before entering the magnetic field:
$v= \sqrt{ \frac{2 e \Delta V}{m} } = \sqrt{ \frac{2(1.6 \cdot 10^{-19}C)(780 V)}{9.1 \cdot 10^{-31} kg} }=1.66 \cdot 10^7 m/s$

Now the electron enters the magnetic field. The Lorentz force provides the centripetal force that keeps the electron in circular orbit:
$evB=m \frac{v^2}{r}$
where B is the intensity of the magnetic field and r is the orbital radius. Since the radius is r=25 cm=0.25 m, we can re-arrange this equation to find B:
$B= \frac{mv}{er}= \frac{(9.1 \cdot 10^{-31}kg)(1.66 \cdot 10^7 m/s)}{(1.6 \cdot 10^{-19}C)(0.25 m)} =3.8 \cdot 10^{-4} T$

3 0

4 years ago

A 1980-kg car is traveling with a speed of 15.5 m/s. What is the magnitude of the horizontal net force that is required to bring

Dennis_Churaev [7]

Answer: 6067.5 N

Explanation:

Work = Change in Energy. To start, all of the energy is kinetic energy, so find the total KE using: KE = 1/2(m)(v^2). Plug in 1980 kg for m and 15.5 m/s for v and get KE = 237847.5 J.

Now, plug this in for work: Work = Force * Distance; so, divide work by distance to get 6067.5 N.

5 0

3 years ago

A man on the moon throws a ball vertically upwards and it is noticed that the ball travels 3.0m less in the fifth second of its

sdas [7]

<h2>Acceleration due to gravity in moon is 1.5 m/s²</h2>

Explanation:

We have equation of motion s = ut + 0.5 at²

Here the ball travels 3 m less distance in fifth second compared to third second.

That is

s₃ = s₅ + 3

Now we have

Distance traveled in third second, s₃ = u x 3 - 0.5 x g x 3² - u x 2 - 0.5 x g x 2²

s₃ = u - 2.5 g

Also

Distance traveled in fifth second, s₅ = u x 5 - 0.5 x g x 5² - u x 4 - 0.5 x g x 4²

s₅ = u - 4.5 g

That is

u - 2.5 g = u - 4.5 g + 3

2 g = 3

g = 1.5 m/s²

Acceleration due to gravity in moon = 1.5 m/s²

8 0

3 years ago

A 2-kg box sits on a horizontal table. the force of friction between the box and the table is 10 n. the box is pushed to the rig

dangina [55]

By Newton's 2nd law of motion, F = ma, where F is force, m is mass, and a is acceleration.

Rearranging this equation to find acceleration would give us:
a = F/m

The horizontal force to the right is 10N, because the box is pushed to the right with a force of 20N, and the friction force of 10N opposes that, so:
20N - 10N = 10N

The mass is 2kg.

Putting these values into the equation gives us:
a = F/m
= 10/2
= 5ms^-2

The acceleration of the box is 5ms^-2

6 0

4 years ago