You have a working electrical parallel circuit with three light bulbs, then 1 bulb burns

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

ale4655 [162]

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.

7 0

3 years ago

SP 15The magnetic field in a region of space is measured to be:This field is known to be caused by a cluster of long-straight wi

tino4ka555 [31]

Answer:

i = 0.477 10⁴ B

the current flows in the counterclockwise

Explanation:

For this exercise let's use the Ampere law

∫ B . ds = μ₀ I

Where the path is closed

Let's start by locating the current vines that are parallel to the z-axis, so it must be exterminated along the x-axis and as the specific direction is not indicated, suppose it extends along the y-axis.

From BiotSavart's law, the field must be perpendicular to the direction of the current, so the magnetic field must go in the x direction.

We apply the law of Ampere the segment parallel to the x-axis is the one that contributes to the integral, since the other two have an angle of 90º with the magnetic field

Segment on the y axis

L₀ = (y2-y1)

L₀ = 3-0 = 3 cm

Segment on the point x = 2 cm

L₁ = 3-0

L₁ = 3cm

B L = μ₀ I

B 2L = μ₀ I

i = 2 L B /μ₀

i= 2 0.03 / 4π 10⁻⁷ B

i = 4.77 10⁴ B

The current is perpendicular to the magnetic field whereby the current flows in the counterclockwise

8 0

3 years ago

What happens when calcium reacts with chlorine?

choli [55]

Answer: They create calcuim chloride, CaCl2

7 0

3 years ago

Which of the following is not a subatomic particle?

Fynjy0 [20]

D. Nucleus because it is not a part of the group.

7 0

3 years ago

A woman can row a boat at 5.60 km/h in still water. (a) If she is crossing a river where the current is 2.80 km/h, in what direc

katrin2010 [14]

Answer:

a) θ=210°, b) t=1.155hr, c) t=1.333hr, d) t=1.333hr, e) θ=180° (straight across), f) t=1hr.

Explanation:

So, the very first thing we nee to do when solving this problem is draw a diagram that represents it. In the attached picture I show a diagram for each part of this problem.

part a)

So, for her to move in a direction directly opposite her starting point, the x-component of her velocity must be de same as the velocity of the river in the opposite direction. We can use this fact to find the angle we need. If we analize the triangle I drew in the diagram, we can ses that:

$cos \theta = \frac {V_{river}}{V_{boat}}$

When solving for theta, we get that:

$\theta =cos^{-1} ( \frac {V_{river}}{V_{boat}})$

so now we can substitute the corresponding values:

$\theta =cos^{-1} ( \frac {2.80km/hr}{5.60km/hr}})$

Which yields:

$\theta = 60^{o}$

but we are measuring the angle relative to the line perpendicular to the river, positive if down the river. So we need to subtract the angle from 270° so we get:

θ=270°-60°=210°

part b)

for part b, we need to find what the y-component for the velocity of the boat is for an angle of 210° as shown in the problem, so we get that:

$V_{y}=5.60km/hr*cos(210^{o})$

$V_{y}=-4.85km/hr$

The woman will head in a negative 5.60km distance from one side to the other, so we get that the time it takes her to go to the other side of the river is:

$t=\frac{y}{V_{y}}$

$t=\frac{5.60km}{4.85km/hr}=1.155hr$

part c)

In order to find the time it takes her to travel 2.80km down and up the river, we need to find the velocities she will have in both directions. First, down stream:

$V_{ds}=V_{river}+V{boat}$

$V_{ds}=2.80km/hr+5.60km/hr=8.40km/hr$

and now up stream:

$V_{us}=V_{boat}-V{river}$

$V_{us}=5.60km/hr-2.80km/hr=2.80km/hr$

Once we got these two velocities we will now need to find the time to take each trip:

time down stream:

$t_{ds}=\frac{x}{v_{ds}}$

$t_{ds}=\frac{2.80km}{8.40km/hr}=0.333hr$

and the time up stream:

$t_{us}=\frac{x}{v_{us}}$

$t_{us}=\frac{2.80km}{2,80km/hr}=1hr$

so the total time will be:

$t_{ds}+t_{us}=0.333hr+1hr=1.333hr$

d) the time it takes the boat to go upstream and then downstream for the same distance is the same as the time we got on part c, since both times will be the same but they will come in different order, but their sum will be just the same:

t=1.333hr

e) For her to cross the river faster, she must row in a 180° direction (this is in a direction straight accross the river) that way she will use all her velocity to move across the river. (Even though she will move a certain distance horizontally and will not reach a point opposite to the starting point.)

f) In order to find the time it takes her to get to the other side, we need to divide the distance into the velocity of the boat.

$t=\frac{d}{v_{boat}}$

$t=\frac{5.60km}{5.60km/hr}$

so

t= 1hr

4 0

3 years ago

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