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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.

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In a collision, a 25.0 kg mass moving at 3.0 m/s transfers all of its momentum to a 5.0 kg mass.

nadezda [96]

Answer:

Explanation:

The momentum of the 25 kg mass is

$p=mv$

$p=25kg*3m/s= 75kg*m/s$

If this whole momentum of the object is transferred to the 5.0 kg object then according to the law of conservation of momentum, the momentum of the 25.0 kg object must be transferred to the 5.0 kg object:

$75kg*m/s = 5.0kg*v$

$v=\dfrac{75}{5}$

$\boxed{v=15m/s}$

8 0

3 years ago

A car is moving in uniform circular motion. If the cars speed were to double to keep the car moving with the same radius the acc

Stells [14]

Answer:

The centripetal acceleration would increase by a factor of 4

Correct choice: B.

Explanation:

Circular Motion

The circular motion is described when an object rotates about a fixed point called center. The distance from the object to the center is the radius. There are other magnitudes in the circular motion like the angular speed, tangent speed, and centripetal acceleration. The formulas are:

$v_t=w\ r$

$\displaystyle a_c=\frac{v_t^2}{r}$

If the speed is doubled and the radius is the same, then

$\displaystyle a_c=\frac{(2v_t)^2}{r}$

$\displaystyle a_c=4\frac{v_t^2}{r}$

The centripetal acceleration would increase by a factor of 4

Correct choice: B.

5 0

3 years ago

A hollow cylinder with an inner radius of 5 mm and an outer radius of 26 mm conducts a 4-A current flowing parallel to the axis

bearhunter [10]

Answer:

B = 38.2μT

Explanation:

By the Ampere's law you have that the magnetic field generated by a current, in a wire, is given by:

$B=\frac{\mu_o I_r}{2\pi r}$ (1)

μo: magnetic permeability of vacuum = 4π*10^-7 T/A

r: distance from the center of the cylinder, in which B is calculated

Ir: current for the distance r

In this case, you first calculate the current Ir, by using the following relation:

$I_r=JA_r$

J: current density

Ar: cross sectional area for r in the hollow cylinder

Ar is given by $A_r=\pi(r^2-R_1^2)$

The current density is given by the total area and the total current:

$J=\frac{I_T}{A_T}=\frac{I_T}{\pi(R_2^2-R_1^2)}$

R2: outer radius = 26mm = 26*10^-3 m

R1: inner radius = 5 mm = 5*10^-3 m

IT: total current = 4 A

Then, the current in the wire for a distance r is:

$I_r=JA_r=\frac{I_T}{\pi(R_2^2-R_1^2)}\pi(r^2-R_1^2)\\\\I_r=I_T\frac{r^2-R_1^2}{R_2^2-R_1^2}$ (2)

You replace the last result of equation (2) into the equation (1):

$B=\frac{\mu_oI_T}{2\pi r}(\frac{r^2-R_1^2}{R_2^2-R_1^2})$

Finally. you replace the values of all parameters:

$B=\frac{(4\pi*10^{-7}T/A)(4A)}{2\PI (12*10^{-3}m)}(\frac{(12*10^{-3})^2-(5*10^{-3}m)^2}{(26*10^{-3}m)^2-(5*10^{-3}m)^2})\\\\B=3.82*10^{-5}T=38.2\mu T$

hence, the magnitude of the magnetic field at a point 12 mm from the center of the hollow cylinder, is 38.2μT

8 0

3 years ago

Which of these describes the part of the water cycle happening at D

tino4ka555 [31]

The answer is Runoff.

7 0

3 years ago

Read 2 more answers

Which best summarizes a concept related to the work-energy theorem?

Maurinko [17]

Answer:

When work is positive, the environment does work on an object.

Explanation:

According to the work-energy theorem, the net work done by the forces on a body or an object is equal to the change produced in the kinetic energy of the body or an object.

The concept that summarizes a concept related to the work-energy theorem is that ''When work is positive, the environment does work on an object.''

4 0

3 years ago