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Blababa [14]
3 years ago
13

When the burner setting is changed to low the burner continues to produce heat?

Physics
1 answer:
Dahasolnce [82]3 years ago
7 0
Yes, the burner continues to produce heat, but not as persistant as the higher setting.
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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 "<em><u>insolation</u></em>" ... 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) = <em>(666 and 2/3) watt/m²</em> .

Note that this is the intensity of the <em><u>incident</u></em> 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
If two objects crash into each other, what happens? Are both objects affected? How?
Gemiola [76]

Explanation:

In a collision between two objects, both objects experience forces that are equal in magnitude and opposite in direction

6 0
2 years ago
How is the acceleration of an object in uniform circular motion constant?
nevsk [136]
In both magnitude and direction since acceleration is a vector quantity
5 0
3 years ago
Read 2 more answers
At what frequency should a 200-turn, flat coil of cross sectional area of 300 cm2 be rotated in a uniform 30-mT magnetic field t
Alexxandr [17]

Answer:

The frequency of the coil is 7.07 Hz

Explanation:

Given;

number of turns of the coil, 200 turn

cross sectional area of the coil, A = 300 cm² = 0.03 m²

magnitude of the magnetic field, B = 30 mT = 0.03 T

Maximum value of the induced emf, E = 8 V

The maximum induced emf in the coil is given by;

E = NBAω

Where;

ω is angular frequency = 2πf

E = NBA(2πf)

f = E / 2πNBA

f = (8) / (2π x 200 x 0.03 x 0.03)

f = 7.07 Hz

Therefore, the frequency of the coil is 7.07 Hz

7 0
3 years ago
g A bowling ball with a mass of 3.86 kg and a radius of 0.161 m starts from rest at a height of 2.5 m and rolls down a 48.4 o sl
Fynjy0 [20]

Answer:

v=1.5m/s

Explanation:

The gravitational potential energy gets transformed into translational and rotational kinetic energy, so we can write mgh=\frac{mv^2}{2}+\frac{I\omega^2}{2}. Since v=r\omega (the ball rolls without slipping) and for a solid sphere I=\frac{2mr^2}{5}, we have:

mgh=\frac{mv^2}{2}+\frac{2mr^2\omega^2}{2*5}=\frac{mv^2}{2}+\frac{mv^2}{5}=\frac{7mv^2}{10}

So our translational speed will be:

v=\sqrt{\frac{10gh}{7}}=\sqrt{\frac{10(9.8m/s^2)(0.161m)}{7}}=1.5m/s

6 0
3 years ago
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