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Marina86 [1]
2 years ago
12

What 3 things use electromagnetic energy in nature?​

Physics
1 answer:
AURORKA [14]2 years ago
8 0
Electromagnetic radiation (energy) is released from the Sun. Three examples would be Light, Ultraviolet Light, Gamma Rays.
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An electron is released a short distance above earth's surface. a second electron directly below it exerts an electrostatic forc
slamgirl [31]

The mass of an electron is 9.109 x 10⁻³¹ kg

The weight of the electron is (mass) x (g) =  8.926 x 10⁻³⁰ Newton

The charge on an electron is -1.602 x 10⁻¹⁹ Coulomb

The repelling force between the two electrons is (K · q₁ · q₂ / r²) =

(8.98755 x 10⁹ N-m²/C²) x (1.602 x 10⁻¹⁹ C)² / D²

In order for the bottom one to just exactly hold the top one up at a distance 'D', the repelling force has to be exactly equal to the weight of the upper electron.

8.926 x 10⁻³⁰ N = (8.98755 x 10⁹ N-m²/C²)·(1.602 x 10⁻¹⁹ C)² / D²

We have to solve THAT ugly mess for ' D '.

Clean up the units first:

Cancel the C² on the right side, then divide each side by Newton:

8.926 x 10⁻³⁰ = (8.98755 x 10⁹ m²) x (1.602 x 10⁻¹⁹)² / D²

Now, let's multiply both sides by (D² x 10²⁹) :

D² x 8.926 x 10⁻¹ = (8.98755 m²) x (1.602)²

Divide each side by (0.8926):

D² = (8.98755 x 1.602²) / (0.8926)  meter²

D² = 25.84 m²

Take the square root of each side:

<em>D = 5.08 meters</em>

I am shocked, impressed, and amazed !

Are you shocked, impressed, or amazed ?

3 0
3 years ago
When observing an object, you determine it is moving at a constant velocity and in a straight line. Which of these BEST describe
Marina86 [1]
The answer would be C. Hope this helps XD
8 0
3 years ago
Read 2 more answers
El Niño occurs every 2 to 12 years. true false
Novosadov [1.4K]
True
It is True I took the test
5 0
2 years ago
Read 2 more answers
a typical cmall flashlight contains two batteries each having na emf of 2.0 v connected in series with a bulb havin ga resistanc
Helen [10]

Answer:

P = 0.25 W

Explanation:

Given that,

The emf of the battry, E = 2 V

The resistance of a bulb, R = 16 ohms

We need to find the power delivered to the bulb. We know that, the formula for the power delivered is given by :

P=\dfrac{V^2}{R}\\\\P=\dfrac{2^2}{16}\\\\=0.25\ W

So, 0.25 W power is delivered to the bulb.

5 0
2 years ago
An ideally efficient heat pump delivers 1000 J of heat to room air at 300 K. If it extracted heat from 260 K outdoor air, how mu
choli [55]

Answer:

Wnet, in, = 133.33J

Explanation:

Given that

Pump heat QH = 1000J

Warm temperature TH= 300K

Cold temperature TL= 260K

Since the heat pump is completely reversible, the combination of coefficient of performance expression is given as,

From first law of thermodynamics,

COP(HP, rev) = 1/(1-TL/TH)

COP(HP, rev) = 1/(1-260/300)

COP(HP, rev) = 1/(1-0.867)

COP(HP, rev) = 1/0.133

COP(HP, rev) = 7.5

The power required to drive the the heat pump is given as

Wnet, in= QH/COP(HP, rev)

Wnet, in = 1000/7.5

Wnet, in = 133.333J. QED

So the 133.33J was the amount heat that was originally work consumed in the transfer.

Extra....

According to the first law, the rate at which heat is removed from the low temperature reservoir is given as

QL=QH-Wnet, in

QL=1000-133.333

QL=866.67J

5 0
3 years ago
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