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dimulka [17.4K]
2 years ago
9

A point source emits electromagnetic energy at a rate of 100 W. The intensity 10 m from the source is

Physics
1 answer:
MariettaO [177]2 years ago
5 0

Answer:

I=0.0795\ W/m^2

Explanation:

Given that,

Power of electromagnetic energy, P = 100 W

We need to find the intensity at a distance of 10 m from the source. Intensity is equal to the power per unit area. So,

I=\dfrac{P}{4\pi r^2}\\\\I=\dfrac{100}{4\pi (10)^2}\\\\I=0.0795\ W/m^2

So, the intensity at a distance of 10 m is 0.0795\ W/m^2

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Sometimes students are taught that "air expands as it is heated". And sometimes they are taught that "as air expands it cools".
hram777 [196]

Answer:

Both are true under specific circumstances. And are related to Boyle's law. volume and pressure in a gas are inversely proportional.

Explanation:

There is a tendency to entropy in our reality, that is, in particular true and visible with gases, they tend to occupy the whole space where they are confined, when we heat a volume of gas, then the movement of the particles and in consequence the pressure of the gas increases and to compensate this the volume tends to be increased too, according to Boyle's law. And the opposite happens when the volume is increased, then the pressure is  relieved and since the particles are further one from each other, then the temperature is lower, and therefore it cools down.

7 0
3 years ago
The work function for a metal surface is 4.98 eV. What is the largest wavelength of light in nm that will produce photoelectrons
bulgar [2K]

Answer:\lambda =248.99 nm

Explanation:

Given

Work function\left ( \phi \right )=4.98\approx 1.602\times 10^{-19}\times 4.98

h=6.626\times 10^{-34} J

c=2.998\times 10^8

\phi =\frac{hc}{\lambda }

\lambda =\frac{hc}{\phi }

\lambda =\frac{6.626\times 10^{-34}\times 2.998\times 10^8}{4.98\times 1.602\times 10^{-19}}

\lambda =248.99 nm

3 0
3 years ago
The total resistance of a 15-ohm, an 65-ohm and a 35-ohm resistor connected in parallel ​
yanalaym [24]

Answer:

I think its 9.0397 Ohms

Explanation:

take the reciprocal of all the resistances: 1/15, 1/65, 1/35

then add them: = 151/1365

then reciprocal the answer: =1365/151

And chuck it on a calculator: =9.04 Ohms

I think this is right but I'm not entirely sure. Tell me if I'm right by the way!

3 0
3 years ago
Explain how evaporation cools a liquid​
MAVERICK [17]

When evaporation occurs liquid absorbs heat from the surroundings to get converted to its vapour form as a result, there is an overall decrease in the heat leading to cooling of the liquid.

Hope that this was helpful :)

8 0
2 years ago
How much heat (in kJ) is required to warm 13.0 g of ice, initially at -12.0 ∘C, to steam at 109.0 ∘C?
Sunny_sXe [5.5K]
<h2>Answer:</h2>

39.699 kJ

<h2>Explanation:</h2>

In this situation, there are a few transformations as follows;

(i) Heat required to warm the ice from -12°C to its melting point.

(ii) Heat required to melt the ice.

(iii) Heat required to boil the melted ice to boiling point (i.e to steam)

(iv) Heat required to vapourize the water

(v) Heat required to heat the steam from 100°C to 109.0°C

The sum of all the heat processes gives the heat required to warm the ice to steam;

<h3><em>Calculate each of these heat processes</em></h3>

<em>From (i);</em>

Let the heat required to warm the ice from -12.0°C to its melting point (0°C) be Q₁.

Q₁ = m x c x ΔT        -----------------------(i)

Where;

m = mass of ice = 13.0g

c = specific heat capacity of ice = 2.09 J/g°C

ΔT = final temperature - initial temperature = 0°C - (-12°C) = 12°C

Substitute these values into equation (i) as follows;

Q₁ = 13.0 x 2.09 x 12 = 326.04 J

<em>From (ii);</em>

Let the heat required to melt the ice be Q₂. This heat is called the heat of fusion and it is given by;

Q₂ = m x L        -----------------------(ii)

Where;

m = mass of ice = 13.0g

L = latent heat of fusion of ice = 333.6 J/g

Substitute these values into equation (ii) as follows;

Q₂ = 13.0 x 333.6

Q₂ = 4336.8 J

<em>From (iii);</em>

Let the heat required to boil the melted ice from 0°C to boiling point of 100°C be Q₃.

Q₃ =  m x c x ΔT        -----------------------(i)

Where;

m = mass of melted ice (water) which is still 13.0g

c = specific heat capacity of melted ice (water) = 4.2 J/g°C

ΔT = final temperature - initial temperature = 100°C - 0°C = 100°C

Substitute these values into equation (i) as follows;

Q₁ = 13.0 x 4.2 x 100 = 5460 J

<em>From (iv);</em>

Let the heat required to vaporize the water (melted ice) be Q₄. This heat is called the heat of vaporization and it is given by;

Q₄ = m x L        -----------------------(iv)

Where;

m = mass of ice = 13.0g

L = latent heat of vaporization of water = 2257 J/g

Substitute these values into equation (iv) as follows;

Q₄ = 13.0 x 2257

Q₄ = 29341 J

<em>From (v);</em>

Let the heat required to heat the steam from 100°C to 109°C be Q₅.

Q₅ =  m x c x ΔT        -----------------------(i)

Where;

m = mass of steam which is still 13.0g

c = specific heat capacity of steam = 2.01 J/g°C

ΔT = final temperature - initial temperature = 109.0°C - 100°C = 9°C

Substitute these values into equation (i) as follows;

Q₅ = 13.0 x 2.01 x 9 = 235.17J

<em>Finally:</em>

<em>Sum all the heat values together;</em>

Q = Q₁ + Q₂ + Q₃ + Q₄ + Q₅

Q = 326.04 + 4336.8 + 5460 + 29341 + 235.17

Q = 39699.01 J

Q = 39.699 kJ

Therefore, the amount of heat (in kJ) required is 39.699

7 0
2 years ago
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