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nika2105 [10]
3 years ago
8

The description for a certain brand of house paint claims a coverage of 475 ft²/gal.

Physics
1 answer:
AlexFokin [52]3 years ago
8 0

Answer:

(a) 11.66 square meters per liter

(b) 11657.8 per meters

(c) 0.00211 gal per square feet

Explanation:

(a) 475ft^2/gal = 475ft^2/gal × (1m/3.2808ft)^2 × 1gal/3.7854L = 11.66m^2/L

(b) 475ft^2/gal = 475ft^2/gal × (1m/3.2808ft)^2 × 264.17gal/1m^3 = 11657.8/m

(c) Inverse of 475ft^2/gal = 1/475ft^3/gal = 0.00211gal/ft^3

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If the Sun subtends a solid angle Ω on the sky, and the flux from the Sun just above the Earth’s atmosphere, integrated over all
Arada [10]

Answer:

A)Ω = 7.8 × 10^−5 steradians.

B) TE = 5800K

C) fλ(λ1) = (π ^2 ) /ΩBλ(T)

Explanation:

A) First of all, if we assume that the Sun emits isotropically at a luminosity (L⊙) , the flux at a given distance R from the sun would be f(d) = L⊙/ (4πd^2)

The ratio of flux at the solar photosphere to the flux at the Earth’s atmosphere would be: F⊙/{f(d⊙)} = (R⊙)^2 / (d⊙)^2

Now if we think of this relationship of the flux and the earth as a conical pattern, we'll deduce that the solid angle subtended by the sun at Earth’s surface to be;

Ω = π[(R⊙)^2 / (d⊙)^2]

Combining this with the ratio earlier gotten, well arrive at;

F⊙ = {f(d⊙ )π} /Ω

Now let's express The radius of the sun (R) in terms of its angular diameter (2α) and this gives;

R⊙ ≈ αd⊙

Now combining this with the equation for Ω earlier, we get;

Ω ≈ πα^2

So, = π((0.57/2π) /180)^2 = 7.8 × 10^−5 steradians.

B) from Stefan-Boltzmann Law,

F⊙ = σ(TE)^4

From the beginning, we know that;

F⊙ = {f(d⊙ )π} /Ω

And so replacing that in the stephan boltzmann law, we get ;

{f(d⊙ )π} /Ωσ = (TE)^4

So, (TE)^4 = {π (1.4 kWm^(−2))} / [(7.8 × 10^(−5 ) steradians x (5.66961 × 10^(−8))]

In stephan boltzmann law, σ = 5.66961 × 10^(−8)

And so, TE is approximately 5800K.

C) In order to relate fλ(λ1) with T, let's assume the sun’s surface to be an isotropically emitting blackbody, i.e its specific intensity is Iλ = Bλ(T). Hence, the flux at Sun’s surface for a given wavelength would be;

Fλ(λ1) = πBλ(T)

Now, if we combine this with the expression of F⊙ gotten earlier, well get the relation;

fλ(λ1) = (π ^2 ) /ΩBλ(T)

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Which statement represents a healthy choice for a pregnant woman? A mother eats twice as much to support her growing fetus. A mo
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Answer:

A mother with gestational diabetes decides to limit her sugar intake.

Explanation:

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3 years ago
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Please answer of this question​
Bogdan [553]

Answer:

\frac{\pi }{15}m

Explanation:

At 10am, the minute hand and hour hand are ' 2 hours apart', since the minute hand is at 12pm and hour hand is at 10am.

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2 years ago
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Mashcka [7]

Answer:

3. When the number of turns, N is doubled, the strength of the electromagnet is also doubled

4. Doubling the voltage, doubles the strength of the electromagnet

5. The number of paper clips a 7.5 V battery would pick is approximately 28 paper clips

The number of paper clips a 7.5 V battery would pick is 59 paperclips

6. For the 50-coil electromagnet, the average number of paper clips a 1 V battery would pick is approximately 7 paperclips

For the 50-coil electromagnet, the average number of paper clips a 1 V battery would pick is 16 paperclips

Explanation:

3. The Magnetomotive Force, MMF = The Number of Turns on the Coil, N × The Current I Flowing in the Coil, I

∴ MMF = N × I

When the number of turns, N is doubled, the magnetomotive force, MMF is also doubled, and the strength of the electromagnet is doubled

4. Given that the voltage, V applied to the coil = The current, I flowing × The resistance, R of the coil, we have

V = I × R

Therefore, for a given constant resistance in the coil, doubling the voltage, doubles the current and therefore doubles the strength of the electromagnet

5. The average slope for the 25-coil electromagnet = (23 - 12)/(6 - 3) = 3.\bar 6

The number of paper clips a 7.5 V battery would pick = 12 + (7.5 - 3) × 11/3 = 28.5 paperclips  ≈ 28 paper clips

The average slope for the 50-coil electromagnet = (48 - 26)/(6 - 3) = 7.\bar 3

The number of paper clips a 7.5 V battery would pick = 26 + (7.5 - 3) × 22/3 = 59 paperclips

6. The slope calculated from a start point of approximately 0.4 V, is given as follows;

The slope for the 25-coil electromagnet = (12 - 6)/(3 - 0.4) = 30/13

Therefore, for the 25-coil electromagnet,  the average number of paper clips a 1 V battery would pick = 6 + (1 - 0.4) × 30/13) = 96/13 ≈ 7 paperclips

The slope for the 50-coil electromagnet = (26 - 13)/(3 - 0.4) = 5

Therefore, for the 50-coil electromagnet,  the average number of paper clips a 1 V battery would pick = 13 + (1 - 0.4) × 5 = 16 paperclips

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