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Luda [366]
3 years ago
15

A 1500 kg car skids to a halt on a wet road where μk = 0.51. how fast was the car traveling if it leaves 66-m-long skid marks?

Physics
1 answer:
djverab [1.8K]3 years ago
3 0

We got coefficient of friction uk = 0.51<span>
Mass m = 1500 kg 
normal force on the car = weight of the car = 1500 * 9.8 N = 14700 N 
and the force of friction on the car, f = uk * normal force = 0.51 * 14700 N = 7497 N 
taking Acceleration a = -f/m = -7497/1500 = -4.998 m/s^2 
Displacement s = 66 m 
Final velocity v = 0 
Initial velocity u = ? 
v^2 = u^2 + 2as 
<span>0 = u^2 - 2 * 4.998 * 66
<span>u = 25.69 m/s </span></span></span>

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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
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Answer:

A)Ω = 7.8 × 10^−5 steradians.

B) TE = 5800K

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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;

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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.

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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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