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Black_prince [1.1K]
3 years ago
6

Two identical silver spheres of mass m and radius r are placed a distance R (sphere 1) and 2R (sphere 2) from the Sun, respectiv

ely. The ratio of the gravitational force exerted by the Sun on sphere 1 to the pressure of solar radiation on sphere 1 is T1; the ratio for sphere 2 is T2. The ratio of T2 to T1 is ___.
Physics
1 answer:
lys-0071 [83]3 years ago
3 0

Answer:

The ratio of T2 to T1 is 1.0

Explanation:

The gravitational force exerted on each sphere by the sun is inversely proporational to the square of the distance between the sun and each of the spheres.

Provided that the two spheres have the same radius r, the pressure of solar radiation too, is inversely proportional to the square of the distance of each sphere from the sun.

Let F₁ and F₂ = gravitational force of the sun on the first and second sphere respectively

P₁ and P₂ = Pressure of solar radiation on the first and second sphere respectively

M = mass of the Sun

m = mass of the spheres, equal masses.

For the first sphere that is distance R from the sun.

F₁ = (GmM/R²)

P₁ = (k/R²)

T₁ = (F₁/P₁) = (GmM/k)

For the second sphere that is at a distance 2R from the sun

F₂ = [GmM/(2R)²] = (GmM/4R²)

P₂ = [k/(2R)²] = (k/4R²)

T₂ = (F₂/P₂) = (GmM/k)

(T₁/T₂) = (GmM/k) ÷ (GmM/k) = 1.0

Hope this Helps!!!

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Light of wavelength 476.1 nm falls on two slits spaced 0.29 mm apart. What is the required distance from the slits to the screen
stich3 [128]

Answer:

The distance is D  =  2.6 \ m

Explanation:

From the question we are told that

    The wavelength of the light is  \lambda  =  476.1 \ nm  =  476.1 *10^{-9} \ m

      The  distance between the slit is  d =  0.29 \  mm  =  0.29 *10^{-3} \ m

       The  between the first and second dark fringes is  y =  4.2 \ mm  =  4.2 *10^{-3} \ m

Generally  fringe width is mathematically represented as

       y  =  \frac{\lambda * D }{d}

Where D is the distance of the slit to the screen

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        D  =  \frac{y *  d}{\lambda }

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       D  =  \frac{ 4.2 *10^{-3} *   0.29 *10^{-3}}{ 476.1 *10^{-9} }

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A certain field line diagram illustrates the electric field due to three particles that carry charges 5.0 μC, -3.0 μC, and -2.0
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6

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Number of lines emanate from + 5 micro coulomb is 15 .

They terminates at negative charges that means at - 3 micro coulomb and - 2 micro Coulomb.

the electric field lines terminates at - 3 micro Coulomb and - 2 micro Coulomb is in the ratio of 3 : 2.

So the lines terminating at - 3 micro coulomb

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So the lines terminating at - 2 micro coulomb

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So, the number of filed lines terminates at - 2 micro Coulomb are 6.

4 0
3 years ago
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