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Illusion [34]
2 years ago
8

Planets are not uniform inside. Normally, they are densest at the center and have decreasing density outward toward the surface.

Model a spherically symmetric planet, with the same radius as the earth, as having a density that decreases linearly with distance from the center. Let the density be 1.60 x 10^4 kg/m^3 at the center and 2100 kg/m^3 at the surface.
What is the acceleration due to gravity at the surface of this planet?
Physics
1 answer:
SSSSS [86.1K]2 years ago
7 0

Answer:

a = 9.94 m/s²

Explanation:

given,

density at center= 1.6 x 10⁴ kg/m³

density at the surface = 2100 Kg/m³

volume mass density as function of distance

\rho(r) = ar^2 - br^3

r is the radius of the spherical shell

dr is the thickness

volume of shell

dV = 4 \pi r^2 dr

mass of shell

dM = \rho(r)dV

\rho = \rho_0 - br

now,

dM = (\rho_0 - br)(4 \pi r^2)dr

integrating both side

M = \int_0^{R} (\rho_0 - br)(4 \pi r^2)dr

M = \dfrac{4\pi}{3}R^3\rho_0 - \pi R^4(\dfrac{\rho_0-\rho}{R})

M = \pi R^3(\dfrac{\rho_0}{3}+\rho)

we know,

a = \dfrac{GM}{R^2}

a = \dfrac{G( \pi R^3(\dfrac{\rho_0}{3}+\rho))}{R^2}

a =\pi RG(\dfrac{\rho_0}{3}+\rho)

a =\pi (6.674\times 10^{-11}\times 6.38 \times 10^6)(\dfrac{1.60\times 10^4}{3}+2.1\times 10^3)

a = 9.94 m/s²

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A parallel plate capacitor stores charge, and thus, stores energy in the form of electric potential energy. The total energy sto
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The charges on the plates,

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3 years ago
4. Lead has a density of 11.5g/cmº. A rectangular block of lead measures 7cm x5cmx2cm.
lisov135 [29]

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(a) 70cm³

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3 0
3 years ago
A slit has a width of W1 = 4.4 × 10-6 m. When light with a wavelength of λ1 = 487 nm passes through this slit, the width of the
Vitek1552 [10]

Answer:

The width of the central bright fringe on the screen is observed to be unchanged is 4.48*10^{-6}m

Explanation:

To solve the problem it is necessary to apply the concepts related to interference from two sources. Destructive interference produces the dark fringes.  Dark fringes in the diffraction pattern of a single slit are found at angles θ for which

w sin\theta = m\lambda

Where,

w = width

\lambda =wavelength

m is an integer, m = 1, 2, 3...

We here know that as sin\theta as w are constant, then

\frac{w_1}{\lambda_1} = \frac{w_2}{\lambda_2}

We need to find w_2, then

w_2 = \frac{w_1}{\lambda_1}\lambda_2

Replacing with our values:

w_2 = \frac{4.4*10^{-6}}{487}496

w_2 = 4.48*10^{-6}m

Therefore the width of the central bright fringe on the screen is observed to be unchanged is 4.48*10^{-6}m

3 0
3 years ago
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