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expeople1 [14]
3 years ago
14

If a planet has the same mass as the earth, but has twice the radius, how does the surface gravity, g, compare to g on the surfa

ce of the earth
Physics
1 answer:
shepuryov [24]3 years ago
3 0

Answer:

The surface gravity g of the planet is 1/4 of the surface gravity on earth.

Explanation:

Surface gravity is given by the following formula:

g=G\frac{m}{r^{2}}

So the gravity of both the earth and the planet is written in terms of their own radius, so we get:

g_{E}=G\frac{m}{r_{E}^{2}}

g_{P}=G\frac{m}{r_{P}^{2}}

The problem tells us the radius of the planet is twice that of the radius on earth, so:

r_{P}=2r_{E}

If we substituted that into the gravity of the planet equation we would end up with the following formula:

g_{P}=G\frac{m}{(2r_{E})^{2}}

Which yields:

g_{P}=G\frac{m}{4r_{E}^{2}}

So we can now compare the two gravities:

\frac{g_{P}}{g_{E}}=\frac{G\frac{m}{4r_{E}^{2}}}{G\frac{m}{r_{E}^{2}}}

When simplifying the ratio we end up with:

\frac{g_{P}}{g_{E}}=\frac{1}{4}

So the gravity acceleration on the surface of the planet is 1/4 of that on the surface of Earth.

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When temp is below 0 what process causes atoms to begin to clump
podryga [215]
Hi welcome to Brainly!
I believe the correct answer to your question is the Bose-Einstein condensation

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6 0
3 years ago
Suppose that the average speed (vrms) of carbon dioxide molecules (molar mass 44.0 g/mol) in a flame is found to be 2.67 105 m/s
34kurt

Answer:

T = 1.26 \times 10^8 K

Explanation:

As we know that rms speed of ideal gas is given by the formula

v_{rms} = \sqrt{\frac{3RT}{M}}

here we know that

v_{rms} = 2.67 \times 10^5 m/s

molecular mass of gas is given as

M = 44 g/mol = 0.044 kg/mol

now from above formula we have

2.67\times 10^5 = \sqrt{\frac{3(8.31)T}{0.044}}

now we have

T = 1.26 \times 10^8 K

7 0
4 years ago
A 2.8-carat diamond is grown under a high pressure of 58 × 10 9 N / m 2 .
Rzqust [24]

Answer:

a)  ΔV = 2,118 10⁻⁸ m³   b)  ΔR= 0.0143 cm

Explanation:

a) For this part we use the concept of density

    ρ = m / V

As we are told that 1 carat is 0.2g we can make a rule of proportions (three) to find the weight of 2.8 carats

    m = 2.8 Qt (0.2 g / 1 Qt) = 0.56 g = 0.56 10-3 kg

   

    V = m / ρ

    V = 0.56 / 3.52

    V = 0.159 cm3

We use the relation of the bulk module

    B = P / (Δv/V)

    ΔV = V P / B

    ΔV = 0.159 10⁻⁶ 58 10⁹ /4.43 10¹¹

    ΔV = 2,118 10⁻⁸ m³

b) indicates that we approximate the diamond to a sphere

    V = 4/3 π R³

For this part let's look for the initial radius

    R₀ = ∛ ¾ V /π

    R₀ = ∛ (¾ 0.159 /π)

    R₀ = 0.3361 cm

Now we look for the final volume and with this the final radius

    V_{f} = V + ΔV

    V_{f} = 0.159 + 2.118 10⁻²

    V_{f} = 0.18018 cm3

    R_{f} = ∛ (¾ 0.18018 /π)

    R_{f} = 0.3504 cm

The radius increment is

    ΔR = R_{f} - R₀

    ΔR = 0.3504 - 0.3361

    ΔR= 0.0143 cm

4 0
3 years ago
Where on this diagram does the ball have the highest point of gravitational potential energy?
mixer [17]
It should be at the very top since it has more space to fall which gives it more potential energy
3 0
3 years ago
How much work is required to lift a 10-newton weight from 4.0 meters to 40 meters above the surface of Earth?
kumpel [21]
<h3>Answer : 360J</h3>

<h3>Way to do : </h3>

s = 40m - 4m = 36m

W = F × s

= 10N × 36m = 360J

<h3>A bit of explanation : </h3>

W = Work (J)

F = Force / weight (N)

s = distance (m)

6 0
3 years ago
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