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Volgvan
3 years ago
5

What is the free-fall acceleration at the surface of the jupiter?

Physics
1 answer:
Rufina [12.5K]3 years ago
4 0

Answer:

24.79 m/s2

Explanation:

Let us assume that there is an object with a mass of 'm' on the

 

Jupiter. Jupiter will attract this object:

mg_j=G\frac{mM_j}{r_j^{2} }

g_j=G\frac{M_j}{r_j^{2} }

g_j=6.67*10^{-11} \frac{1.9*10^{27} }{71492000^{2} }

g_j=24.79m/s^{2}

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If it takes 200 joules of energy to lift a bucket of water 3 meters in 2 seconds, how much power would be required to do the sam
Harlamova29_29 [7]

200 joules of work energy are involved.  That's all we need to know to answer the question.  Once we know that 200 joules of work energy are involved, we don't care what was lifted, or how far, or how long it took, or how many people worked on it, or how much they were paid, or what was the distribution of their gender identities, or the ethnic diversity among the team. or what day each of them celebrates as their sabbath.  Any other information besides the 200 joules is only there to distract us, and see whether we're paying attention.

Power = (work or energy) / (time to do the work or move the energy)

Power = (200 joules) / (5 seconds)

<em>Power = 40 watts</em>

3 0
3 years ago
What’s the velocity of a sound wave traveling through air at a temperature of 18°C (64.4°F)?
Anna11 [10]

Answer:

342 m/s

Explanation:

The velocity of sound in air is approximated as:

v ≈ 331.4 + 0.6 T

where v is the velocity in m/s and T is the temperature in Celsius.

At T = 18:

v ≈ 331.4 + 0.6 (18)

v ≈ 342.2

The velocity is approximately 342 m/s.

4 0
3 years ago
Read 2 more answers
Two objects of the same size are both perfect blackbodies. One has a temperature of 3000 K, so its frequency of maximum emission
bija089 [108]

Answer:

a) The colder body (3000k), b) hearter body c) 12000K body

Explanation:

This exercise should know the power emitted by the objects and the distribution of this emission in the energy spectrum, for this we will use Stefan's laws and that of Wien's displacement

Stefan's Law                     P = σ A e T⁴

Wien displacement law   λ T = 2,898 10⁻³ m K

Let's calculate the power emitted for each object.

As they are perfect black bodies e = 1, they also indicate that they have the same area

T = 3000K

       P₁ = σ A T₁⁴

T = 12000K

       P₂ = σ A T₂⁴

       P₂ / P₁ = T₂⁴ / T₁⁴

       P₂ / P₁ = (12000/3000)⁴

       P₂ / P₁ = 256

This indicates that the hottest body emission is 256 times the coldest body emission.

Let's calculate the maximum emission wavelength

Body 1

T = 3000K

       λ T = 2,898 10-3

       λ₁ = 2.89810-3 / T

       λ₁ = 2,898 10-3 / 3000

       λ₁ = 0.966 10-6 m

      λ₁ = 966 nm

T = 12000K

      λ₂ = 2,898 10-3 / 12000

      λ₂ = 0.2415 10-6 m

      λ₂ = 214 nm

a) The colder body (3000k) emits more light in the infrared, since the emission of the hot body is at a minimum (emission tail)

b) The two bodies have emission in this region, the body of 3000K in the part of rise of the emission and the body to 12000K in the descent of the emission even when this body emits 256 times more than the other, so this body should have the highest broadcast in this area

c) The emission of the hottest 12000K body is mainly in UV

d) The hottest body emits more energy in UV and visible

e) No body has greater emission in all zones

5 0
3 years ago
A 3 kg rubber block is resting on wet concrete. The coefficient of static friction is 0.3. What is the minimum force that must b
mrs_skeptik [129]

Answer:

You would have to find the friction force of the rubber block which would be found with the equation of Normal force (mass*gravity) times cooeficient of friction which would give 8.82 N for the amount of friction and because you need more force than 8.82 N (assuming gravity is 9.8)

8 0
2 years ago
Explain how the fountain pen is filled up with ink​
Leno4ka [110]

Answer: A piston-filling fountain pen has a piston — just like in a car — inside the barrel. This piston goes down to expel air or ink and then back up, pulling ink into the barrel. The typical process is very simple, assuming the pen is clean and dry: Push the piston down, expelling any air in the barrel

8 0
2 years ago
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