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AnnZ [28]
2 years ago
15

The solar mass of the Sun is 1. The orbital period of Jupiter is 11. 9 Earth years. What is the distance between Jupiter and the

Sun? 5. 2 AU 41 AU 52 AU 410 AU.
Physics
1 answer:
cestrela7 [59]2 years ago
6 0

The distance between Jupiter and the sun is 5.2 AU.

According to Kepler's third law, the square of the period of revolution of planets is proportional to the cube of their mean distances from the sun. From this; T^2 = r^3.

Now, we are told that the orbital period (T) is 11. 9 Earth years. We have to make the distance the subject of the formula.

r =T^2/3

r = (11.9)^2/3

r = 5.2 AU

Learn more: brainly.com/question/15207516

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A particular balloon can be stretched to a maximum surface area of 1257 cm2. The balloon is filled with 3.1 L of helium gas at a
chubhunter [2.5K]

Answer:

The ballon will brust at

<em>Pmax = 518 Torr ≈ 0.687 Atm </em>

<em />

<em />

Explanation:

Hello!

To solve this problem we are going to use the ideal gass law

PV = nRT

Where n (number of moles) and R are constants (in the present case)

Therefore, we can relate to thermodynamic states with their respective pressure, volume and temperature.

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2} --- (*)

Our initial state is:

P1 = 754 torr

V1 = 3.1 L

T1 = 294 K

If we consider the final state at which the ballon will explode, then:

P2 = Pmax

V2 = Vmax

T2 = 273 K

We also know that the maximum surface area is: 1257 cm^2

If we consider a spherical ballon, we can obtain the maximum radius:

R_{max} = \sqrt{\frac{A_{max}}{4 \pi}}

Rmax = 10.001 cm

Therefore, the max volume will be:

V_{max} = \frac{4}{3} \pi R_{max}^3

Vmax = 4 190.05 cm^3 = 4.19 L

Now, from (*)

P_{max} = P_1 \frac{V_1T_2}{V_2T_1}

Therefore:

Pmax= P1 * (0.687)

That is:

Pmax = 518 Torr

6 0
3 years ago
A rocket consists of a shuttle and a fuel tank. The rocket flies through the air with a momentum of 180000kgm/s at a velocity of
babunello [35]

Answer:

Answer is 1000kg

Hope it help

6 0
2 years ago
A 8.4-mH inductor carries a current I = Imaxsin ωt, with Imax = 4.00 A and f = ω/2π = 60.0 Hz. What is the self-induced emf as a
Aleks04 [339]

Answer:

E= -3.166 cosωt   V

Explanation:

Given that

I = Imax sinωt

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Imax= 4 A

f = ω/2π = 60.0 Hz

ω = 120π  rad/s

We know that self induce E given as

E=-L\dfrac{dI}{dt}

\dfrac{dI}{dt}= Imax \ \omega\ cos\omega t

E=-L\times Imax \ \omega\ cos\omega t

E=-8.4\times 120\times \pi \ cos\omega t

E= -3166.72 cosωt  m V

E= -3.166 cosωt   V

This is the induce emf.

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