How do storms on Jupiter differ from storm systems on Earth?

Physics

1 answer:

Vladimir [108]2 years ago

8 0

Answer:

As, Jupiter are one of the largest planet in the solar system and the large amount of the mass of the jupiter are consisted with the gases. The storm tracks in the symmetrical path at the proper latitude in the system. But the storm tracks on the earth in the system where planet are highly variable.

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After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 55.0 cm. She finds the pendulum m

forsale [732]

Answer:

Acceleration due to gravity will be $g=5.718m/sec^2$

Explanation:

We have given length of pendulum l = 55 cm = 0.55 m

It is given that pendulum completed 100 swings in 145 sec

So time taken by pendulum for 1 swing $=\frac{145}{100}=1.45sec$

We have to find the acceleration due to gravity at that point

We know that time period of pendulum;um is given by

$T=2\pi \sqrt{\frac{l}{g}}$

So $1.45=2\times 3.14\times \sqrt{\frac{0.55}{g}}$

$\sqrt{\frac{0.55}{g}}=0.230$

Squaring both side

${\frac{0.3025}{g}}=0.0529$

$g=5.718m/sec^2$

So acceleration due to gravity will be $g=5.718m/sec^2$

3 0

3 years ago

Saturn moves in an orbit around the Sun with radius 10 AU. How many degrees does it move on the Celestial in one year? (Hint: Ca

Lana71 [14]

Answer:

B. About 12 degrees

Explanation:

The orbital period is calculated using the following expression:

T = 2π*( $\sqrt{\frac{r^3}{Gm}}$ )

Where r is the distance of the planet to the sun, G is the gravitational constant and m is the mass of the sun.

Now, we don't actually need to solve the values of the constants, since we now that the distance from the sun to Saturn is 10 times the distance from the sun to the earth. We now this because 1 AU is the distance from the earth to the sun.

Now, we divide the expression used to calculate the orbital period of Saturn by the expression used to calculate the orbital period of the earth. Notice that the constants will cancel and we will get the rate of orbital periods in terms of the distances to the sun:

$\frac{Tsaturn}{Tearth}$ = $\sqrt{\frac{rSaturn^3}{rEarth^3} } = \sqrt{10^3}}$