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ollegr [7]
3 years ago
6

Jupiter’s Great Red Spot rotates completely every six days. If the spot is circular (not quite true, but a reasonable approximat

ion for this matter) and 26,000 km in diameter, what are the wind speeds (i.e. how fast the gas is moving) at the outer edges ofthe storm (in km/hr)? (Hints: Recall the definition of speed. Think about how far a cloud on the edge of the storm must travel.)
Physics
1 answer:
artcher [175]3 years ago
5 0

Answer:

v = 567.2 km/h

Explanation:

As we know that if Jupiter Rotate one complete rotation then it means that the it will turn by 360 degree angle

so here the distance covered by the point on its surface in one complete rotation is given by

distance = 2\pi r

distance = \pi D

now we will have the time to complete the rotation given as

t = 6 days

t = 6 (24 h) = 144 h

now the speed is given by

speed = \frac{distance}{time}

speed = \frac{\pi D}{t}

speed = \frac{\pi(26000 km)}{144}

v = 567.2 km/h

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4) (5 points) Given are the magnitudes and orientations (with respect to x-axis) of 3
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Expand each vector into their component forms:

\vec A=(4.5\,\mathrm N)(\cos\theta_A\,\vec\imath+\sin\theta_A\,\vec\jmath)=(2.58\,\vec\imath+3.69\,\vec\jmath)\,\mathrm N

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3 years ago
A father fashions a swing for his children out of a long rope that he fastens to the limb of a tall tree. As one of the children
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Answer:

The centripetal acceleration of the child at the bottom of the swing is 15.04 m/s².

                     

Explanation:

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the density of a steel is 7.8.find the mass of steel cube of side 10 cm. find volume of steel if the mass is 8kg​
lukranit [14]

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The density of the steel   =  7.8

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<u>(1)The mass of steel cube :</u>

We know that,

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