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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.)

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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In the same configuration of the previous problem 3, four long straight wires are perpendicular to the page, and their cross sec

faust18 [17]

Complete Question:

In the same configuration of the previous problem 3, four long straight wires are perpendicular to the page, and their cross sections form a square of edge length a = 13.5 cm. Each wire carries 7.50 A, and the currents are out of the page in wires 1 and 4 and into the page in wires 2 and 3.

a) Draw a diagram in a (x,y) plane of the four wires with wire 4 perpendicular to the origin. Indicate the current's directions.

b) Draw a diagram of all magnetic fields produced at the position of wire 3 by the other three currents.

c) Draw a diagram of all magnetic forces produced at the position of wire 3 by the other three currents.

d) What are magnitude and direction of the net magnetic force per meter of wire length on wire 3?

Answer:

force, 1.318 ₓ 10⁻⁴

direction, 18.435°

Explanation:

The attached file gives a breakdown step by step solution to the questions

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3 years ago

A uniform disk with a 25 cm radius swings without friction about a nail through the rim. If it is released from rest from a posi

ValentinkaMS [17]

Answer:

Explanation:

During the swing , the center of mass will go down due to which disc will lose potential energy which will be converted into rotational kinetic energy

mgh = 1/2 I ω² where m is mass of the disc , h is height by which c.m goes down which will be equal to radius of disc , I is moment of inertia of disc about the nail at rim , ω is angular velocity .

mgr = 1/2 x ( 1/2 m r²+ mr²) x ω²

gr = 1/2 x 1/2 r² x ω² + 1/2r² x ω²

g = 1 / 4 x ω² r + 1 / 2 x ω² r

g = 3 x ω² r/ 4

ω² = 4g /3 r

= 4 x 9.8 / 3 x .25

= 52.26

ω = 7.23 rad / s .

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3 years ago

when a circular plate of metal is heated in an oven, its radius increases at .03 cm/min, at what rate is the area increasing whe

Alinara [238K]

Answer:

Rate of change of area will be $9.796cm^2/min$

Explanation:

We have given rate of change of radius $\frac{dr}{dt}=0.03cm/min$

Radius of the circular plate r = 52 cm

Area is given by $A=\pi r^2$

So $\frac{dA}{dt}=2\pi r\frac{dr}{dt}$

Puting the value of r and $\frac{dr}{dt}$

$\frac{dA}{dt}=2\times 3.14\times 52\times 0.03=9.796cm^2/min$

So rate of change of area will be $9.796cm^2/min$

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3 years ago

At an altitude of 5000 m the rocket's acceleration has increased to 6.9 m/s2 . What mass of fuel has it burned?

sergey [27]

1) Initial upward acceleration: $6.0 m/s^2$

2) Mass of burned fuel: $0.10\cdot 10^4 kg$

Explanation:

1)

There are two forces acting on the rocket at the beginning:

- The force of gravity, of magnitude $F_g = mg$ , in the downward direction, where

$m=1.9\cdot 10^4 kg$ is the rocket's mass

$g=9.8 m/s^2$ is the acceleration of gravity

- The thrust of the motor, T, in the upward direction, of magnitude

$T=3.0\cdot 10^5 N$

According to Newton's second law of motion, the net force on the rocket must be equal to the product between its mass and its acceleration, so we can write:

$T-mg=ma$ (1)

where a is the acceleration of the rocket.

Solving for a, we find the initial acceleration:

$a=\frac{T-mg}{m}=\frac{3.0\cdot 10^5-(1.9\cdot 10^4)(9.8)}{1.9\cdot 10^4}=6.0 m/s^2$

2)

When the rocket reaches an altitude of 5000 m, its acceleration has increased to

$a'=6.9 m/s^2$

The reason for this increase is that the mass of the rocket has decreased, because the rocket has burned some fuel.

We can therefore rewrite eq.(1) as

$T-m'g=m'a'$

where

$m'$ is the new mass of the rocket

Re-arranging the equation and solving for m', we find

$m'=\frac{T}{g+a}=\frac{3.0\cdot 10^5}{9.8+6.9}=1.8\cdot 10^4 kg$

And since the initial mass of the rocket was

$m=1.9 \cdot 10^4 kg$

This means that the mass of fuel burned is

$\Delta m = m-m'=1.9\cdot 10^4 - 1.80\cdot 10^4 = 0.10\cdot 10^4 kg$

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