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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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Olegator [25]
The acceleration of the particle as a function of time t is
a(t) = t + t^2
The velocity of the particle at time t is the integral of the acceleration:
v(t) =  \int {a(t)} \, dt =  \frac{t^2}{2} +  \frac{t^3}{3}  + C
where the constant C can be found by requiring that the velocity at time t=0 is v=3:
v(0) = 3
and we find C=v_0=3
so the velocity is
v(t)=3+ \frac{t^2}{2}+ \frac{t^3}{3}

The position of the particle at time t is the integral of the velocity:
x(t)=\int {v(t) } \,dt = 3t +  \frac{t^3}{6}+ \frac{t^4}{12}   +D
where D can be found by requiring that the initial position at time t=0 is zero:
x(0)=0
from which we find D=0, so 
 x(t)=3t + \frac{t^3}{6}+ \frac{t^4}{12} 

To solve the problem, now we just have to substitute t=5 into x(t) and v(t) to find the position and the velocity of the particle at t=5.

The position is:
x(5)=3(5) +  \frac{5^3}{6}+ \frac{5^4}{12}=87.92
and the velocity is:
v(5) = 3+ \frac{5^2}{2}+ \frac{5^3}{3}=57.17
6 0
3 years ago
van object moving with uniform acceleration has a velocity of 10.0 cm/s in the positive x-direction when its x-coordinate is 3.0
creativ13 [48]

Initial velocity of object vi=10.0 cm/s

initial position fo vector of the object is xi=3.09 cm

Final position of vector xf=-5.00cm

then displacement of object s = xf-xi=-5.00-3.09=-8.09cm

time t=2.55 s

s=vit+1/2at2

-5.00 = 11*2.55+1/2*a2.552

a = (-5.00 - 10*2.55*2)/2.552 = 2.94 cm

Acceleration is 2.94 cm.

<h3>What is acceleration?</h3>

Speed increase is the name we provide for any cycle where the speed changes. Since speed is a speed and a bearing, there are simply two different ways for you to speed up: change your speed or shift your course or change both. In mechanics, speed increase is the pace of progress of the speed of an item concerning time. Speed increases are vector amounts. The direction of an item's speed increase is given by the direction of the net power following up on that article. An item's typical speed increase throughout some stretch of time is its adjustment of speed separated by the term of the period. Numerically, quick speed increase, in the meantime, is the constraint of the typical speed increase over a little time period. In the terms of analytics, immediate speed increase is the subordinate of the speed vector concerning time.

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4 0
2 years ago
A boat traveling across a river has a resultant velocity of 10 km/h and travels 34 degrees with respect to the shore. A) What is
vodka [1.7K]

Answer:

a) 1.55 m/s

b) 2.3 m/s

Explanation:

We know that the boat travels across the river, if we define the river as the x-axis, then the velocity of the boat is only on the y-axis.

Then we can write the velocity of the boat in still water as:

S = (0, B)

Now, when the boat is on the river, the velocity of the boat will be equal to the velocity of the boat in still water plus the velocity of the river.

The velocity of the river is:

v = (R, 0).

Then the velocity of the boat in that river is:

V' = (0, B) + (R, 0) = (R, B)

Now, we know that the velocity of the boat is 10km/h, and it travels at an angle of 34° with respect to the shore.

We can use the Pythagoreans theorem to write the components of this velocity as:

x-axis component = 10km/h*cos(34°) = 8.29 km/h

y-axis component = 10km/h*sin(34°) =  5.59 km/h

Then the velocity of the boat can be written in components as:

velocity = ( 8.29 km/h,  5.59 km/h)

And we knew that the velocity of the boat was written as  (R, B)

Then we must have:

R = 8.29 km/h

B = 5.59 km/h

a) The speed of the boat in m/s:

We know that the speed of the boat is 5.59 km/h.

First, we know that:

1km = 1000m, then:

5.59 km/h = 5.59*(1000m)h = 5,590 m/h

And we know that:

1h = 3600s

Then we can write:

5,590 m/h = 5,590 m/(3600s) = 1.55 m/s

b) The speed of the river in m/s:

We know that the speed of the river is 8.29 km/h

Using the same reasoning as above, we can do the change of units as follows:

8.29 km/h = 8.29 (1000m)/(3600s) = 2.3 m/s

6 0
3 years ago
A 0.50-kg mass attached to the end of a string swings in a vertical circle (radius 2.0 m). When the mass is at the highest point
il63 [147K]

Answer:

31.1 N

Explanation:

m = mass attached to string = 0.50 kg

r = radius of the vertical circle = 2.0 m

v = speed of the mass at the highest point = 12 m/s

T = force of the string on the mass attached.

At the highest point, force equation is given as

T + mg =\frac{mv^{2}}{r}

Inserting the values

T + (0.50)(9.8) =\frac{(0.50)(12)^{2}}{2}

T = 31.1 N

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3 years ago
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