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telo118 [61]
3 years ago
11

Ganymede is one of the many moons of Jupiter. It is nearly spherical in shape. It is larger than the planet Mercury and slightly

smaller than the planet Mars. If it is so large compared with the bodies around it, why is it called a moon and not a planet?
Physics
2 answers:
lesya [120]3 years ago
5 0
Moons revolve around a planet, and planets revolve around the sun. Ganymede is considered a moon because it revolves around the planet Jupiter, therefore, it is a moon. :) I hope this helps! 
Karo-lina-s [1.5K]3 years ago
5 0

Answer:

C. it moves in an orbit around Jupiter

Explanation:

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dedylja [7]

Answer:

200

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3 years ago
Benita is studying the erosion of soil after a heavy rainfall. She observes that rainwater washes away very little soil from are
yanalaym [24]

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yes bc it washed it away so

Explanation:

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4 0
3 years ago
Electric charge is distributed over the disk x2 + y2 ≤ 4 so that the charge density at (x, y) is rho(x, y) = 4x + 4y + 4x2 + 4y2
maw [93]

Answer:

Q=185.84C

Explanation:

We have to take into account the integral

Q=\int \rho dV

In this case we have a superficial density in coordinate system.

Hence, we have for R: x2 + y2 ≤ 4

Q=\int_{-2}^2\int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}\rho dydx

but, for symmetry:

Q=4\int_0^2\int_0^{\sqrt{4-x^2}}\rho dydx\\\\Q=4\int_0^2\int_0^{\sqrt{4-x^2}}(4x+4y+4x^2+4y^2) dydx\\\\Q=4\int_0^{2}[4x\sqrt{4-x^2}+2(4-x^2)+4x^2\sqrt{4-x^2}+\frac{4}{3}(4-x^2)^{3/2}]dx\\\\Q=4[46.46]=185.84C

HOPE THIS HELPS!!

8 0
3 years ago
if the displacement of a body is proportional to the square of time, state the nature of motion of the body ​
Liula [17]

Explanation:

If the displacement of an object is proportional to the square of the time taken then the body is moving with uniformly accelerated motion as it will follow Newton's second equation of motion for a particular initial velocity, which can be given by, s=ut+21at2.

hope this is helpful to you

3 0
3 years ago
When a rocket is 4 kilometers high, it is moving vertically upward at a speed of 400 kilometers per hour. At that instant, how f
Y_Kistochka [10]

Answer:

The angle of elevation of the rocket is increasing at a rate of 48.780º per second.

Explanation:

Geometrically speaking, the distance between the rocket and the observer (r), measured in kilometers, can be represented by a right triangle:

r = \sqrt{x^{2}+y^{2}} (1)

Where:

x - Horizontal distance between the rocket and the observer, measured in kilometers.

y - Vertical distance between the rocket and the observer, measured in kilometers.

The angle of elevation of the rocket (\theta), measured in sexagesimal degrees, is defined by the following trigonometric relation:

\tan \theta = \frac{y}{x} (2)

If we know that x = 5\,km, then the expression is:

\tan \theta = \frac{y}{5}

And the rate of change of this angle is determined by derivatives:

\sec^{2}\theta \cdot \dot \theta = \frac{1}{5}\cdot \dot y

\frac{\dot \theta}{\cos^{2}\theta} = \frac{\dot y}{5}

\frac{\dot \theta\cdot (25+y^{2})}{25} = \frac{\dot y}{5}

\dot \theta = \frac{5\cdot \dot y}{25+y^{2}}

Where:

\dot \theta - Rate of change of the angle of elevation, measured in sexagesimal degrees.

\dot y - Vertical speed of the rocket, measured in kilometers per hour.

If we know that y = 4\,km and \dot y = 400\,\frac{km}{h}, then the rate of change of the angle of elevation is:

\dot \theta = 48.780\,\frac{\circ}{s}

The angle of elevation of the rocket is increasing at a rate of 48.780º per second.

3 0
3 years ago
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