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3 years ago

Find the orbital speed v for a satellite in a circular orbit of radius R.Express the orbital speed in terms of G, M, and R.

Physics

1 answer:

AlekseyPX3 years ago

5 0

<h2>Answer: $V=\sqrt{G\frac{M}{R}}$ </h2>

The velocity of a satellite describing a circular orbit is constant and defined by the following expression:

$V=\sqrt{G\frac{M}{R}}$ (1)

Where:

$G$ is the gravity constant

$M$ the mass of the massive body around which the satellite is orbiting

$R$ the radius of the orbit (measured from the center of the planet to the satellite).

Note this orbital speed, as well as orbital period, does not depend on the mass of the satellite. It depends on the mass of the massive body.

In addition, this orbital speed is constant because at all times both the kinetic energy and the potential remain constant in a circular (closed) orbit.

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You work for the city water department and need to pump 3400 liters/minute of water from a tank at ground level into a vented (i

umka21 [38]

Answer:

$P_E=46.2778\ kW$

$v=1.804\ m.s^{-1}$

Explanation:

Given:

flow rate of water, $\dot{V}=3400\ L.min^{-1}=3.4\ m^3.min^{-1}$

∵Density of water is 1 kg per liter

∴mass flow rate of water, $\dot{m}=3400\ kg.min^{-1}$

height of pumping, $h=75\ m$

efficiency of motor drive, $\eta=0.9$

diameter of pipe, $D =0.2\ m$

Now the power required for pumping the water at given conditions:

$P=\dot{m}.g.h$

$P=\frac{3400}{60} \times 9.8\times 75$

$P=41650\ W$

Hence the electric power required:

$P_E \times \eta=P$

$P_E \times 0.9=41650$

$P_E=46.2778\ kW$

Flow velocity is given as:

$v=\dot{V}\div a$

where: a = cross sectional area of flow through the pipe

$v=\frac{3.4}{60}\div (\pi.\frac{0.2^2}{4} )$

$v=1.804\ m.s^{-1}$

3 0

2 years ago

The density of aluminum is 2.7 g/cm3. A metal sample has a mass of 52.0 grams and a volume of 17.1 cubic centimeters. Could the

Fudgin [204]

Answer:
__________________________________________________
No; the sample could not be aluminum;
since the density of aluminum, " 2.7 g/cm³ " , is NOT close enough to the density of the sample, " 3.04 g/cm³ " .
________________________________________________
Explanation:
________________________________________________
Density is expressed as "mass per unit volume" ;

in which:
"mass, "m", is expressed in units of "g" (grams); and:
"Volume, "V", is expressed in units of "cm³ " (such as in this problem); or in units of "mL" ;
__________________________________________________
{Note the exact conversion: " 1 cm³ = 1 mL " .}.
__________________________________________________
The formula for density: D = m/V ;

Given: The density of aluminum is: 2.7 g/cm³.

Given: A sample has a mass of 52.0 g ; and Volume of 17.1 cm³ ; could it be aluminum?
_________________________________________________________
Let us divide the mass of the sample by the volume of the sample;
by using the formula:
___________________________________________
D = m / V ;

and see if the value is at, or very close to "2.7 g/cm³ ".

If it is, then it could be aluminum.
____________________________________________________
The density for the sample:

D = (52.0 / 17.1) g/cm³ = 3.0409356725146199 g/cm³ ;
→round to "3 significant figures" ;
= 3.04 g/cm³ .
_______________________________________________
No; the sample could not be aluminum; since the density of aluminum,
"2.7 g/cm³ " is NOT close enough to the density of the sample,
"3.04 g/cm³ " .
____________________________________________________

5 0

3 years ago

If you pour different liquids into a graduated cylinder, which layer would be at the bottom?

klasskru [66]

the one that is more dense!

5 0

3 years ago

What is the acceleration of a 10 kg mass pushed by a 5 N force?

yanalaym [24]

$g-\ gravitational \ acceleration \\ g= \frac{G}{m} = \frac{5N}{10kg}= \\ g=0,5 \frac{N}{kg}$