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3 years ago

What is the shape of the orbit of satellites

2 answers:

8 0

Every closed orbit is an ellipse.

If the big body just bends the small body's path but doesn't capture it, then the small body's path is a hyperbola.

dem82 [27]3 years ago

4 0

Answer:

The shape will be elliptical hopefully this helps!

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Why is it expensive to bring electricity into urban areas?

motikmotik

Because of the pole and the generator you would have to biuld

8 0

3 years ago

Read 2 more answers

Two satellites are in circular orbits around a planet that has radius 9.00x10^6 m. One satellite has mass 53.0 kg , orbital radi

Andreas93 [3]

Answer:

The orbital speed of this second satellite is 5195.16 m/s.

Explanation:

Given that,

Orbital radius of first satellite $r_{1}= 8.20\times10^{7}$

Orbital radius of second satellite $r_{2}=7.00\times10^{7}\ m$

Mass of first satellite $m_{1}=53.0\ kg$

Mass of second satellite $m_{2}=54.0\ kg$

Orbital speed of first satellite = 4800 m/s

We need to calculate the orbital speed of this second satellite

Using formula of orbital speed

$v=\sqrt{\dfrac{GM}{r}}$

From this relation,

$v_{1}\propto\dfrac{1}{\sqrt{r}}$

Now, $\dfrac{v_{1}}{v_{2}}=\sqrt{\dfrac{r_{2}}{r_{1}}}$

$v_{2}=v_{1}\times\sqrt{\dfrac{r_{1}}{r_{2}}}$

Put the value into the formula

$v_{2}=4800\times\sqrt{\dfrac{ 8.20\times10^{7}}{7.00\times10^{7}}}$

$v_{2}=5195.16\ m/s$

Hence, The orbital speed of this second satellite is 5195.16 m/s.

4 0

3 years ago

If you act without reason or sound judgement, people will describe you as __________.

EastWind [94]

Answer:

irrational

Explanation:

8 0

2 years ago

Practical steam engines utilize 450ºC steam, which is later exhausted at 270ºC.

Naily [24]

(a) 0.249 (24.9 %)

The maximum efficiency of a heat engine is given by

$\eta = 1-\frac{T_C}{T_H}$

where

Tc is the low-temperature reservoir

Th is the high-temperature reservoir

For the engine in this problem,

$T_C = 270^{\circ}C+273=543 K$

$T_H = 450^{\circ}C+273=723 K$

Therefore the maximum efficiency is

$\eta = 1-\frac{T_C}{T_H}=1-\frac{543}{723}=0.249$

(b-c) 0.221 (22.1 %)

The second steam engine operates using the exhaust of the first. So we have:

$T_H = 270^{\circ}C+273=543 K$ is the high-temperature reservoir

$T_C = 150^{\circ}C+273=423 K$ is the low-temperature reservoir

If we apply again the formula of the efficiency

$\eta = 1-\frac{T_C}{T_H}$

The maximum efficiency of the second engine is

$\eta = 1-\frac{T_C}{T_H}=1-\frac{423}{543}=0.221$

8 0

3 years ago

Question:<br> the distance of the car can move by velocity 15 m/s for time 5s is :<br> Physics

tangare [24]

Answer:

c) 2.05

Explanation:

7 0

2 years ago