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3 years ago

What universal force is primarily responsible for keeping satellites in orbit around Earth?

Physics

2 answers:

g100num [7]3 years ago

6 0

<span>A) electromagnetic force </span>

xeze [42]3 years ago

3 0

Your answer to the question is C

You might be interested in

Can anyone ask me how many protons, neutrons, and electrons I need and just explain where to put them.

Aliun [14]

Answer:

on the first shell (ring) there will be 2 electrons

and on the 2nd shell there will be only one electron

while in the nucleus (the middle of the diagram) there will be 4 neutrons and 3 protons

Explanation:

you can see the picture attached

8 0

3 years ago

How to describe electric current as the flow of electrons

givi [52]

An electric current is a flow of electric charge. In electric circuits this charge is often carried by moving electrons in a wire. It can also be carried by ions in an electrolyte, or by both ions and electrons such as in an ionised gas (plasma).

4 0

3 years ago

A force parallel to the x-axis acts on a particle moving along the x-axis. This force produces potential energy U(x) given by U(

noname [10]

Answer:

Explanation:

U(x)=αx⁴

Force F = - dU / dx

= 4αx³

= 4 x .75 x³

= 3 x³

Force when x = .8 m

F = 3 (0.8)³

= 1.54 N

8 0

3 years ago

A TV with a power rating of 267 W uses 1.9 kWh in one day. How many hours was the TV during that day?

Aleonysh [2.5K]

1.9kWh/0.267kW = 7.11 hours

4 0

3 years ago

Consider a 30-cm-diameter hemispherical enclosure. The dome is maintained at 600 K, and heat is supplied from the dome at a rate

SIZIF [17.4K]

Answer:

$\epsilon_2=0.098$

Explanation:

Diameter $d=30cm=0.3m$

Temperature $T=600k$

Rate of supply $r=65W$

Emissivity of base surface $\in_b =0.55$

Temperature at base $T_b=400k$

Generally the equation for Area of base surface is mathematically given by

$A_b=\frac{\pi}{4}d^2$

$A_b=\frac{\pi}{4}0.3^2$

$A_b=0.0707m^2$

Generally the equation for Area of Hemispherical dome is mathematically given by

$A_h=\frac{\pi}{2}d^2$

$A_h=\frac{\pi}{2}0.3^2$

$A_h=0.1414m^2$

Since base is a flat surface

$F_{11}+F_{12}=1$

$F_{11}=0$

Therefore

$F_{12}=1$

$A_b=0.0707m^2$

Generally the equation for Net rate of radiation heat transfer between two surfaces is mathematically given by

$Q_{21}=-Q_{12}$

$Q_{21}=\frac{\sigma(T_1^4-T_2^4)}{\frac{1-\epsilon}{A_b\epsilon_1} +\frac{1}{A_bF_{12}} +\frac{1-\epsilon_2}{A_h*\epsilon_2} }$

Where

$\sigma=5.67*10^{-8}$

Therefore

$65=\frac{(5.67*10^{-8}(400^4-600^4))}{\frac{1-0.55}{0.0707*0.55}+\frac{1}{0.0707}+\frac{1-\epsilon_2}{0.1414*\epsilon_2}}$

$\epsilon_2=0.098$

$\epsilon_2 \approx 0.1$

Therefore the emissivity of the dome is

$\epsilon_2=0.098$

3 0

3 years ago