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Nuetrik [128]
2 years ago
9

If fire needs oxygen to burn, where does the sun get oxygen if there is no oxygen in space?​

Physics
2 answers:
Inessa [10]2 years ago
6 0

The sun is not a burning fire.

It's much much much hotter than that.

The sun's energy is the result of continuous nuclear fusion in it's core. We know how to do that on Earth, but the only thing we've been able to use it for so far is hydrogen bombs and other thermonuclear weapons.

elena55 [62]2 years ago
5 0

Answer:

through nuclear fusion

Explanation:

The burning of the sun is not chemical combustion. It is nuclear fusion. This combustion releases energy which we experience as the heat and light given off by the flame.

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In the voltage multiplier experiment, why not use a zener diode?​
Svet_ta [14]

Answer:

<em>A voltage multiplier is an electrical circuit that converts AC electrical power from a lower voltage to a higher DC voltage, typically using a network of capacitors and diodes.</em>

6 0
2 years ago
Has anyone done the Physical Science B- Forces and Motion Unit Test?? If so could you please help me
Sedbober [7]
I haven't, but I'm sure @Porshia has!
4 0
3 years ago
How far from Earth must a space probe be along a line toward the Sun so that the Sun's gravitational pull on the probe balances
Tatiana [17]

Answer:

258774.9441 m

Explanation:

x = Distance of probe from Earth

y = Distance of probe from Sun

Distance between Earth and Sun = x+y=149.6\times 10^6\ m

G = Gravitational constant

M_s = Mass of Sun = 1.989\times 10^{30}

M_e = Mass of Earth = 5.972\times 10^{24}\ kg

According to the question

\frac{GM_sm}{x^2}=\frac{GM_em}{y^2}\\\Rightarrow \frac{M_s}{x^2}=\frac{M_e}{y^2}\\\Rightarrow x=\sqrt{\frac{M_s\times y^2}{M_e}}\\\Rightarrow x=\sqrt{\frac{1.989\times 10^{30}\times y^2}{5.972\times 10^{24}}}\\\Rightarrow x=577.10852y

x+y=149.6\times 10^6\\\Rightarrow 577.10852y+y=149.6\times 10^6\\\Rightarrow 578.10852y=149.6\times 10^6\\\Rightarrow y=\frac{149.6\times 10^6}{578.10852}\\\Rightarrow y=258774.9441\ m

The probe should be 258774.9441 m from Earth

7 0
3 years ago
A golf ball is launched horizontally at a speed of 11 meters per second and a high of 6.4 m above the ground. How long will it t
Serjik [45]

The answer is A.

t = ( 2h / g )^1/2 = ( 2 x 6.4 / 9.8 )^1/2 = 1.14s

8 0
2 years ago
Consider a sealed 20 cm high electronic box whose base dimensions are 40cm x 40cm placed in a vacuum chamber. The emissivity of
Genrish500 [490]

Answer:

T_{surr}=296.289\ K

In Celsius:

T_{surr}=296.289-273\\T_{surr}=23.289^oC

Explanation:

The formula we are going to use is:

\dot Q_{rad}=\epsilon\sigma A_s(T_s^4-T_{surr}^4)

Where:

ε is the emissivity

σ is the Stefan constant

T_s is the final temperature of surrounding surfaces

T_{surr} is the required temperature

A_s is the are of surrounding surface

Calculating The area:

A_s=(0.4)(0.4)+4(0.4)(0.2)\\A_s=0.48\ m^2

σ= 5.67*10^{-8}\ W/m^2.K^4

ε =0.95

T_s=55+273

T_s=328 K

\dot Q_{rad=100 W

100=0.95(5.67*10^{-8})(0.48)(328^4-T_{surr}^4)\\3867693926=(328^4-T_{surr}^4)\\T_{surr}^4=7706623130\\T_{surr}=296.289\ K

In Celsius:

T_{surr}=296.289-273\\T_{surr}=23.289^oC

8 0
3 years ago
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