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-Dominant- [34]
2 years ago
13

If the sun is 400 times bigger than the moon, how couild the moon possibly cover the sun during a solar eclipse?​

Physics
2 answers:
Vilka [71]2 years ago
6 0

Answer:

It's all about perspective. The moon is far closer to the Earth than the Sun is, so they appear roughly the same size. If they were closer to each other, then obviously the moon wouldn't be large enough to cover a substantial amount of the sun's light. But given the huge distance both between them and the moon and the earth, to us they look relatively the same.

AlekseyPX2 years ago
3 0

Explanation:

the Moon passes between Earth and the Sun Even though the Moon is much smaller than the Sun, because it is just the right distance away from Earth, the Moon can fully block the Sun's light from Earth's perspective This completely blocks out the Sun's light

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A sample of n2 gas occupies a volume of 746 ml at stp. What volume would n2 gas occupy at 155 ◦c at a pressure of 368 torr?
musickatia [10]

Answer:

2.41 L

Explanation:

We can solve the problem by using the ideal gas equation, which can be rewritten as:

\frac{p_1 V_1}{T_1}=\frac{p_2 V_2}{T_2}

where we have:

p_1 = 1.01\cdot 10^5 Pa (initial pressure is stp pressure)

V_1 = 746 mL = 0.746 L = 7.46\cdot 10^{-4}m^3 is the initial volume

T_1 = 0^{\circ}=273 K is the initial temperature (stp temperature)

p_2 = 368 torr = 4.9\cdot 10^4 Pa is the final pressure

V_2 = ? is the final volume

T=155^{\circ}=428 K is the final temperature

By substituting the numbers inside the formula and solving for V2, we find the final volume:

V_2 = \frac{p_1 V_1 T_2}{T_1 p_2}=\frac{(1.01\cdot 10^5 Pa)(7.46\cdot 10^{-4} m^3)(428 K)}{(273 K)(4.9\cdot 10^4 Pa)}=2.41\cdot 10^{-3} m^3

which corresponds to 2.41 L.

7 0
3 years ago
A satellite is launched to orbit the Earth at an altitude of 3.25 107 m for use in the Global Positioning System (GPS). Take the
Korolek [52]

Answer:Orbital period =21.22hrs

Explanation:

given that

mass of earth M = 5.97 x 10^24 kg

radius of a satellite's orbit, R=  earth's radius + height of the satellite

6.38X 10^6 +  3.25 X10^7 m =3.89 X 10^7m

Speed of satellite, v= \sqrt GM/R

where G = 6.673 x 10-11 N m2/kg2

V= \sqrt (6.673x10^-11 x 5.97x10^ 24)/(3.89 X 10^ 7m)

V =10,241082.2

v= 3,200.2m/s

a) Orbital period

\sqrt GM/R = \frac{2\pi r}{T}

V= \frac{2\pi r}{T}

T= 2 \pi r/ V

= 2 X 3.142 X 3.89 X 10^7m/ 3,200.2m/s

=76,385.1 s

60 sec= 1min

60mins = 1hr

76,385.1s =hr

76,385.1/3600=21.22hrs

3 0
3 years ago
How would you differentiate the theoretical and actual velocities of PE and KE in falling objects?
Solnce55 [7]

Answer: you've been trolled

Explanation:

8 0
3 years ago
"the time range specified for a historical search defines the" _______________ .
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The answer is period. This is the date range for which you ask the system to return the events depending on the search keys, and keywords you have entered.

5 0
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________ energy is the energy something contains due to its position above the Earth.
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d. Gravitational potential energy is the energy something contains due to its position above the Earth.

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