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sergeinik [125]
3 years ago
5

Suppose each object emitted a burst of light right now.Rank the objects from left to right based on the amount of time itwould t

ake this light to reach Earth, from longest time to shortesttime:_____.
-Star on the far side of the Andromeda Galaxy
-Star on the near side of Andromeda Galaxy
-Star on far side of Milky Way Galaxy
-Star near center of the Milky Way Galaxy
-Orion Nebulla
-Alpha Centauri
-Pluto
-Sun
Physics
1 answer:
Wittaler [7]3 years ago
4 0

Answer:

the order of arrival is from highest to lowest

star other side of Andromeda> star near side of andromeda> other side of milky way   > center of the milky way> nevulosa orion> Pluto> Sum

Explanation:

The light that comes from stars and galaxies travels in a vacuum so its speed is constant and with a value of c = 3 108 m / s, so time will be directly proportional to the distance to the object

             x = c t

the order of arrival is from highest to lowest

star other side of Andromeda> star near side of andromeda> other side of milky way   > center of the milky way> nevulosa orion> Pluto> Sum

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What step does the Rin PRICES stand for? Why is this step important?
posledela

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about 4 km

Explanation:

15 minutes is a quarter of an hour, so you divide 16km by 4 to get your answer

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2 years ago
A system gains 1500 J of heat, while the internal energy of the system increases by 4500 J and the volume decreases by . Assume
Assoli18 [71]

Answer:

Hence the pressure is 3\times 10^5 Pa

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Given data

Q=1500 J   system gains heat

ΔV=- 0.010 m^3     there is a decrease in volume

ΔU= 4500 J        internal energy decrease

We know work done is

W= Q- ΔU

=1500-4500= -3000 J

The change in the volume at constant pressure is

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there fore P = W/ΔV= -3000/-0.01= 3×10^5

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3 0
3 years ago
If a proton and an electron are released when they are 7.00×10−10 m apart (typical atomic distances), find the initial accelerat
Ugo [173]

Answer:

The acceleration of the proton is 2.823 x 10¹⁷ m/s²

The acceleration of the electron is 5.175 x 10²⁰ m/s²

Explanation:

Given;

distance between the electron and proton, r = 7 x 10⁻¹⁰ m

mass of proton, m_p = 1.67 x 10⁻²⁷ kg

mass of electron, m_e = 9.11 x 10⁻³¹ kg

The attractive force between the two charges is given by Coulomb's law;

F = \frac{k(q_p)(q_e)}{r^2}

where;

k is Coulomb's constant = 9 x 10⁹ Nm²/c²

F = \frac{k(q_p)(q_e)}{r^2} \\\\F = \frac{(9*10^9)(1.602*10^{-19})(1.602*10^{-19})}{(7*10^{-10})^2} \\\\F = 4.714 *10^{-10} \ N

Acceleration of proton is given by;

F = ma

F = m_pa_p\\\\a_p = \frac{F}{m_p}\\\\a_p = \frac{4.714*10^{-10}}{1.67*10^{-27}}\\\\a_p = 2.823 *10^{17} \ m/s^2

Acceleration of the electron is given by;

F = m_ea_e\\\\a_e = \frac{F}{m_e}\\\\a_e = \frac{4.714*10^{-10}}{9.11*10^{-31}}\\\\a_e = 5.175 *10^{20} \ m/s^2

7 0
3 years ago
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