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sammy [17]
3 years ago
12

What is the approximate number of galaxies in the known universe, assuming that the milky way has an average mass and that all o

f the mass in the known universe is in galaxies? the mass of the milky way is on the order of 1042 kg, and the mass of the known universe is about about 1053 kg?
Physics
1 answer:
igor_vitrenko [27]3 years ago
5 0
I think that the correct reporting of the given values are 10⁴² kg for the Milky Way and 10⁵³ kg for the whole universe. That would be logical in that case. Since it is mentioned that each galaxy has the same mass as the Milky Way, then the number of galaxies in the universe are:

Number of galaxies = 10⁵³ kg/10²⁴ kg = 10¹¹ or 100 trillion galaxies 
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An object of mass m travels along the parabola yequalsx squared with a constant speed of 5 ​units/sec. What is the force on the
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Explanation:

The object is moving along the parabola y = x² and is at the point (√2, 2).  Because the object is changing directions, it has a centripetal acceleration towards the center of the circle of curvature.

First, we need to find the radius of curvature.  This is given by the equation:

R = [1 + (y')²]^(³/₂) / |y"|

y' = 2x and y" = 2:

R = [1 + (2x)²]^(³/₂) / |2|

R = (1 + 4x²)^(³/₂) / 2

At x = √2:

R = (1 + 4(√2)²)^(³/₂) / 2

R = (9)^(³/₂) / 2

R = 27 / 2

R = 13.5

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F = m v² / r

F = m (5)² / 13.5

F = 1.85 m

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3 years ago
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castortr0y [4]
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3 years ago
The electric force between two charged objects of charge +Q0 that are separated by a distance R0 is F0. The charge of one of the
vovikov84 [41]

Answer:

F=3F_o(\frac{R_o}{R_{o2}})^2

Explanation:

This problem is approached using Coulomb's law of electrostatic attraction which states that the force F of attraction or repulsion between two point charges, Q_1 and Q_2 is directly proportional to the product of the charges and inversely proportional to the square of their distance of separation R.

F=\frac{kQ_1Q_2}{R^2}..................(1)

where k is the electrostatic constant.

We can make k the subject of formula  as follows;

k=\frac{FR^2}{Q_1Q_2}...........(2)

Since k is a constant, equation (2) implies that the ratio of the product of the of the force and the distance between two charges to the product of charges is a constant. Hence if we alter the charges or their distance of separation and take the same ratio as stated in equation(2) we will get the same result, which is k.

According to the problem, one of the two identical charges was altered from Q_o to 3Q_o and their distance of separation from R_o to R_{o2}, this also made the force between them to change from F_o to F_{o2}. Therefore as stated by equation (2), we can write the following;

\frac{F_oR_o^2}{Q_o*Q_o}=\frac{F_{o2}R_{o2}^2}{3Q_o*Q_o}.............(3)

Therefore;

\frac{F_oR_o^2}{Q_o^2}=\frac{F_{o2}R_{o2}^2}{3Q_o^2}.............(4)

From equation (4) we now make the new force F_{o2} the subject of formula as follows;

{F_oR_o^2}*{3Q_o^2}=F_{o2}R_{o2}^2*{Q_o^2}

Q_o then cancels out from both side of the equation, hence we obtain the following;

3{F_oR_o^2}=F_{o2}R_{o2}^2.............(4)

From equation (4) we can now write the following;

F_{o2}=\frac{3F_oR_o^2}{R_{o2}^2}

This could also be expressed as follows;

F_{o2}=3F_o(\frac{R_o}{R_{o2}})^2

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Reason. Evaporation is a process in which molecules from surface of liquid convert into gas.

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