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Snezhnost [94]
3 years ago
12

A distant galaxy is determined to be 150 million light years distant and moving away from us; using the Hubble law determine its

velocity in terms
Physics
1 answer:
dlinn [17]3 years ago
4 0

Your question kind of petered out there towards the end and you didn't specify
the terms, so I'll pick my own.

The "Hubble Constant" hasn't yet been pinned down precisely, so let's pick a
round number that's in the neighborhood of the last 20 years of measurements:

             <em>70 km per second per megaparsec</em>.

We'll also need to know that 1 parsec = about 3.262 light years.

So the speed of your receding galaxy is

         (Distance in LY) x (1 megaparsec / 3,262,000 LY) x (70 km/sec-mpsc) =

              (150 million) x  (1 / 3,262,000) x (70 km/sec) =

                                 <em>3,219 km/sec  </em>in the direction away from us (rounded)

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Find the GCF 12, 18, and 84
ahrayia [7]
To find the answer, plot down the factors for every number.

12:   1, 2 ,3 ,4, 6, 12

18:   1, 2, 3, 6, 9, 18

84:  1, 2, 3, 4, 6, 7, 12

If you noticed, the number that was common to the 3 numbers, were 1, 2, 3, and 6
And 6 is the bigger number
So 6 is your GCF
5 0
3 years ago
Read 2 more answers
Space-faring astronauts cannot use standard weight scales (since they are constantly in free fall) so instead they determine the
valentinak56 [21]

Answer:

ma = 48.48kg

Explanation:

To find the mass of the astronaut, you first calculate the mass of the chair by using the information about the period of oscillation of the empty chair and the spring constant. You use the following formula:

T=2\pi\sqrt{\frac{m_c}{k}}     (1)

mc: mass of the chair

k: spring constant = 600N/m

T: period of oscillation of the chair = 0.9s

You solve the equation (1) for mc, and then you replace the values of the other parameters:

m_c=\frac{T^2k}{4\pi^2}=\frac{(0.9s)^2(600N/m)}{4\pi^2}=12.31kg    (2)

Next, you calculate the mass of the chair and astronaut by using the information about the period of the chair when the astronaut is sitting on the chair:

T': period of chair when the astronaut is sitting = 2.0s

M: mass of the astronaut plus mass of the chair = ?

T'=2\pi\sqrt{\frac{M}{k}}\\\\M=\frac{T'^2k}{4\pi^2}=\frac{(2.0s)^2(600N/m)}{4\pi^2}\\\\M=60.79kg (3)

Finally, the mass of the astronaut is the difference between M and mc (results from (2) and (3)) :

m_a=M-m_c=60.79kg-12.31kg=48.48kg

The mass of the astronaut is 48.48 kg

3 0
3 years ago
Methods to reduce the frictional force between the an object and surface which it is in contact.
telo118 [61]
Use of lubricant
Use of ball bearers
Use of streamlined body
Use of graphite
7 0
3 years ago
What is the (magnitude of the) centripetal acceleration (as a multiple of g=9.8~\mathrm{m/s^2}g=9.8 m/s ​2 ​​ ) towards the Eart
Wittaler [7]

Answer:

The centripetal acceleration as a multiple of g=9.8 m/s^{2} is 1.020x10^{-3}m/s^{2}

Explanation:

The centripetal acceleration is defined as:

a = \frac{v^{2}}{r}  (1)

Where v is the velocity and r is the radius

Since the person is standing in the Earth surfaces, their velocity will be the same of the Earth. That one can be determined by means of the orbital velocity:

v = \frac{2 \pi r}{T}  (2)

Where r is the radius and T is the period.

For this case the person is standing at a latitude 71.9^{\circ}. Remember that the latitude is given from the equator. The configuration of this system is shown in the image below.

It is necessary to use the radius at the latitude given. That radius can be found by means of trigonometric.

\cos \theta = \frac{adjacent}{hypotenuse}

\cos \theta = \frac{r_{71.9^{\circ}}}{r_{e}} (3)

Where r_{71.9^{\circ}} is the radius at the latitude of 71.9^{\circ} and r_{e} is the radius at the equator (6.37x10^{6}m).

r_{71.9^{\circ}} can be isolated from equation 3:

r_{71.9^{\circ}} = r_{e} \cos \theta  (4)

r_{71.9^{\circ}} = (6.37x10^{6}m) \cos (71.9^{\circ})

r_{71.9^{\circ}} = 1.97x10^{6} m

Then, equation 2 can be used

v = \frac{2 \pi (1.97x10^{6} m)}{24h}

Notice that the period is the time that the Earth takes to give a complete revolution (24 hours), this period will be expressed in seconds for a better representation of the velocity.

T = 24h . \frac{3600s}{1h} ⇒ 84600s

v = \frac{2 \pi (1.97x10^{6} m)}{84600s}

v = 146.31m/s

Finally, equation 1 can be used:

a = \frac{(146.31m/s)^{2}}{(1.97x10^{6}m)}

a = 0.010m/s^{2}

Hence, the centripetal acceleration is 0.010m/s^{2}

To given the centripetal acceleration as a multiple of g=9.8 m/s^{2}​ it is gotten:

\frac{0.010m/s^{2}}{9.8 m/s^{2}} = 1.020x10^{-3}m/s^{2}

6 0
3 years ago
What is the weight of a column of air with cross-sectional area 4. 5 m^2 extending from earth's surface to the top of the atmosp
sineoko [7]

The weight of a column of air with cross-sectional area 4. 5 m^2 extending from earth's surface to the top of the atmosphere is, 4.56*10^5N.

To find the answer, we have to know about the pressure.

<h3>How to find the weight of a column of air?</h3>
  • As we know that the expression of pressure as,

                 P=\frac{F}{A}

where; F is the force, here it is equal to the weight of the air column, and A is the area of cross section.

  • It is given that, the air column is extending from earth's surface to the top of the atmosphere, thus, the pressure will be atmospheric pressure,

             P=1atm=1.013*10^5Pascals

  • From this, the value of weight will be,

            F=mg=P*A=1.013*10^5*4.5=4.56*10^5N

Thus, we can conclude that, the weight of a column of air with cross-sectional area 4. 5 m^2 extending from earth's surface to the top of the atmosphere is, 4.56*10^5N.

Learn more about the pressure here:

brainly.com/question/12830237

#SPJ4

4 0
1 year ago
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