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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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PHYSICS 50 POINTS PLEASE HELP
tangare [24]

Answer:

One way to look at Newton’s three laws of motion is this:

The third law states what forces are. That is, all forces are interactions between two different objects. If one object is interacting with another, then equal and opposite forces act on each object. So no force acts alone. When you exert a force on something, it is exerting the identical force back on you.

The first and second laws deal with the consequences of the forces that act on an object. The first law says that in the absence of a net force on an object, it simply continues doing whatever it was already doing. If it is at rest, it will remain at rest. If it is in motion, it will continue with that same motion - at constant speed and in the direction it was already traveling.

The second law says what happens if there is a net force on the object. In that case, the object accelerates - either by changing its speed, its direction, or both - in proportion and in the direction of the net force that acts on it. The amount of acceleration depends the object’s mass. That is, the larger the mass the smaller the acceleration for a given net force. The first and second laws can be summarized in the mathematical expression

F = ma

where F is the vector sum of all the forces that act on the object at any given moment (i.e., the net force), m is the mass of the object, and a is the acceleration of the object due to the net force at that moment - and is always in the same direction of the net force.

And notice that in a way, the first law is then “contained” within the second. That is, if the net force is zero on an object, then so is the acceleration. That is, either the object is (still) at rest or, if already in motion, the velocity didn’t change, in either case, the acceleration was zero.

Explanation:

4 0
2 years ago
Read 2 more answers
A sealed tank containing seawater to a height of 10.5 mm also contains air above the water at a gauge pressure of 2.95 atmatm. W
weqwewe [10]

Answer:

The water is flowing at the rate of 28.04 m/s.

Explanation:

Given;

Height of sea water, z₁ = 10.5 m

gauge pressure, P_{gauge \ pressure} = 2.95 atm

Atmospheric pressure, P_{atm} = 101325 Pa

To determine the speed of the water, apply Bernoulli's equation;

P_1 + \rho gz_1 + \frac{1}{2}\rho v_1^2 = P_2 + \rho gz_2 + \frac{1}{2}\rho v_2^2

where;

P₁ = P_{gauge \ pressure} + P_{atm \ pressure}

P₂ = P_{atm}

v₁ = 0

z₂ = 0

Substitute in these values and the Bernoulli's equation will reduce to;

P_1 + \rho gz_1 + \frac{1}{2}\rho v_1^2 =  P_2 + \rho gz_2 + \frac{1}{2}\rho v_2^2\\\\P_1 + \rho gz_1 + \frac{1}{2}\rho (0)^2 =  P_2 + \rho g(0) + \frac{1}{2}\rho v_2^2\\\\P_1 + \rho gz_1 =  P_2 + \frac{1}{2}\rho v_2^2\\\\P_{gauge} + P_{atm} + \rho gz_1 = P_{atm} + \frac{1}{2}\rho v_2^2\\\\P_{gauge} +  \rho gz_1 =  \frac{1}{2}\rho v_2^2\\\\v_2^2 = \frac{2(P_{gauge} +  \rho gz_1)}{\rho} \\\\v_2 = \sqrt{ \frac{2(P_{gauge} +  \rho gz_1)}{\rho} }

where;

\rho is the density of seawater = 1030 kg/m³

v_2 = \sqrt{ \frac{2(2.95*101325 \ + \  1030*9.8*10.5 )}{1030} }\\\\v_2 = 28.04 \ m/s

Therefore, the water is flowing at the rate of 28.04 m/s.

7 0
3 years ago
The most common units for expressing the density of a substance are the g/cm3.
Luden [163]
True.

Density = mass / volume,  Unit = g / cm³.

This is a common unit because of its affiliation with the SI unit and because that also our popular liquid which is water = 1 g/cm³  
6 0
3 years ago
In a choir practice room, two parallel walls are 5.70 m apart. The singers stand against the north wall. The organist faces the
ololo11 [35]

Answer:

4.98 m

Explanation:

Given that

Width of the mirror, d = 0.6 m

Organist distance to the mirror, s = 0.78 m

Distance between the singer and the organist, S = 5.7 + 0.78 = 6.48 m

Width of north wall, D?

Using the simple relationship

D/S = d/s, on rearranging

D = dS /s

D = (0.6 * 6.48) / 0.78

D = 3.888 / 0.78

D = 4.98 m

Therefore, we can conclude that the Width of north wall is 4.98 m

7 0
2 years ago
A 1-megabit computer memory chip contains many 27 fF capacitors. Each capacitor has a plate area of 3.09 × 10−11 m 2 . Determine
valentina_108 [34]

Answer:

Plate separation of each capacitor is 101.132 °A

Explanation:

The formula to calculate the capacitance in empty space as a function of distance (square parallel plates) is:

C=\epsilon_{0}\frac{Area}{distance}

clearing for distance:

distance=\epsilon_0 \frac{Area}{Capacitance} \\\\\epsilon_0=8.8542(10)^{-12}C^2/Nm\\\\Area=3.09(10)^{-11}m^2\\Capacitance=27(10)^{-15}F\\\\Replacing\\\\distance=\frac{8.8542(10)^{-12}*3.09(10)^{-11}}{27(10)^{-15}} =1.0133(10)^{-8}m\\\\In A\\distance= 101.132 A

8 0
3 years ago
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