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Firlakuza [10]
2 years ago
7

Why do some quasars have red shifts greater than 1?.

Physics
1 answer:
slega [8]2 years ago
8 0

Answer:

They are very distant.

Explanation:

With relativistic red shifts that take into account dilation of space-time, as Einstein predicted.

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A student uses the circuit shown to determine the resistance of two identical resistors.
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Answer:

B. 0.552

Explanation:

To find the resistance in the circuit above, u simply divide the current in the circuit by the voltage to get the resistance.

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3 years ago
Consider the same 70kg/686N student on the surface on another planet from the table above: Jupiter. Tompared to the gravitationa
Yuliya22 [10]

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Dont mind me

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3 years ago
These graphs show measurements of the density of air as different sound waves pass a single point.
Ymorist [56]

Answer: D!

It is the option with the greatest amplitude.

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3 years ago
Read 2 more answers
One isotope of bromine has an atomic mass of 78.92amu and a relative abundance of 50.69%. The other major isotope of bromine has
castortr0y [4]

Answer:

The average atomic mass is 79.91 amu.

Explanation:

Since  

Atomic mass can be find by Multiplying the relative abundance of each isotope by its atomic mass, then add them together to get the atomic mass of the element.

so

Atomic  mass = (0.5069)(78.92 amu) + (0.4931)(80.92 amu)

                       =79.91 amu

So the Atomic mass of the  bromine is 79.91amu.

8 0
3 years ago
Positive Charge Q is distributed uniformly along the x-axis from x=0 to x=a. A positive point charge q is located on the positiv
deff fn [24]

Answer:

 electric field E = - k Q (1 /r(r-a)), force    F = - k Q qo / r (r-a) and force for r>>a    F ≈ - k Q qo / r²

Explanation:

You are asked to find the electric field of a continuous charge distribution, so we must use the equation

       

           E = k ∫dp /r²

Where k is the Coulomb constant that is worth 8.99 10⁹ N m² / C², r is the distance between the load distribution and the test charge, in this case everything is on the X axis.

We must find the charge differential (dq), let's use that uniformly distributed and create a linear charge density

          λ = q / x

As it is constant, we can write it based on differentials

         λ = dq / dx

         dq = λ dx

We already have all the terms, let's  integrate enter its limits, lower the distance from the left end of the distribution to the test charge (x = r) and the upper limit that is the distance from the left end of distribution to the test load ( x = r - a) where r> a

         E = k ∫ λ dx / x²

         E = k la (- 1 / x)

Let's get the negative sign from the parentheses

         E = - k λ (1 / x)

         E = - k λ (1 /(r-a)  -1 /r) = - k λ [a / r (r-a)]

Let's change the charge density with the value of the total charge λ = Q / a

         E = - k Q/a  [a / r (r-a)]

         E = - k Q (1 /r(r-a))

b) We calculate the force.  

         F = E qo

         F = - k Q qo / r (r-a)

c) the force for charge porbe very far r >> a. In this case we can take r from the parentheses and neglect (a/r)

         F = - k Qqo / r² (1 -  a/r)

         F ≈ - k Q qo / r²

6 0
3 years ago
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