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bixtya [17]
3 years ago
13

ou are out stargazing with your 13.4-cm telescope. You point your telescope at an interesting formation in the sky, which you th

ink is a binary star system. A binary star system consists of two stars in orbit around each other. You guess that the average wavelength coming from the stars is 518 nm. What is the minimum angular separation between the two stars required for your telescope to resolve the two stars of the binary system
Physics
1 answer:
Alinara [238K]3 years ago
8 0

Answer:

θ = 4.716 10⁻⁶ rad

Explanation:

In order for the releases to be considered separate, they must meet the Rayleigh criterion that establishes that the maximum diffraction of one star must coincide with the first minimum of the diffraction pattern of the second star.

We use the diffraction equation for a slit

            a sin θ = m λ

The minimum occurs at m = 1

             sin θ = λ / a

Since the angles in these systems are very small, we can approximate the sine to its angle in radians

             θ = λ / a

The telescope has a circular aperture whereby polar cords should be used, which introduces a constant number

           θ = 1.22 λ / a

Let's calculate

          θ = 1.22 518 10⁻⁹ / 13.4 10⁻²

          θ = 4.716 10⁻⁶ rad

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One ring of radius a is uniformly charged with charge +Q and is placed so its axis is the x-axis. A second ring with charge –Q i
kati45 [8]

Answer:

The force exerted on an electron is 7.2\times10^{-18}\ N

Explanation:

Given that,

Charge = 3 μC

Radius a=1 m

Distance  = 5 m

We need to calculate the electric field at any point on the axis of a charged ring

Using formula of electric field

E=\dfrac{kqx}{(a^2+x^2)^{\frac{3}{2}}}\hat{x}

E_{1}=\dfrac{kqx}{(a^2+x^2)^{\frac{3}{2}}}\hat{x}

Put the value into the formula

E_{1}=\dfrac{9\times10^{9}\times3\times10^{-6}\times5}{(1^2+5^2)^{\frac{3}{2}}}

E_{1}=1.0183\times10^{3}\ N/C

Using formula of electric field again

E_{2}=\dfrac{kqx}{(a^2+x^2)^{\frac{3}{2}}}\hat{x}

Put the value into the formula

E_{2}=\dfrac{9\times10^{9}\times(-3\times10^{-6})\times5}{((0.5)^2+5^2)^{\frac{3}{2}}}

E_{2}=-1.064\times10^{3}\ N/C

We need to calculate the resultant electric field

Using formula of electric field

E=E_{1}+E_{2}

Put the value into the formula

E=1.0183\times10^{3}-1.064\times10^{3}

E=-0.045\times10^{3}\ N/C

We need to calculate the force exerted on an electron

Using formula of electric field

E = \dfrac{F}{q}

F=E\times q

Put the value into the formula

F=-0.045\times10^{3}\times(-1.6\times10^{-19})

F=7.2\times10^{-18}\ N

Hence, The force exerted on an electron is 7.2\times10^{-18}\ N

8 0
3 years ago
A force of 50 newtons causes a sled to accelerate at a rate of 5 meters per second. What is the mass of the sled.
notka56 [123]
F=ma
50=m(5)
m=10kg
hence,ans is B
5 0
3 years ago
Read 2 more answers
What force is necessary to accelerate a 5.0 kg mass from rest to a final velocity of 10.0 m/s in 5.0 s?
vesna_86 [32]

Answer:

10 N

Explanation:

F = ma = m(Δv/t) = 5.0(10.0 - 0)/5.0 = 10 N

4 0
3 years ago
If two people standing on a scooter board push off of each other what happens ( Newton’s 3rd law )
Keith_Richards [23]

Answer:They will go off in opposite directions with the same force.

Explanation:

Newton's Third Law states that for every action, there is an equal but opposite reaction. If two people are standing on a scooter and push off each other (following the Law), it should come to the conclusion that they could go off in opposite with the same force.

7 0
3 years ago
A circle has an initial radius of 50ft when the radius begins decreasing at a rate of 2ft/s. what is the rate of change of the a
valkas [14]
The area of the circle with radius r is
A = πr²

The rate of change of area with respect to time is
\frac{dA}{dt} = \frac{dA}{dr} . \frac{dr}{dt} =2 \pi r. \frac{dr}{dt}

The rate of change of the radius is given as
\frac{dr}{dt} =-2 \,  \frac{ft}{s}
Therefore
\frac{dA}{dt} =-4 \pi r \,  \frac{ft^{2}}{s}

When r = 10 ft, obtain
\frac{dA}{dt}|_{r=10 \, ft} = -40 \pi  \,  \frac{ft^{2}}{s}

Answer: - 40π ft²/s (or - 127.5 ft²/s)
7 0
3 years ago
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