If this decay has a half-life of 10.2 years, what mass of 60.8g carbon-14 will remain after 20.4 years

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

Alinara [238K]

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

8 0

3 years ago

Suppose you are driving a car, and a cup of coffee is on the seat beside you. Choose the frame with respect to which the cup is

Reil [10]

We want to find the frame of reference (viewpoint) where the cup of coffee inside your car moves the fastest.

The correct option will be A, the astronaut in the ISS.

---------------------------------

Let's see how to get the correct option, first, remember that the frame of reference means the "viewpoint" that you are taking. So we need to find the viewpoint where the cup of coffee moves faster.

Now just let's analyze all the given options to see in which one the cup moves fastest.

A) Here the astronaut sees the car moving, so here the cup has the speed of the car, and the astronaut is also moving. Remember that the International Space Station orbits with a speed of 7.6km/s (this is really fast). So in the frame of reference of the astronaut, he/she is at a stop, so this velocity is assigned to all the other objects. Thus, he sees the cup of coffee moving really fast.

B) In it's own frame of reference, the cup does not move.

C) The observer will see the car moving, and the cup is inside the car, so in this frame of reference the velocity of the cup is the same as the one of the car.

D) The incoming car has a given velocity V', from this frame, the car with the cup of coffee will have its own velocity plus the V' of the incoming car, so from this frame the velocity of the cup of coffee is larger than the velocity of the car with the cup.

E) You are driving the car with the cup of coffee, so the cup of coffee is always next to you, so in this frame, the cup does not move.

Concluding, is easy to see that the frame of reference where the cup moves the fastest is in the first one, for the nature of the astronaut's frame of reference velocity, he/she will see the cup moving the fastest.

If you want to learn more, you can read:

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8 0

2 years ago

A Ferris wheel with a diameter of 35 m starts from rest and achieves its maximum operational tangential speed of 2.3 m/s in a ti

sveticcg [70]

First calculate the radius 35/2=17.5m

5 0

3 years ago

Which describes the molecular motion of a solid nickel marble rolled across the floor?

Gemiola [76]

I can't decide between A and B, but B seems more likely to me. Even though the molecules don't look like they're moving, the area of contact is slightly more compressed.

7 0

4 years ago

Please help!!! I have a physics exam tomorrow and I just can't wrap my head around this one!

Ainat [17]

The total electrostatic force on charge A is $28 \mu N$

Explanation:

The magnitude of the electrostatic force between two charges is given by Coulomb's law:

$F=k\frac{q_1 q_2}{r^2}$

where:

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q_1, q_2$ are the two charges

r is the separation between the two charges

Here we have three positively charged particles A,B and C, located at the following positions:

$x_A = 0\\x_B = 10 m\\x_C = 20 m$

The magnitudes of the three charges are:

$q_A = q_B = q_C = 0.5 \mu C = 0.5\cdot 10^{-6}C$

The force exerted by B on A is to the left (because the force between two positive charges is repulsive), and the force exerted by C on A is also to the left (also repulsive). Therefore, the net force on A is just the sum of the two forces exerted by charges B and C:

$F_A = F_{BA} + F_{CA} = k\frac{q_B q_A}{(x_B-x_A)^2}+k\frac{q_C q_A}{(x_C-x_A)^2}=\\=(8.99\cdot 10^9) \frac{(0.5\cdot 10^{-6})^2}{(10)^2}+(8.99\cdot 10^9) \frac{(0.5\cdot 10^{-6})^2}{(20)^2}=2.8\cdot 10^{-5} N = 28 \mu N$

Learn more about electric force:

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#LearnwithBrainly

3 0

3 years ago