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azamat
2 years ago
15

What is the diameter of the earth? Explain how long it will take to measure the entire sun.

Physics
2 answers:
andreev551 [17]2 years ago
5 0

Answer:

Distance between earth and sun, around 150 million km, is defined as one Astronomical Unit (AU). The radius of the Sun, Run is around 700,000 km. The orbital speed of Earth is around 30 km/s.

Explanation:

jok3333 [9.3K]2 years ago
5 0
You know that the Earth goes around the Sun, once per year. Let's suppose that the path the Earth takes, its "orbit", is a circle. If the Earth's orbit is perfectly circular, then that means the Earth is always the same distance from the Sun. And if it's always the same distance away, this means that the Sun would always appear to have the same size in the sky. Another way to say this is that the Sun's angular diameter would appear to be constant.
Investigation:

How can we test this hypothesis? Start by grabbing four pictures of the Sun, taken a few months apart. (Ask yourself: Why four? Why not just any two, six months apart?) Load the four pictures into your browser, or your image processor (NIH Image, ImagePC, etc.), and LOOK. Is the Sun the same size in all four images?
There are a couple of ways to compare the relative sizes. Most straightforward is to make a ruler out of a piece of paper and hold it up to the computer screen; use the same ruler to measure all four images. If you have an image processor like NIH Image or ImagePC, another way is to put the cursor on the left- and then right-hand edges of the Sun, and then use the printout of the cursor position to count how many pixels across is the Sun.

More Technical: Also in an image processor like NIH Image or ImagePC, you can use the "Image Math" selection under the "Process" menu. Choose one image (say, January) as the 'standard' and then take turns subtracting the other images away from the 'standard.' What do you see?

Discussion:

If the Sun is the same diameter in every image, then the comparison of the four images should reveal only where sunspots have appeared/disappeared, but nothing significant about the Sun's edge. But if the solar diameter has changed in the months between when the images were taken, you should be able to detect this with your paper ruler. If you try the More Technical exercise, subtracting one image from another, then the result of the subtraction should show a ring, an annulus, that corresponds to the change in angular diameter. An example of this subtraction is provided at URL:
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3 years ago
A particle is confined to the x-axis between x = 0 and x = 1 nm. The potential energy U = 0 inside this region and U is infinite
DiKsa [7]

Answer:

Explanation:

According to heisenberg uncertainty Principle

Δx Δp ≥ h / 4π , where Δx  is uncertainty in position , Δp is uncertainty in momentum .

Given

Δx = 1 nm

Δp ≥ h /1nm x  4π

≥ 6.6 x 10⁻³⁴ / 10⁻⁹ x  4 π

≥  . 5254 x ⁻²⁵

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Two wires, both with current out of the page, are next to one another. The wire on the left has a current of 1 A and the wire on
Soloha48 [4]

Answer:

C. The left wire attracts the right wire and exerts as much force as the right wire does.

Explanation:

To know what is the answer you first take into account the magnetic field generated by each current, for a distance of d:

B_1=\frac{\mu_oI_1}{2\pi d}=\frac{\mu_o}{2\pi d}(1A)\\\\B_2=\frac{\mu_oI_2}{2\pi d}=\frac{\mu_o}{2\pi d}(2A)=2B_1\\\\

Next, you use the formula for the magnetic force produced by the wires:

\vec{F_B}=I\vec{L}\ X \vec{B}

if the direction of the L vector is in +k direction, the first wire produced a magnetic field with direction +y, that is, +j and the second wire produced magnetic field with direction -y, that is, -j (this because the direction of the magnetic field is obtained by suing the right hand rule). Hence, the direction of the magnetic force on each wire, produced by the other one is:

\vec{F_{B1}}=I_1L\hat{k}\ X\ B_2(-\hat{j})=I_1LB_2\hat{i}=(2A^2)\frac{L\mu_o}{2\pi d}\hat{i}\\\\\vec{F_{B2}}=I_2L\hat{k}\ X\ B_2(\hat{j})=I_2LB_1\hat{i}=-(2A^2)\frac{L\mu_o}{2\pi d}\hat{i}

Hence, due to this result you have that:

C. The left wire attracts the right wire and exerts as much force as the right wire does.

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