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Klio2033 [76]
3 years ago
9

How do you know the earth is rotating on its axis

Physics
1 answer:
cupoosta [38]3 years ago
4 0

Answer:

Read below!

Explanation:

You can watch the sun wheel across the sky during the day, and the stars at night. Focus a telescope on any star besides the north star--especially southern stars--and you can watch them drift across your field of view.  

An alternative explanation is that all the stars are painted on (or holes in) some canopy that rotates around the earth. This explanation does not account for the motion of the "wanderers," or planets, as the Greeks called them, or for the path of the moon among the stars.  

As we know the stars are massive bodies of significant and varying distance to the earth, the notion they all swing around us in unison seems highly implausible

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JulijaS [17]

Part a)

Flow rate is defined as rate of volume flow

it is determined by

Q = \frac{dV}{dt}

now if the radius of pipe is reduced then we assume here that liquid flow is ideal flow here and there is no change in the density of liquid.

So here we know that since mass is always conserved

so

\frac{dm}{dt}_{in} = \frac{dm}{dt}_{out}

so we have

\rho\frac{dV}{dt}_{in} = \rho\frac{dV}{dt}_{out}

\frac{dV}{dt}_{in} = \frac{dV}{dt}_{out}

now we can say from above equation that there is no effect on the flow rate is we change the radius of pipe

Part b)

now in order to find the speed of flow'

\frac{dV}{dt}_{in} = \frac{dV}{dt}_{out}

A_{in}v_{in} = A_{out}v_{out}

\pi r^2 v_{in} = \pi (\frac{r}{n})^2 v_{out}

v_{in} = \frac{v_{out}}{n^2}

so final speed will be

v_{out} = n^2 v_{in}

here we have n = 3

v_{out} = 9* v_{in}

so flow speed will be 9 times more than initial speed

6 0
3 years ago
A space station consists of two donut-shaped living chambers, A and B, that have the radii shown in the drawing. As the station
tensa zangetsu [6.8K]
The angle subtended by the arc of movement of both chambers is the same. Thus:
S = r∅, for chamber A:

2.3 x 10² = 3.2 x 10² ∅
∅ = 0.72 rad

∅ is the same for B so:
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6 0
3 years ago
How do carbon 12 and carbon 13 differ?
astraxan [27]

Answer:

A) & B)

Explanation:

First, the numbers 12 and 13 represent the <em>atomic mass number</em> of the atoms.

  • So, A) is true: thus Carbon12 and Carbon 13 have different mass numbers.

The <em>mass number</em> is equal to the total number of protons and neutrons. Consider that any element has the same number of protons, regardless of the number of neutrons. The number of protons in Carbon is 6.

The amount of neutrons can be calculated by: <em>mass number </em>minus <em>number of protons.</em>

For Carbon 12: 12-6=6

For Carbon 13: 13-6=7

  • B) is true, we just proved they have different amounts of neutrons.

In order for the charge of the atom to be neutral, the amount of electrons must be equal to the number of protons (as they have opposite charges). And we now know that the number of protons in Carbon12 and Carbon13 are always the same

  • C) is false, the number of electrons is the same in both atoms

The atomic charges in both are neutral, due to the fact that they have the same amount of protons and electrons in both cases. Is only the neutrons (thus the mass numbers)that change

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4 0
3 years ago
The first and second coils have the same length, and the third and fourth coils have the same length. They differ only in the cr
stealth61 [152]

Answer:

\frac{R_2}{R_1}=\frac{A_1}{A_2}\\\frac{R_4}{R_3}=\frac{A_3}{A_4}

Explanation:

The resistance of a conductor is directly proportional to its length and is inversely proportional to its cross-sectional area, this dependence is given by:

R=\frac{\rho L}{A}

\rho is the material's resistance, L is the legth and A is the cross-sectional area.

For the first and second coils, we have:

R_1=\frac{\rho L}{A_1}\\R_2=\frac{\rho L}{A_2}\\\rho L=R_1A_1\\\rho L=R_2A_2\\R_1A_1=R_2A_2\\\frac{R_2}{R_1}=\frac{A_1}{A_2}

For the third and fourth coils, we have:

R_3=\frac{\rho L'}{A_3}\\R_4=\frac{\rho L'}{A_4}\\\rho L'=R_3A_3\\\rho L'=R_4A_4\\R_3A_3=R_4A_4\\\frac{R_4}{R_3}=\frac{A_3}{A_4}

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