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3 years ago

Explain why your senses would lead you to believe that the Earth is center of the universe

Physics

1 answer:

ASHA 777 [7]3 years ago

5 0

<u>Answer:</u>

This is due to the apparent movement of the celestial bodies, seen from an observer centered on the Earth. If we did not have knowledge of the advances of science and the latest discoveries, we would think that the Earth is the center of the universe.

For example, for an observer on Earth as a reference system, the Sun moves every day from east to west, as do the planets and all the stars that we observe at night, which leads the observer to believe that those in motion are these bodies outside in the sky while the earth is still and fixed in the universe.

Keep in mind that this theory of geocentrism was widely accepted for a long time in antiquity, until the astronomer Nicolaus Copernicus proposed a completely different and opposite idea, heliocentrism (universe centered on the Sun).

However, at present it is known and accepted that the universe has no center, that the Earth revolves around the Sun and that in reality all the bodies of the universe are in movement.

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I am trying to find the magnitude of a resultant vector. Do i take inconsideration the negatives when i find the x & y compo

attashe74 [19]

Absolutely ! If you have two vectors with equal magnitudes and opposite
directions, then one of them is the negative of the other. Their correct
vector sum is zero, and that's exactly the magnitude of the resultant vector.

(Think of fifty football players pulling on each end of the rope in a tug-of-war.
Their forces are equal in magnitude but opposite in sign, and the flag that
hangs from the middle of the rope goes nowhere, because the resultant
force on it is zero.)

This gross, messy explanation is completely applicable when you're totaling up
the x-components or the y-components.

4 0

3 years ago

The compressor on an air conditioner draws 35.0 A when it starts up. If the start-up time is 0.59 s, how much charge passes a cr

Elden [556K]

Answer:

is a conditioner es 13.4

6 0

3 years ago

The x component of vector is -27.3 m and the y component is +43.6 m. (a) What is the magnitude of ? (b) What is the angle betwee

lara [203]

Answer:51.44 units

Explanation:

Given

x component of vector is $-27.3\hat{i}$

y component of vector is $43.6\hat{j}$

so position vector is

$r=-27.3\hat{i}+43.6\hat{j}$

Magnitude of vector is

$|r|=\sqrt{27.3^2+43.6^2}$

$|r|=\sqrt{2646.25}$

|r|=51.44 units

Direction

$tan\theta =\frac{43.6}{-27.3}=-1.597$

vector is in 2nd quadrant thus

$180-\theta =57.94$

$\theta =122.06^{\circ}$

4 0

3 years ago

When a firecracker explodes, what types of energy does it give off?

fomenos

The answer is c.
Sound, light and heat energy.

Hope this helped :)

7 0

3 years ago

A uniformly charged, one-dimensional rod of length L has total positive charge Q. Itsleft end is located at x = ????L and its ri

GREYUIT [131]

Answer:

$|\vec{F}| = \frac{1}{4\pi\epsilon_0}\frac{qQ}{L}(\ln(L+x_0)-\ln(x_0))$

Explanation:

The force on the point charge q exerted by the rod can be found by Coulomb's Law.

$\vec{F} = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}\^r$

Unfortunately, Coulomb's Law is valid for points charges only, and the rod is not a point charge.

In this case, we have to choose an infinitesimal portion on the rod, which is basically a point, and calculate the force exerted by this point, then integrate this small force (dF) over the entire rod.

We will choose an infinitesimal portion from a distance 'x' from the origin, and the length of this portion will be denoted as 'dx'. The charge of this small portion will be 'dq'.

Applying Coulomb's Law:

$d\vec{F} = \frac{1}{4\pi\epsilon_0}\frac{qdq}{x + x_0}(\^x)$

The direction of the force on 'q' is to the right, since both charges are positive, and they repel each other.

Now, we have to write 'dq' in term of the known quantities.

$\frac{Q}{L} = \frac{dq}{dx}\\dq = \frac{Qdx}{L}$

Now, substitute this into 'dF':

$d\vec{F} = \frac{1}{4\pi\epsilon_0}\frac{qQdx}{L(x+x_0)}(\^x)$

Now we can integrate dF over the rod.

$\vec{F} = \int{d\vec{F}} = \frac{1}{4\pi\epsilon_0}\frac{qQ}{L}\int\limits^{L}_0 {\frac{1}{x+x_0}} \, dx = \frac{1}{4\pi\epsilon_0}\frac{qQ}{L}(\ln(L+x_0)-\ln(x_0))(\^x)$

4 0

3 years ago