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3 years ago

Which of these best describes how an appropriate star chart is selected to locate objects in the sky?

Physics

2 answers:

lisov135 [29]3 years ago

7 0

Answer:

match the day and time of the year and location

Explanation:

The stars that we see on the sky change with the station of the year, they change their position depending of the day of the year and the location, mainly the latitude since on that depends what "side" of the earth the sky is facing towards, at certain points the north hemisphere sees a different sky than the south, but the east and west as long as they are in the same latitude do see the same sky.

Sergio [31]3 years ago

6 0

The layout of the stars in the sky is determined by the date, time of night, and your location (mainly latitude). So to pick the best star chart, you should go with the one that's closest to the present date and your location, then make allowance for what time it is. Everything in the sky moves about a degree every 4 minutes.

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A car moves in a circular motion and it is subject to a centripetal acceleration of 24 m/s2. If the radius of the circular path

Softa [21]

Answer:

Explanation:

Centripetal acceleration is given by:

$a_{c} = v^{2}/r$

Thus, centripetal acceleration is inversely proportional to the radius. Thus, when radius will double, the centripetal acceleration will be halved.

3 0

3 years ago

If this resolving power is diffraction-limited, to what effective diameter of your eye's optical system does this correspond? us

ZanzabumX [31]

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5 0

3 years ago

A baseball diamond is a square (don’t be fooled that it is usually shown rotated by 45°). Each side of the square is 90 feet lon

gayaneshka [121]

A line from second to home creates a right triangle with 45 degree angles.

This makes the diagonal line the side length of the square multiplied by the sqrt(2)

Second to home = 90sqrt(2) = 127.279 feet ( round answer as needed)

8 0

3 years ago

Which scenario does not describe an example of acceleration?

tekilochka [14]

Answer:

B is the right answer

Explanation:

It is decreasing.

3 0

3 years ago

Force F acts between two charges, q1 and q2, separated by a distance d. If q1 is increased to twice its original value and the d

vodka [1.7K]

Okay, haven't done physics in years, let's see if I remember this.

So Coulomb's Law states that $F = k \frac{Q_1Q_2}{d^2}$ so if we double the charge on $Q_1$ and double the distance to $(2d)$ we plug these into the equation to find

<span> $F_{new} = k \frac{2Q_1Q_2}{(2d)^2}=k \frac{2Q_1Q_2}{4d^2} = \frac{2}{4} \cdot k \frac{Q_1Q_2}{d^2} = \frac{1}{2} \cdot F_{old}$ </span>

So we see the new force is exactly 1/2 of the old force so your answer should be $\frac{1}{2}F$ if I can remember my physics correctly.

8 0

4 years ago

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