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Stolb23 [73]
3 years ago
6

________ occurs when an object in the outer reaches of the solar system passes between earth and a far distant star, temporarily

blocking light from the star.
Physics
1 answer:
murzikaleks [220]3 years ago
6 0
<span>Occulation occurs when an object in outer space reaches the solar system and passes between earth and a far distant star, temporarily blocking light from the star. Occulation is an astronomy term referring to when an object in the foreground blocks the view of an object in the background.</span>
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A runner traveled 7.7 miles an hour and a half what was their average speed
Alenkasestr [34]

Answer:5.13333333...

Explanation: 7.7 is the distance and an hour and a half is the average speed. You would have to divide the total distance by the total time. So it would be 7.7 divided by 1.5 which would equal 5.13333333333...

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Any clue on this one I know it’s not D
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Which month has the longest day and shortest night?
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3 years ago
Height of cannon 5 m, initial speed of projectile 15m/s, angle of launch 0 degrees. What is the range and time in the air? Pleas
Westkost [7]

Answer:

<em>The range is 15.15 m and the time in the air is 1.01 s</em>

Explanation:

<u>Horizontal Motion</u>

When an object is thrown horizontally (with angle 0°) with a speed v from a height h, it follows a curved path ruled exclusively by gravity until it eventually hits the ground.

The range or maximum horizontal distance traveled by the object can be calculated as follows:

\displaystyle d=v\cdot\sqrt{\frac  {2h}{g}}

To calculate the time the object takes to hit the ground, we use the equation below:

\displaystyle t=\sqrt{\frac{2h}{g}}

The cannon is shot from a height of h=5 m with an initial speed of v=15 m/s. The range is calculated below:

\displaystyle d=15\cdot\sqrt{\frac  {2*5}{9.8}}=15*1.01

d = 15.15 m

The time in the air is:

\displaystyle t=\sqrt{\frac{2*5}{9.8}}

t = 1.01 s

The range is 15.15 m and the time in the air is 1.01 s

8 0
3 years ago
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