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Temka [501]
3 years ago
12

A series of parallel linear water wave fronts are traveling directly toward the shore at 15.5 cm/s on an otherwise placid lake.

A long concrete barrier that runs parallel to the shore at a distance of 3.10 m away has a hole in it. You count the wave crests and observe that 75.0 of them pass by each minute, and you also observe that no waves reach the shore at ±62.3cm from the point directly opposite the hole, but waves do reach the shore everywhere within this distance.
Part A

How wide is the hole in the barrier?

Part B

At what other angles do you find no waves hitting the shore?

Enter your answers numerically separated by commas.

Physics
1 answer:
nata0808 [166]3 years ago
7 0

Find the given attachment

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A ball is thrown upward from an initial height of 1.5m the ball reaches a height of 5m then falls to the ground . What Is the di
jeka94

Answer:

The distance traveled by the ball is 8.5 m

Explanation:

Initial height of the ball, h₁ = 1.5 m above the ground

final height of the ball, h₂ = 5m

Upward distance = distance traveled by the ball from a height of 1.5m to 5m = 5m - 1.5m = 3.5 m

Downward distance = distance traveled by the ball from 5m height to the ground =5m - 0 = 5m

Total distance traveled = upward distance + downward distance

Total distance traveled = 3.5 m + 5m = 8.5 m

Therefore, the distance traveled by the ball is 8.5 m

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3 years ago
A student read his right ear against his desk while Teacher talks loudly on the desk and can hear the tapping sound only through
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Answer:

The student hears the wave that is transmitted by the desk

Explanation:

Mechanical waves need a material medium to be able to be transmitted, in the case of sound waves, one of the most common media is air, but it is also transmitted in other media in this case, stationery is transmitted.

The student hears the wave that is transmitted by the desk

The speed of the wave is proportional to the density of the material, so the wave that the student hears arrives much faster through the desk than through the air

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g A ball thrown straight up into the air is found to be moving at 7.94 m/s after falling 2.72 m below its release point. Find th
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The ball has height <em>y</em> and velocity <em>v</em> at time <em>t</em> according to

<em>y</em> = <em>v</em>₀ <em>t</em> - 1/2 <em>g</em> <em>t</em> ²

and

<em>v</em> = <em>v</em>₀ - <em>g t</em>

where <em>v</em>₀ is its initial speed and <em>g</em> = 9.80 m/s² is the magnitude of the acceleration due to gravity.

The ball is falling with a velocity of 7.94 m/s when it's 2.72 m below the release point, which at time <em>t </em>such that

-2.72 m = <em>v</em>₀ <em>t</em> - 1/2 <em>g</em> <em>t</em> ²

-7.94 m/s = <em>v</em>₀ - <em>g t</em>

Solve for <em>t</em> in the second equation:

<em>t </em>= (<em>v</em>₀ + 7.94 m/s)/<em>g</em>

Substitute this into the first equation and solve for <em>v</em>₀ :

-2.72 m = <em>v</em>₀ (<em>v</em>₀ + 7.94 m/s) /<em>g</em> - 1/2 <em>g</em> ((<em>v</em>₀ + 7.94 m/s)/<em>g</em>)²

-2.72 m = <em>v</em>₀²/<em>g</em> + (7.94 m/s) <em>v</em>₀/<em>g</em> - 1/2 (<em>v</em>₀ + 7.94 m/s)²/<em>g</em>

2 (-2.72 m) <em>g</em> = 2<em>v</em>₀² + 2 (7.94 m/s) <em>v</em>₀ - (<em>v</em>₀ + 7.94 m/s)²

2 (-2.72 m) (9.80 m/s²) = 2<em>v</em>₀² + (15.9 m/s) <em>v</em>₀ - (<em>v</em>₀² + (15.9 m/s) <em>v</em>₀ + 63.0 m²/s²)

-53.3 m²/s² = <em>v</em>₀² - 63.0 m²/s²

<em>v</em>₀² = 9.73 m²/s²

<em>v</em>₀ = 3.12 m/s

3 0
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Answer:

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Explanation:

Detailed explanation and calculation is shown in the image below

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