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3 years ago

If the new moon happens on January 15th, what shape will it be on February 6th?

Physics

1 answer:

Jobisdone [24]3 years ago

3 0

-- From January 15 to February 6 is a period of 22 days.

-- The period of the full cycle of moon phases is 29.53 days.

-- So those dates represent (22/29.53) = 74.5% of a full cycle of phases.

-- That's almost exactly 3/4 of a full cycle, so on February 6, the moon would be almost exactly at Third Quarter. That's the left half of a disk (viewed from the northern hemisphere).

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Waves are observed passing under a dock. Wave crests are 8.0 meters apart. The time for a complete wave to pass by is 4.0 second

ki77a [65]

To answer that question, we don't care what the highest and lowest
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often those pass by us.

You said it takes 4 seconds for a complete wave to pass by.
Through the sheer power of intellect, I'm able to take that information
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There's your frequency . . . 1/4 per second, or 0.25 Hz.

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Arisa [49]

The correct one is A

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3 years ago

In a system with only a single force acting upon a body, what is the relationship between the change in kinetic energy and the w

Makovka662 [10]

Answer: W.D = 1/2mv^2

Explanation:

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The work done will be converted to kinetic energy.

W.D = 1/2mv^2

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3 years ago

Which statement describes how globes represent Earth’s surface?

n200080 [17]

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3 years ago

Read 2 more answers

A wheel of radius 25cm has eight spokes. It is mounted on a fixed axle and is rotating at a constant angular speed w. You shoot

spayn [35]

Explanation:

We will assume that the rim of the wheel is also very thin, like the spokes. The distance s between the spokes along the rim is

$s = \frac{1}{8}C = \frac{1}{8}(2\pi)(0.25\:\text{m}) = 0.196\:\text{m}$

The 20-cm arrow, traveling at 6 m/s, will travel its length in

$t = \dfrac{0.2\:\text{m}}{6\:\text{m/s}} = \dfrac{1}{30}\:\text{s}$

The fastest speed that the wheel can spin without clipping the arrow is

$v = \dfrac{s}{t} = \dfrac{0.196\:\text{m}}{\left(\dfrac{1}{30\:\text{s}}\right)} = 5.9\:\text{m/s}$

The angular velocity $\omega$ of the wheel is given by

$\omega = \dfrac{v}{r} = \dfrac{5.9\:\text{m/s}}{0.25\:\text{m}} = 23.6\:\text{rad/s}$

In terms of rev/s, we can convert the answer above as follows:

$23.6\:\dfrac{\text{rad}}{\text{s}}×\dfrac{1\:\text{rev}}{2\pi\:\text{rad}} = 3.8\:\text{rev/s}$

As you probably noticed, I did the calculations based on the assumption that I'm aiming for the edge of the wheel because this is the part of the wheel where a point travels a longer linear distance compared to ones closer to the axle, thus giving the arrow a better chance to pass through the wheel without getting clipped by the spokes. If you aim closer to the axle, then the wheel needs to spin slower to allow the arrow to get through without hitting the spokes.

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3 years ago