What is the phase of the Moon if it . . .

Physics

1 answer:

anzhelika [568]3 years ago

8 0

Answer:

Waxing Gibbous

Third quarter

Waning Gibbous

Explanation:

If moon rises at 3:00 pm then the phase of the moon will be "Waxing Gibbous".

This is because, the moon is actually not fully illuminated but has achieved more than half of its full illumination.

If the moon is highest in the sky at sunrise then the phase of the Moon will be the "Third quarter"

This is because of the fact that at this position moon will rise at midnight, thus it will be at the highest point at the time of the sunrise.

If the moon sets at 10:00 am then the phase of the Moon is "Waning Gibbous"

This is because of the fact that at this position the Moon is moving towards becoming new Moon but at the same time, the moon is illuminated more than its half illumination.

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Which of the following is not a step in the inquiry process?

Contact [7]

Invention I think which is B as they don't invent things in an enquiry

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3 years ago

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1. A block with a mass of 5.0 kg is pushed on a frictionless surface by applying a horizontal force of 80.0 N. The block starts

Snezhnost [94]

Answer:

397 j

Explanation:

Because 5.0kg yuh

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2 years ago

You are climbing in the high sierra when you suddenly find yourself at the edge of a fog-shrouded cliff. to find the height of t

sergey [27]

The total time it takes the sound to reach our on top of the cliff (10.0 s) is actually sum of two times:
- the time it takes the rock to hit the ground starting from the top of the cliff, t1
- the time it takes the sound to reach the top of the cliff starting from the ground, t2
1) The rock moves by uniformly accelerated motion, with constant acceleration $g=9.81 m/s^2$ . Its law of motion is given by $y(t)=h-v_0t-\frac{1}{2}gt^2$ , where $v_0=0$ is the initial velocity of the rock, and h is the height of the cliff. The time t1 is the time at which the rock reaches the ground, so that y(t1)=0, and the equation becomes
$0=h-\frac{1}{2}gt_1^2$
2) The sound moves from the ground to the top of the cliff by uniform motion, with constant speed v=343 m/s. Therefore, the sound covers the distance h (the height of the cliff) in a time t2 given by
$h=vt_2$
3) If we rewrite h in both equations, we can write:
$=\frac{1}{2}gt_1^2=vt_2$ (1)
4) We also know that the sum of t1 and t2 is equal to 10 seconds:
$t_1+t_2=10$
from which we find
$t_2=10-t_1$
if we substitute this into eq.(1), we get
$\frac{1}{2}gt_1^2+vt_1-10v=0$
Numerically:
$4.9t_1^2 + 343 t_1 - 3430=0$
Solving the equation, we find the solution $t_1=8.87 s$ (the other solution is negative, so it does not have physical meaning). As a consequence,
$t_2 = 10-t_1 = 1.13 s$
and the height of the cliff is given by
<span> $h=vt_2=(343 m/s)(1.13 s)=388 m$ </span>