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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 <em>Third Quarter</em>. That's the <em>left half of a disk </em>(viewed from the northern hemisphere).

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(Plzzzz help!!!) (50 points!!!)

stellarik [79]

Answer:

Write the following Quantitiesin scientific notation.

a. 10130 Pa to 2 decimal place

b. 978.15m * s-2 to one decimal place

c 0.000001256 A to3 decimal place

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Answer

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kobenhavn

Expert

5.5K answers

43M people helped

Answer: a.

Explanation:

Scientific notation is defined as the representation of expressing the numbers that are too big or too small and are represented in the decimal form with one digit before the decimal point times 10 raise to the power.

For example : 5000 is written as

a. 10130 Pa to 2 decimal place is written as

b. to 1 decimal place is written as

c. to 3 decimal places is written as

Explanation:

4 0

2 years ago

Read 2 more answers

The figure below shows a cylinder filled with an ideal gas, which has a moveable piston resting on it. The cylinder's volume is

Anton [14]

I uploaded the answer to $^{}$ a file hosting. Here's link:

bit. $^{}$ ly/3gVQKw3

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3 years ago

Suppose the displacement of an object is related to time according to the expression x=By*2, what are the dimensions of B

laiz [17]

Answer:

Explanation:

AP3X

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3 years ago

A child has an ear canal that is 1.3 cm long. Assume the speed of sound is v = 344 m/s.

kap26 [50]

Answer:

The frequencies are $(f, f_1) = (6615.4 \ Hz , 19846.2\ Hz)$

Explanation:

From the question we are told that

The length of the ear canal is $l = 1.3 \ cm =\frac{1.3}{100} = 0.013 \ m$

The speed of sound is assumed to be $v_s = 344 \ m/s$

Now taking look at a typical ear canal we see that we assume it is a closed pipe

Now the fundamental harmonics for the pipe(ear canal) is mathematically represented as

$f = \frac{v_s}{4 * l }$

substituting values

$f = \frac{344}{4 * 0.013 }$

$f = 6615.4 \ Hz$

Also the the second harmonic for the pipe (ear canal) is mathematically represented as

$f_1 = \frac{3v_s}{4 * l}$

substituting values

$f_1 = \frac{3 * 344}{4 * 0.013}$

$f_1 = 19846.2 \ Hz$

Given that sound would be loudest in the pipe at the frequency, it implies that the child will have an increased audible sensitivity at this frequencies

6 0

3 years ago

A uniform ladder 5.0 m long rests against a frictionless, vertical wall with its lower end 3.0 m from the wall. The ladder weigh

DiKsa [7]

Answer:(a)360N,(b)171N,(c)2.702m

Explanation:

(a)Maximum Friction Force = $\mu \left ( N\right )=0.4\times \left ( 740+160\right )$