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3 years ago

Difference between diurnal and annual motion in two points

Physics

1 answer:

Valentin [98]3 years ago

5 0

Answer:

Explanation below:

Explanation:

Annual motion describes the changes in motion of the earth around the sun. Diurnal motion can be better understood as the change in motion caused by Earths rotation at the poles.

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the frequency of the middle b note on a piano is 493.88 hz. what is the wavelength of this note in centimeters? the speed of sou

vekshin1

Answer:

69.5

Explanation:

V=FX

343.06=493.88×X

X=343.06/493.88

X=0.695m

to cm is 69.5cm

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2 years ago

In the diagram below, point A is southwest of point B. Jaleh walks for 400 seconds along path 1, going from point A to point B.

timama [110]

Answer:

Explanation:

velocity = displacement (m) / change of time (s)

velocity = (400 + 300) / (100 + 400)

velocity = 1.4 m/s

4 0

2 years ago

Read 2 more answers

A certain wire has a resistance of 110 Ω. What is the resistance of a second wire, made of the same material, that is 1/4 as lon

Svet_ta [14]

Answer:

New Resistance = 247.5 ohm

Explanation:

Resistance = resistivity * length / area

Since resistivity for the material is constant, resistance is directly proportional to (length/area).

This means that if (length/area) decreases or increases by any ratio, then resistance will increase or decrease by the same ratio.

So let's find the change in length/area

New length = 0.25 old length

New area = (1/9) old area (This is because area equation contains a square of the diameter. if diameter decreases by 1/3, area decreases by (1/3)^2 )

So we now get length /area:

New length / new area = ( 0.25 old length) / (1/9 of old area)

New length / new area = 9*0.25 (old length / old area)

New length / new area = 2.25 (old length / old area)

To get the new resistance, we simply multiply it by the ratio we just found.

This equals:

110 * 2.25 = 247.5 ohm

4 0

3 years ago

Read 2 more answers

An athlete at high performance inhales 4.0L of air at 1 atm and 298 K. The inhaled and exhaled air contain 0.5% and 6.2% by volu

LenaWriter [7]

To solve this problem we will calculate the total volume of inhaled and exhaled water. From the ideal gas equation we will find the total number of moles of water.

An athlete at high performance inhales 4.0L of air at 1atm and 298K.

The inhaled and exhaled air contain 0.5% and 6.2% by volume of water, respectively.

During inhalation, volume of water taken is

$V_i = (4L)(0.5\%)$