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user100 [1]
3 years ago
6

The Köppen classification system is used as a classification system for

Physics
2 answers:
Kobotan [32]3 years ago
7 0
D. Climates.
It is currently the most widely used classification system used to determine the climates of different regions. It consists of several different factor classifications and was created in 1884 by a Russian climatologist Wladimir Köppen.

nirvana33 [79]3 years ago
6 0
It is used to classify the climates around the world

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Why is streak generally a more useful property than color in identifying a mineral?
poizon [28]
You may confuse two minerals by their colors so using streak is another way for showing their "true colors "
3 0
3 years ago
Calculate the net force on particle q1. First, find the direction of the force particle q2 is exerting on particle q1. Is it pus
ValentinkaMS [17]

The net force on particle particle q1 is 13.06 N towards the left.

<h3>Force on q1 due to q2</h3>

F(12) = kq₁q₂/r₂

F(12) = (9 x 10⁹ x 13 x 10⁻⁶ x 7.7 x 10⁻⁶)/(0.25²)

F(12) = -14.41 N  (towards left)

<h3>Force on q1 due to q3</h3>

F(13) = (9 x 10⁹ x 7.7 x 10⁻⁶ x 5.9 x 10⁻⁶)/(0.55²)

F(13) = 1.352 N (towards right)

<h3>Net force on q1</h3>

F(net) = 1.352 N - 14.41 N

F(net) = -13.06 N

Thus, the net force on particle particle q1 is 13.06 N towards the left.

Learn more about force here: brainly.com/question/12970081

#SPJ1

8 0
1 year ago
Please help ASAP. In what direction do you need to apply force to move an object vertically?
PilotLPTM [1.2K]
The answer is up . tylrhscjwizn
5 0
3 years ago
Will give correct answer brainliest<br><br>5 kg m/s<br>8kg m/s<br>80 kg m/s<br>200 kg m/s​
o-na [289]

Answer: Here this will help you..

Explanation:

1 kg-m/s to kilogram-force meter/second = 1 kilogram-force meter/second

5 kg-m/s to kilogram-force meter/second = 5 kilogram-force meter/second

10 kg-m/s to kilogram-force meter/second = 10 kilogram-force meter/second

20 kg-m/s to kilogram-force meter/second = 20 kilogram-force meter/second

30 kg-m/s to kilogram-force meter/second = 30 kilogram-force meter/second

40 kg-m/s to kilogram-force meter/second = 40 kilogram-force meter/second

50 kg-m/s to kilogram-force meter/second = 50 kilogram-force meter/second

75 kg-m/s to kilogram-force meter/second = 75 kilogram-force meter/second

100 kg-m/s to kilogram-force meter/second = 100 kilogram-force meter/second

8 0
2 years ago
A nylon guitar string is fixed between two lab posts 2.00 m apart. The string has a linear mass density of μ=7.20 g/m\mu=7.20~\t
vladimir2022 [97]

Answer:

4.6 m

Explanation:

First of all, we can find the frequency of the wave in the string with the formula:

f=\frac{1}{2L}\sqrt{\frac{T}{\mu}}

where we have

L = 2.00 m is the length of the string

T = 160.00 N is the tension

\mu =7.20 g/m = 0.0072 kg/m is the mass linear density

Solving the equation,

f=\frac{1}{2(2.00 m)}\sqrt{\frac{160.00 N}{0.0072 kg/m}}=37.3 Hz

The frequency of the wave in the string is transmitted into the tube, which oscillates resonating at same frequency.

The n=1 mode (fundamental frequency) of an open-open tube is given by

f=\frac{v}{2L}

where

v = 343 m/s is the speed of sound

Using f = 37.3 Hz and re-arranging the equation, we find L, the length of the tube:

L=\frac{v}{2f}=\frac{343 m/s}{2(37.3 Hz)}=4.6 m

4 0
3 years ago
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