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matrenka [14]
4 years ago
5

I need help with number 8

Physics
1 answer:
Mashutka [201]4 years ago
5 0

Answer:

cxvfbgfdnsdfisblkfuoasegfoucaegru

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KIM [24]
That's the impression you get from the sound, that you call 'high' or 'low', determined by the FREQUENCY of the wave.
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3 years ago
How can an element be identified using its emission spectrum?
posledela
The correct choice is C .

Technically, choice-A is the process involved.  But it's already been done,
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3 years ago
Approximating the eye as a single thin lens 2.70 cm from the retina, find the focal length of the eye when it is focused on an o
Daniel [21]

The focal length will be = 2.67 cm

The distance between the convex lens or a concave mirror and the focal point of a lens or mirror is called the focal length. It is the point where parallel rays of light meet or converge.

given

u (object distance) = 260 cm

v (image distance) = 2.70 cm

f (focal length) = ?

using lens formula

1/f = 1/u + 1/v

    = 1/260 + 1/2.70

    = 2.67 cm

The focal length will be = 2.67 cm

To learn more about focal length here

brainly.com/question/14104969

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5 0
2 years ago
((THIS IS WORTH 50 POINTS))
e-lub [12.9K]

Answer:

sorry I had find answer from everywhere but can't find

Explanation:

can u can send it with some information please

4 0
3 years ago
Read 2 more answers
An object possessing an excess of 6.0x10^6 electrons has a net charge of
guajiro [1.7K]
A single electron has a charge of
e=-1.6 \cdot 10^{19}C
Therefore, if we have an excess of N=6.0\cdot10^6 electrons, the total net charge will be the product between the charge of a single electron and the total number of electrons in excess:
Q=Ne=(6.0\cdot10^6)(-1.6 \cdot 10^{-19}C)=9.6 \cdot 10^{-13}C
7 0
3 years ago
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