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julsineya [31]
3 years ago
10

Which does not contain a lens?

Chemistry
2 answers:
CaHeK987 [17]3 years ago
4 0
I think B or D because an eye obviously doesn't have a lense, mirror..maybe sorry for not giving u straight forward response
atroni [7]3 years ago
3 0
D because the eye has a lense that is a transparent structure behind the iris, the coloured part of the eye. 
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Some common substances used in the laboratory are listed in the table. the chemical formulas of the substances are also listed b
Helga [31]
Its 5 i think

but dont take my word on it

7 0
3 years ago
A __________ has distinct properties and composition that never vary.
erik [133]
<h2>Answer </h2>

Matter

<u>Explanation </u>

A matter has distinct properties and composition that never vary. The matter is anything which possesses mass and occupies space. The matter has different characteristics and a distribution that does not change from unit to unit. Materials which cannot be disintegrated into simpler materials. Each is made of only one kind of atom in subatomic level. For example, a chair, table, and similar everything that has mass and occupies space is matter.

3 0
2 years ago
Read 2 more answers
An explanation for a broad of observation, facts, and tested hypothses is called a
nikitadnepr [17]

Answer:

that's is a theory

Explanation:

hope it may help you

4 0
2 years ago
Is this correct please help?
iren [92.7K]

Answer:

\boxed {\tt Number \ and \ unit}

Explanation:

The two requirements for a measurement are a <u>number</u> and a <u>unit.</u>

For example, here is a measurement:

38.6 cm

The <u>number</u> is 38.6 and the <u>unit</u> is cm, or centimeters.

Therefore, both <em>number </em>and <em>unit</em> are correct.

7 0
3 years ago
Do you expect a strong chemical bond to have a higher or lower frequency than a weak bond?
Llana [10]

Answer:

Higher frequency  

Explanation:

We can imagine a chemical bond between two atoms as if it were two balls connected by a spring.

According to Hooke's Law, the stretching frequency f is

f = \dfrac{1}{2 \pi}\sqrt{ \dfrac {k}{\mu}}

where µ is the reduced mass of the system

\mu = \dfrac{m_{1} + m_{2}}{m_{1}m_{2} }

The strength of the bond is analogous to k, the force constant of the spring. Then,

f \propto \sqrt{k}

Thus, the stronger the bond, the greater the frequency of vibration.

6 0
3 years ago
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