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d1i1m1o1n [39]
3 years ago
5

In chemical reactions, energy is absorbed or released in the form of:

Physics
1 answer:
dsp733 years ago
6 0

Answer:

all chemical reaction involved in early enough energy used to break bonds in reactants and energy released when you want from product

Edith a remake reaction absorb energy and epidemic reaction release energy flow of conversion of energy states that matter cannot be created or destroyed .

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What’s Chemical energy
AVprozaik [17]

Answer:

Chemical energy is energy that chemical substance to undergoes a chemical reaction to make a new substance.

Explanation:

<h3>hope this makes sense and helps</h3>
5 0
3 years ago
Read 2 more answers
The cornea behaves as a thin lens of focal length approximately 1.80 {\rm cm}, although this varies a bit. The material of which
spayn [35]

Answer:

The height of the image will be "1.16 mm".

Explanation:

The given values are:

Object distance, u = 25 cm

Focal distance, f = 1.8 cm

On applying the lens formula, we get

⇒  \frac{1}{v} -\frac{1}{u} =\frac{1}{f}

On putting estimate values, we get

⇒  \frac{1}{v} -\frac{1}{(-25)} =\frac{1}{1.8}

⇒  \frac{1}{v} =\frac{1}{1.8} -\frac{1}{25}

⇒  v=1.94 \ cm

As a result, the image would be established mostly on right side and would be true even though v is positive.

By magnification,

m=\frac{v}{u} and m=\frac{h_{1}}{h_{0}}

⇒  \frac{v}{u} =\frac{h_{1}}{h_{0}}

⇒  \frac{1.94}{25}=\frac{{h_{1}}}{15}

⇒  {h_{1}}=1.16 \ mm

8 0
3 years ago
2.
saul85 [17]

Answer:

<u>B</u>

Explanation:

Planets have different year lengths because it depends how far they revolve from a celestial body. Each planet has its own orbital period. Planets closer to the star will have a lower orbital period compared to the ones that lie far away from it.

4 0
1 year ago
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Which phase of matter contains particles that split into ions and electrons?
Lemur [1.5K]

Answer:

Plasma

Explanation:

7 0
2 years ago
Starting from zero, the electric current takes 2 seconds to reach half its maximum possible value in an RL circuit with a resist
Leno4ka [110]

Answer:

time=4s

Explanation:

we know that in a RL circuit with a resistance R, an inductance L and a battery of emf E, the current (i) will vary in following fashion

i(t)=\frac{E}{R}(1-e^\frac{-t}{\frac{L}{R}}), where imax=\frac{E}{R}

Given that, at i(2)=\frac{imax}{2} =\frac{E}{2R}

⇒\frac{E}{2R}=\frac{E}{R}(1-e^\frac{-2}{\frac{L}{R}})

⇒\frac{1}{2}=1-e^\frac{-2}{\frac{L}{R}}

⇒\frac{1}{2}=e^\frac{-2}{\frac{L}{R}}

Applying logarithm on both sides,

⇒log(\frac{1}{2})=\frac{-2}{\frac{L}{R}}

⇒log(2)=\frac{2}{\frac{L}{R}}

⇒\frac{L}{R}=\frac{2}{log2}

Now substitute i(t)=\frac{3}{4}imax=\frac{3E}{4R}

⇒\frac{3E}{4R}=\frac{E}{R}(1-e^\frac{-t}{\frac{L}{R}})

⇒\frac{3}{4}=1-e^\frac{-t}{\frac{L}{R}}

⇒\frac{1}{4}=e^\frac{-t}{\frac{L}{R}}

Applying logarithm on both sides,

⇒log(\frac{1}{4})=\frac{-t}{\frac{L}{R}}

⇒log(4)=\frac{t}{\frac{L}{R}}

⇒t=log4\frac{L}{R}

now subs. \frac{L}{R}=\frac{2}{log2}

⇒t=log4\frac{2}{log2}

also log4=log2^{2}=2log2

⇒t=2log2\frac{2}{log2}

⇒t=4

5 0
3 years ago
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