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3 years ago

10

What do you make when you dissolve sugar in water?

1 answer:

AlladinOne [14]3 years ago

5 0

It also takes energy to break the hydrogen bonds in water that must be disrupted to insert one of these sucrose molecules into solution. Sugar dissolves in water because energy is given off when the slightly polar sucrose molecules form intermolecular bonds with the polar water molecules.

I hope this helped :)

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The area of the effort and load of the Piston of a hydrolic are 0.5 and 5m respectively. If a force of 100 Newton is applied on

Yanka [14]

Answer:

Explanation:

Using Pascal laws, which states that pressure are the input equals the pressure at the output.

Pressure is given as force/area

P1=P2

Then,

F1/A1=F2/A2

Cross multiply

F1A2=F2A1

Given that

Ae=0.5m² area of effort

Al=5m² area of load

Fl=? Force if load

Fe= 100N. Force of effort

Then applying pascal

Fl/Al=Fe/Ae

Fl/5=100/0.5

FL/5=200

Fl=200×5

Fl=1000N

The first safety load is 1000N

6 0

3 years ago

The drawing shows three particles far away from any other objects and located on a straight line. The masses of these particles

belka [17]

Answer:

$F_a=5.67\times 10^{-5}\ N$

<u /> $F_b=3.49\times 10^{-5}\ N$

$F_c=9.16\times 10^{-5}\ N$

Explanation:

Given:

mass of particle A, $m_a=363\ kg$

mass of particle B, $m_b=517\ kg$

mass of particle C, $m_c=154\ kg$

All the three particles lie on a straight line.

Distance between particle A and B, $x_{ab}=0.5\ m$

Distance between particle B and C, $x_{bc}=0.25\ m$

Since the gravitational force is attractive in nature it will add up when enacted from the same direction.

<u>Force on particle A due to particles B & C:</u>

$F_a=G. \frac{m_a.m_b}{x_{ab}^2} +G. \frac{m_a.m_c}{(x_{ab}+x_{bc})^2}$

$F_a=6.67\times 10^{-11}\times (\frac{363\times 517}{0.5^2}+\frac{363\times 154}{(0.5+0.25)^2})$

$F_a=5.67\times 10^{-5}\ N$

<u>Force on particle C due to particles B & A:</u>

<u /> $F_c=G.\frac{m_c.m_b}{x_{bc}^2} +G.\frac{m_c.m_a}{(x_{ab}+x_{bc})^2}$ <u />

$F_c=6.67\times 10^{-11}\times (\frac{154\times 517}{0.25^2}+\frac{154\times 363}{(0.25+0.5)^2} )$

$F_c=9.16\times 10^{-5}\ N$

<u>Force on particle B due to particles C & A:</u>

<u /> $F_b=G.\frac{m_b.m_c}{x_{bc}^2} -G.\frac{m_b.m_a}{x_{ab}^2}$ <u />

<u /> $F_b=6.67\times 10^{-11}\times (\frac{517\times 154}{0.25^2}-\frac{517\times 363}{0.5^2} )$ <u />

<u /> $F_b=3.49\times 10^{-5}\ N$ <u />

3 0

3 years ago

All the steps of the scientific method rely on valid observations. <br> true or false

butalik [34]

Well, basically because the observations can help you out during the experiment.

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What activities should you avoid to prevent static discharge while working on a computer?

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3 years ago

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alina1380 [7]

I believe it is A. positive acceleration since it is speeding up

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3 years ago