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ozzi
3 years ago
9

Which is present when iodine changes from brown to blue or purple?

Physics
1 answer:
vladimir1956 [14]3 years ago
5 0


If iodine is added to a starch solution, they react with each other and the iodine  darkens to an almost pitch black.

however, if iodine is added to a solution containing no starch, it will show up only as an extremely pale brown. almost colorless and hardly visible.

when following the changes in some inorganic oxidation reduction reactions, iodine may be used as an indicator to follow the changes of iodide ion and iodine element. soluble starch solution is added. only iodine element in the presence of iodide ion will give the characteristic blue black color. neither iodine element alone nor iodide ions alone will give the color result.

hope this answer really helps your question :)

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Newton's Law of Gravitation says that the magnitude F of the force exerted by a body of mass m on a body of mass M is F = GmM r2
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<h2>Answers:</h2>

<h2>(a) </h2>

According to Newton's Law of Gravitation, the Gravity Force is:

F=\frac{GMm}{{r}^{2}}     (1)

This expression can also be written as:

F=GMm{r}^{-2}    (2)

If we derive this force F respect to the distance r between the two masses:

\frac{dF}{dr}dFdr=\frac{d}{dr}(GMm{r}^{-2})dr     (3)

Taking into account GMm are constants:

\frac{dF}{dr}dFdr=-2GMm{r}^{-3}     (4)

Or

\frac{dF}{dr}dFdr=-2\frac{GMm}{{r}^{3}}     (5)

<h2> (b) dF/dr represents the rate of change of the force with respect to the distance between the bodies.  </h2><h2 />

In other words, this means how much does the Gravity Force changes with the distance between the two bodies.

More precisely this change is inversely proportional to the distance elevated to the cubic exponent.

As the distance increases, the Force decreases.

<h2>(c) The minus sign indicates that the bodies are being forced in the negative direction.  </h2>

This is because Gravity is an attractive force, as well as, a central conservative force.

This means it does not depend on time, and both bodies are mutually attracted to each other.

<h2>(d) </h2>

In the first answer we already found the decrease rate of the Gravity force respect to the distance, being its unit N/km:

\frac{dF}{dr}dFdr=-2\frac{GMm}{{r}^{3}}     (5)

We have a force that decreases with a rate 1 \frac{dF_{1}}{dr}dFdr=4N/km when r=20000km:

4N/km=-2\frac{GMm}{{(20000km)}^{3}}     (6)

Isolating -2GMm:

-2GMm=(4N/km)({(20000km)}^{3})     (7)

In addition, we have another force that decreases with a rate 2 \frac{dF_{2}}{dr}dFdr=X when r=10000km:

XN/km=-2\frac{GMm}{{(10000km)}^{3}}     (8)

Isolating -2GMm:

-2GMm=X({(10000km)}^{3})     (9)

Making (7)=(9):

(4N/km)({(20000km)}^{3})=X({(10000km)}^{3}       (10)

Then isolating X:

X=\frac{4N/km)({(20000km)}^{3}}{{(10000km)}^{3}}  

Solving and taking into account the units, we finally have:

X=-32N/km>>>>This is how fast this force changes when r=10000 km

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A stationary 500 kg tank fires a 20 kg miegile at 100 m/s. What is the velocity of the tank after the missile is fired? Assume t
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Answer:

v₁ = 4 [m/s].

Explanation:

This problem can be solved by using the principle of conservation of linear momentum. Where momentum is preserved before and after the missile is fired.

P=m*v

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P = linear momentum [kg*m/s]

m = mass [kg]

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(m_{1}*v_{1})=(m_{2}*v_{2})

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m₁ = mass of the tank = 500 [kg]

v₁ = velocity of the tank after firing the missile [m/s]

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v₂ = velocity of the missile after firing = 100 [m/s]

(500*v_{1})=(20*100)\\v_{1}=2000/500\\v_{1}=4[m/s]

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