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3 years ago

This is a sign of a chemical reaction that involves a new color being created during the reaction.

Physics

1 answer:

dedylja [7]3 years ago

4 0

Answer:

The five conditions of chemical change: color change, formation of a precipitate, formation of a gas, odor change, temperature change.

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Two charges separated by one meter exert a 9 N force on each other. If the charges are pushed to a 3 meter separation, the force

tamaranim1 [39]

Answer:

False

Explanation:

The formula of force that exists between two charges is expressed as;

F = kq1q2/r²

If two charges separated by one meter exert a 9 N force on each other, the;

9 = kq1q2/1²

9 = kq1q2 ..... 1

If the charges are pushed to a 3 meter separation, then;

F = kq1q2/3²

F = kq1q2/9 .... 2

Divide both equations;

9/F = (kq1q2)/ kq1q2/9

9/F = kq1q2 * 9/ kq1q2

9/F = 9

F = 9/9

F = 1N

Hence if the charges are pushed to a 3 meter separation, then the force on EACH charge will be 1N. Hence the answer is False

3 0

3 years ago

Five race cars speed toward the finish line at the Jasper County Speedway. The table lists each car’s speed in meters/second. If

Anarel [89]

I think the answer would be Car C.

7 0

3 years ago

Read 2 more answers

If the distance from a light source triples, how does light intensity change? The intensity will be 3x greater. The intensity wi

Tcecarenko [31]

Answer:

The intensity will be 1/9 as much.

Explanation:

The intensity of the light or any source is inversely related to the square of the distance.

$I\alpha \frac{1}{r^{2} }$

Now according to the question the distance is increased by three times than,

$\frac{I_{2} }{I_{1} }=\frac{r_{1}^{2} }{r_{2}^{2} }$

Therefore,

$\frac{I_{2} }{I_{1} }=\frac{r_{1}^{2} }{(3r_{1})^{2} }\\\frac{I_{2} }{I_{1} }=\frac{1}{9} \\{I_{2}=\frac{1}{9}{I_{1} }$

Therefore the intensity will become 1/9 times to the initial intensity.

3 0

3 years ago

Fill in the blank: An object is most likely to sink in water if _________________.

drek231 [11]

High density
random words to fill up 20 character minimum for answering questions :P

4 0

3 years ago

Read 2 more answers

A scientist wants to model an internal organ with connective tissue as a mass on a spring. The mass of the organ is 2.0 kg, and

vichka [17]

Answer:

Spring constant, k = 5483.11 N/m

Explanation:

It is given that,

Mass of the organ, m = 2 kg

The natural period of oscillation is, T = 0.12 s

Let k is the spring constant for the spring in the scientist's model. The period of oscillation is given by :

$T=2\pi\sqrt{\dfrac{m}{k}}$

$k=\dfrac{4\pi^2 m}{T^2}$

$k=\dfrac{4\pi^2 \times 2\ kg}{(0.12\ s)^2}$

k = 5483.11 N/m

So, the spring constant for the spring in the scientist's model is 5483.11 N/m.

5 0

3 years ago