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faltersainse [42]

4 years ago

8

The formation of a new substance shows that what change occurs

1 answer:

ycow [4]4 years ago

7 0

A chemical change might also result in the formation of a precipitate, such as the appearance of a cloudy material when dissolved substances are mixed. Rotting, burning, cooking, and rusting are all further types of chemical changes because they produce substances that are entirely new chemical compounds.

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Illustrated here are three different<br> of carbon. They vary by the number of

Basile [38]

D) isotopes; neutrons

Explanation:

The diagram illustrates the three different isotopes of carbon. They vary by the number of neutrons.

Isotopy is the existence of two or more atoms of the same element having the same atomic number but different mass numbers due to the differences in the number of neutrons in their various nuclei.

Isotopes of elements have the same electronic configuration
They share the same chemical properties
They differ in masses due the number of neutrons they contain.

Learn more:

Isotopes brainly.com/question/1915462

#learnwithBrainly

5 0

4 years ago

How can I covert these 2,34 meters in decimeters<br>

zysi [14]

Answer:

If you meant 2.34, 2.34 meters = 23.4 decimeters.

Formula: multiply the value in meters by the conversion factor '10'.

So, 2.34 meters = 2.34 × 10 = 23.4 decimeters.

Hope that helps. x

4 0

3 years ago

Read 2 more answers

This should be pretty easy i think.

timama [110]

Answer:

its a sure for not but i think its a 192583

3 0

2 years ago

Use the values from PRACTICE IT to help you work this exercise. Suppose the same two vehicles are both traveling eastward, the c

Mariulka [41]

Answer:

A. $v_{3}=12.17m/s$

B. $v_{car}=6.3m/s\\v_{truck}=-6.3m/s$

C. ΔK $=-4.13x10^3J$

Explanation:

From the exercise we know that the car and the truck are traveling eastward. I'm going to name the car 1 and the truck 2

$v_{1}=5.79m/s\\m_{1}=102kg\\v_{2}=18.5m/s\\m_{2}=103kg$

A. Since the two vehicles become entangled the final mass is:

$m_{3}=102kg+103kg=205kg$

From linear momentum we got that:

$p_{1}=p_{2}$

$m_{1}v_{1}+m_{2}v_{2}=m_{3}v_{3}$

$v_{3}=\frac{m_{1}v_{1}+m_{2}v_{2}}{m_{3} }=\frac{(102kg)(5.79m/s)+(103kg)(18.5m/s)}{(205kg)}$

$v_{3}=12.17m/s$

B. The change in velocity of both vehicles are:

For the car

$v_{car}=v_{f}-v_{o}=12.17m/s-5.79m/s=6.38m/s$

For the truck

$v_{truck}=12.17m/s-18.5m/s=-6.3m/s$

C. The change in kinetic energy is:

ΔK= $K_{2}-K_{1} =\frac{1}{2}m_{3}v_{3}^{2}-(\frac{1}{2}m_{1}v_{1}^{2}+\frac{1}{2}m_{2}v_{2}^{2})$

ΔK= $\frac{1}{2}(205)(12.17)^{2}-(\frac{1}{2}(102)(5.79)^{2}+\frac{1}{2}(103)(18.5)^{2})=-4.13x10^{3}J$

ΔK $=-4.13x10^{3}J$

6 0

3 years ago

Please help me with this

Firlakuza [10]

Answer:

can't see anything sorry can't help

7 0

3 years ago

Read 2 more answers