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mrs_skeptik [129]
3 years ago
15

Classify the following matter

Physics
1 answer:
AURORKA [14]3 years ago
7 0

Answer:

Woah

Explanation:

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An unbalanced force of 40. newtons keeps a
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(4) 4.0 m/s is the correct answer





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Which has more significant figures 3.456 or 2.0002?
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The second one because the zeroes in the middle count. 2.0002
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You are working at a company that manufactures electrical wire. Gold is the most ductile of all metals: it can be stretched into
devlian [24]

Answer:R=1.424 M\Omega

Explanation:

Given

mass of gold m=1.6 gm

Length of wire L=2.2 km

Resistivity of gold \rho =2.44\times 10^{-8}

density of gold =19.3\times 10^3 kg/m^3

and mass=volume\times density

1.6\times 10^{-3}=volume\times 19.3\times 10^3

volume=8.29\times 10^{-8} m^3

And Resistance R=\frac{\rho L}{A}

also be written as

R=\frac{\rho L^2}{V}

where L=length

V=volume

\rho =resistivity\ of\ gold

R=\frac{2.44\times 10^{-8}\times (2200)^2}{8.29 \times 10^{-8}}

R=1.42\times 10^{6} \Omega

R=1.424 M\Omega

5 0
3 years ago
When you look out the window you can see constantly moving air. Air is cycled in order to reach equilibrium. Answer the followin
svet-max [94.6K]
So base on your question about the moving air and its is cycled in order to reach equilibrium the answer of the questions are the following 
A. what are the two major characteristic that change when air is heated and cooled? 
-temperature and molecular structure/ phase of the matter

<span>B. how does these characteristics change the movement of air? 
</span>-if the air has lower temp the possibility that it will become heavier than with the higher
<span>C. what is the name of the cycle that develop during these changes? Air cycle</span>
<span>D. why do the changes you describe in part B provide an example of a cycle? because of its molecular activity will increase</span>
5 0
3 years ago
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A 1,500 kg car’s speed changes from 30 m/s to 15 m/s after the brakes are applied. Calculate the work done onto the car from the
HACTEHA [7]

The work done onto the car is 506,250 J

The work done on a system implies an increase in the internal energy of the system as a result of some forces acting on the system from the outside.

From the parameters given:

  • The mass of the car = 1500 kg
  • The initial speed = 30 m/s
  • The final speed = 15 m/s

The work done onto the car refers to the change in the kinetic energy (i.e. ΔK.E)

\mathbf{=\dfrac{1}{2} mv_1^2 -\dfrac{1}{2} mv_2^2}

\mathbf{=\dfrac{1}{2} m(v_1^2 - v_2^2)}

\mathbf{=\dfrac{1}{2} \times 1500 \times (30^2 - 15^2)}

= 506,250 J

Therefore, we can conclude that the work done on the car is 506,250 J

Learn more about work done here:

brainly.com/question/18762601

7 0
3 years ago
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