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Anna007 [38]
4 years ago
13

Explain how if a solid, liquid, and gas are put into individual containers how they fill the container.

Physics
1 answer:
Sav [38]4 years ago
8 0

Answer:

serial in which container is filled

Solid -base of container

Liquid- above solid

Gas- above liquid

Explanation:

If any mixture of  matter in different state (that solid , liquid or gas )are kept in any container, then substance with higher density will be settled at lowest surface first and similarly the substance with lowest density will be at upper part of container.

In the given container we have to keep solid, liquid and gas

  • sold has the highest density,
  • gas the lowest density and
  • liquid has the density higher than gas but less than solid.

based on this

solid will be at surface of container

above sold will be liquid

above liquid will be presence of Gas

serial in which container is filled

Solid -base of container

Liquid- above solid

Gas- above liquid

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Vector A has magnitude 8.00 mm and is in the xy-plane at an angle of 127∘ counterclockwise from the +x–axis (37∘ past the +y-axi
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-75.35°

Explanation:

Let C be the sum of the two vectors A and B. Hence, we can write the following  

A_{x} +B_{x} =C_{x} ......(1)\\A_{y} +B_{y} =C_{y} ......(2)

but since the vector C is in the -y direction, C_{x}  = 0 and C_{y} = —12 m.  

Thus  

B_{x} =-A_{x} =-[-Acos(180-127)]=(8)*cos(53)\\B_{x} =4.81m

similarly, we can determine B_{y} by rearranging equation (1)  

 B_{y} =C_{y} -A_{y} =-12m-[(8)*sin(53)\\B_{y} =-18.4m

so the magnitude of B is

B=\sqrt{B_{x}^2+B_{y}^2  } \\B=19m

Finally, the direction of B can be calculated as follows  

Ф=tan^{-1} (\frac{B_{y} }{B_{x} } )\\=-75.35

hence the vector B makes an angle of 75.35 clockwise with + x axis

8 0
4 years ago
A point charge with charge q1 = 4.00 μC is held stationary at the origin. A second point charge with charge q2 = -4.40 μC moves
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Answer:

W=0.94J

Explanation:

Electrostatic potential energy is the energy that results from the position of a charge in an electric field. Therefore, the work done to move a charge from point 1 to point 2 will be the change in electrostatic potential energy between point 1 and point 2.

This energy is given by:

U=\frac{K\left |q_1 \right |\left |q_2 \right |}{r}\\

So, the work done to move the chargue is:

W=U_1-U_2\\W=\frac{K\left |q_1 \right |\left |q_2 \right |}{r_1}-\frac{K\left |q_1 \right |\left |q_2 \right |}{r_2}\\r_1=\sqrt{((0.155 m)^2+0 m)^2}=0.115m\\r_2=\sqrt{((0.245 m)^2+(0.270 m)^2}=0.365m\\W=K\left |q_1 \right |\left |q_2 \right |(\frac{1}{r_1}-\frac{1}{r_2})\\W=8.99*10^9\frac{Nm^2}{c^2}(4.00*10^{-6}C)(4.40*10^{-6}C)(\frac{1}{0.115m}-\frac{1}{0.365})\\W=0.94J

The work is positive since the potential energy in 1 is greater than 2.

5 0
4 years ago
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