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4 years ago

The field lines around one end of a bar magnet are shown below.

Chemistry

2 answers:

bulgar [2K]4 years ago

7 0

Answer:

It is a south pole because the field lines enter the magnet at this end.

Explanation:

jus took the test :)

Vladimir [108]4 years ago

6 0

Answer:

<em>The correct option here would be the third one because it is true. </em>

Explanation:

The diagram shows a bar magnet that has lines that are parallel to each other and are pointing towards one side of the rectangle.

The side of rectangle is South Pole because the arrows that are shown in the diagram are nothing but field lines and these lines tend to start from North Pole and the arrow shows moving inside that point hence that point it would be South Pole showing field lines ending at it.

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$\\ \star\sf\longmapsto Volume=\dfrac{Mass}{Density}$

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$\\ \star\sf\longmapsto Volume=14.32mL$

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Read 2 more answers

Calculate the hight of a column of liquid glycerol in meters required to exert the same pressure as 3.02 m if CCl4?

katovenus [111]

Answer:

Approximately $3.81\; \rm m$ .

Explanation:

Look up the density $\rho$ of carbon tetrachloride, $\rm CCl_4$ , and glycerol:

Density of carbon tetrachloride: approximately $1.59\times 10^{3}\; \rm kg \cdot m^{-3}$ .
Density of glycerol: approximately $1.26\times 10^{3}\; \rm kg \cdot m^{-3}$ .

Let $g$ denote the gravitational field strength. (Typically $g \approx 9.81\; \rm N \cdot kg^{-1}$ near the surface of the earth.) For a column of liquid with a height of $h$ , if the density of the liquid is $\rho$ , the pressure at the bottom of the column would be:

$P = \rho\cdot g \cdot h$ .

The pressure at the bottom of this carbon tetrachloride column would be:

$\begin{aligned} P &= \rho \cdot g \cdot h \\ & \approx 1.59\times 10^{3} \; \rm kg \cdot m^{-3} \times 9.81\; \rm N \cdot kg^{-1} \times 3.02 \; \rm m \approx 4.71 \times 10^{4} \; \rm N \cdot m^{-2} \end{aligned}$ .

Rearrange the equation $P = \rho\cdot g \cdot h$ for $h$ :

$\displaystyle h = \frac{P}{\rho \cdot g}$ .

Apply this equation to calculate the height of the liquid glycerol column:

$\begin{aligned}h &= \frac{P}{\rho \cdot g} \\ &\approx \frac{4.71 \times 10^{4}\; \rm N \cdot m^{-2}}{1.26 \times 10^{3}\; \rm kg \cdot m^{-3} \times 9.81\; \rm N \cdot kg^{-1}} \approx 3.81\; \rm m\end{aligned}$ .

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