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1 year ago

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A person walks due south from point A for 500 yards and then due west for 300 yards, arriving at point B. Answer the following q

uestions using complete sentences. 2 points for each correct answer and 2 points for each correct explanation. i. What is the person's displacement from the starting point

1 answer:

Pepsi [2]1 year ago

5 0

The total displacement of the person walking from point A to point B is 300 yards.

As shown in the figure we can conclude that the required method to calculate the total displacement is the Pythagoras theorem.

<h3>Pythagoras theorem in brief :</h3>

According to the Pythagorean Theorem, the square that represents the hypotenuse, or side of a right triangle that faces the right angle, is equal to the total of the squares on the triangle's legs.(or, in popular algebraic notation, $a^2 + b^2 = c^2$ ).

<h3>Calculation: </h3>

Let,

a = 500

b= 300

Hence by using Pythagoras' theorem

Total displacement of the person = $\sqrt{500^{2} + 300^{2} }$ = $\sqrt{900000}$ = $300$

Thus the total displacement of the person from starting point is 300 yards.

Learn more about the displacement examples here:

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In this problem, we have two electrons, so:

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(b) $4.19\times 1065\ Pa$

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<u>Given:</u>

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$T_2$ = The second temperature of air inside the tire = $46^\circ C =(273+46)\ K= 319\ K$

$T_3$ = The third temperature of air inside the tire = $85^\circ C =(273+85)\ K=358 \ K$

$V_1$ = The first volume of air inside the tire

$V_2$ = The second volume of air inside the tire = $30\% V_1 = 0.3V_1$

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<u>Assume:</u>

$P_2$ = The second pressure of air inside the tire

$P_3$ = The third pressure of air inside the tire
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Since the amount pof air inside the tire remains the same, this means the number of moles of air in the tire will remain constant.

Using ideal gas equation, we have

$PV = nRT\\\Rightarrow \dfrac{PV}{T}=nR = constant\,\,\,(\because n,\ R\ are\ constants)$

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Using the above equation for this part of compression in the air, we have

$\therefore \dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}\\\Rightarrow P_2 = \dfrac{V_1}{V_2}\times \dfrac{T_2}{T_1}\times P_1\\\Rightarrow P_2 = \dfrac{V_1}{0.3V_1}\times \dfrac{319}{283}\times 1.01325\times 10^5\\\Rightarrow P_2 =3.81\times 10^5\ Pa$

Hence, the pressure in the tire after the compression is $3.81\times 10^5\ Pa$ .

Part (b):

Again using the equation for this part for the air, we have

$\therefore \dfrac{P_2V_2}{T_2}=\dfrac{P_3V_3}{T_3}\\\Rightarrow P_3 = \dfrac{V_2}{V_3}\times \dfrac{T_3}{T_2}\times P_2\\\Rightarrow P_3 = \dfrac{V_2}{1.02V_2}\times \dfrac{358}{319}\times 3.81\times 10^5\\\Rightarrow P_3 =4.19\times 10^5\ Pa$

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