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Komok [63]
3 years ago
10

Part B: Quantitative

Physics
1 answer:
Feliz [49]3 years ago
4 0

Answer:

(a)

\triangle v=-8\ m/s\\\triangle mv=-56\ kg.m/s

(b)

5600 N

Explanation:

Change in velocity, \triangle v is expressed as

\triangle v= v_f -v_i

Where v represent the velocity and subscripts f and i represent final and initial respectively. Given that the ball finally comes to rest, its final velocity is taken as 0. Substituting 0 for final velocity and the given figure of 8 m/s for initial velocity then the change in velocity is given by

\triangle v=0-8=-8\ m/s

To find the change in momentum, m\triangle v then we substitute 7 kg for m and -8 m/s for \triangle v therefore \triangle\ v=7 Kg\times -8 m/s=-56\ Kg.m/s

(b)

The impact force, F is equivalent to ma where m is the mass of the ball and a is the acceleration. Here, acceleration is given by dividing the change in velocity by time ie

a=\frac {\triangle v}{t}=\frac { v_f -v_i}{t}

Substituting t with 0.05 s then a=\frac {\triangle v}{t}=\frac { v_f -v_i}{t}=\frac {-8}{0.01}=-800 m/s^{2}

Since F=ma then substituting m with 7 Kg we get that F=7*-800=-5600 N

Therefore, the impact force is equivalent to 5600 N

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<u>Given:</u>

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  • n = number of moles of air

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)

Part (a):

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

Hence, the pressure in the tire after the car i driven at high speed is 4.19\times 10^5\ Pa.

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