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PSYCHO15rus [73]
3 years ago
12

WILL GIVE BRAINLIEST!!!

Physics
1 answer:
kumpel [21]3 years ago
7 0

Answer:

72.53 mi/hr

Explanation:

From the question given above, the following data were obtained:

Vertical distance i.e Height (h) = 8.26 m

Horizontal distance (s) = 42.1 m

Horizontal velocity (u) =?

Next, we shall determine the time taken for the car to get to the ground.

This can be obtained as follow:

Height (h) = 8.26 m

Acceleration due to gravity (g) = 9.8 m/s²

Time (t) =?

h = ½gt²

8.26 = ½ × 9.8 × t²

8.26 = 4.9 × t²

Divide both side by 4.9

t² = 8.26 / 4.9

Take the square root of both side by

t = √(8.26 / 4.9)

t = 1.3 s

Next, we shall determine the horizontal velocity of the car. This can be obtained as follow:

Horizontal distance (s) = 42.1 m

Time (t) = 1.3 s

Horizontal velocity (u) =?

s = ut

42.1 = u × 1.3

Divide both side by 1.3

u = 42.1 / 1.3

u = 32.38 m/s

Finally, we shall convert 32.38 m/s to miles per hour (mi/hr). This can be obtained as follow:

1 m/s = 2.24 mi/hr

Therefore,

32.38 m/s = 32.38 m/s × 2.24 mi/hr / 1 m/s

32.38 m/s = 72.53 mi/hr

Thus, the car was moving at a speed of

72.53 mi/hr.

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Answer:

(a) 3.81\times 10^5\ Pa

(b) 4.19\times 1065\ Pa

Explanation:

<u>Given:</u>

  • T_1 = The first temperature of air inside the tire = 10^\circ C =(273+10)\ K =283\ K
  • T_2 = The second temperature of air inside the tire = 46^\circ C =(273+46)\ K= 319\ K
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<u>Assume:</u>

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  • 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)

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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