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elena55 [62]
3 years ago
5

Robebobeccoba fires a bullet from a gun while aiming at a target 149 m away. If the

Physics
1 answer:
Snezhnost [94]3 years ago
5 0

Answer:

329.6 m/s

Explanation:

The bullet has a projectile motion, consisting of two separate motions:

- A horizontal motion with constant velocity

- A free fall motion (constant acceleration) in the vertical direction

The bullet misses the target vertically by 1 m: this means that 1 m is the vertical dispalcement of the bullet. So we can find the total time of the flight by using the suvat equation

s=ut+\frac{1}{2}at^2

where

s = 1 m is the vertical displacement

u = 0 is the initial vertical velocity

a=g=9.8 m/s^2 is the acceleration due to gravity

t is the time

Solving for t,

t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2(1)}{9.8}}=0.452 s

The horizontal motion is at constant speed, which is given by

v_x = \frac{d}{t}

where

d = 149 m is the distance covered by the bullet

t = 0.452 s is the time taken

Substituting,

v_x = \frac{149}{0.452}=329.6 m/s

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The portion of earth underneath the crust containing the asthenosphere and the mesosphere is the
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Read 2 more answers
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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
  • 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
  • V_3 = The third volume of air inside the tire = 2\%V_2+V_2= 102\%V_2=1.02V_2
  • P_1 = The first pressure of air inside the tire = 1.01325\times 10^5\ Pa

<u>Assume:</u>

  • P_2 = The second pressure of air inside the tire
  • P_3 = The third pressure of air inside the tire
  • 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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