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dmitriy555 [2]
3 years ago
10

ACCORDING TO NEWTON'S THIRD LAW EVERY ACTION HAS EQUAL AND OPPOSITE REACTION BUT THEN WHY DON'T WE FLY WHEN WE FART??​

Physics
1 answer:
pav-90 [236]3 years ago
7 0

Answer:

Your fart only has so much force, not nearly enough to launch you into oblivion. Your fart and you still exert a force onto each other, so I guess, hypothetically, you could fly if you really, really try hard enough. Just make sure you don't try too hard and prolapse as a result :)

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An automobile tire is inflated with air originally at 10.0°C and normal atmospheric pressure. During the process, the air is com
solong [7]

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.

8 0
3 years ago
A car accelerates from rest, and travels 400 m in 3.5 seconds. If
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Answer:

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3 0
3 years ago
I need the answer asap everyone have a good day bye
Slav-nsk [51]

Im pretty sure its A cuz is closer to the earth.

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During the first stage of a multistage rocket, a satellite is:
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B. sent through the atmosphere 

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Imagine you are riding on a yacht in the ocean and traveling at 20 mph. You then hit a golf ball at 100 mph from the deck of the
Whitepunk [10]

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As per classical mechanics, the velocity of light will be different in different frame of reference. The light moves in the ether medium which exists every where in the entire universe.

Let us consider a body which moves with a velocity v. Let light is coming along the direction of the body. As per classical mechanics,the velocity of light with respect to the body will be [ c-v].

Let us consider that light is coming from opposite direction. Hence, the velocity of light with respect to the observer will be c+v.

From above we see that velocity of light is different in both the cases which is wrong.

As per Einstein's special theory of relativity, the velocity of light will be same in every frame of reference i.e c=300000 km/s.

As per the question ,the space craft is moving with a velocity 0.1 c.

We are asked to calculate the velocity of the light with respect to an observer present in Mars.

Considering Einstein's theory of relativity, the velocity of light will be c [300000 km/s] with respect to the person in Mars.

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