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3 years ago

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The gage pressure of an automobile tire is measured to be 210kPa before a trip and 220kPa after the trip at a location where the

atmospheric pressure is 95kPa. Assuming the volume of the tire remains constant and the air temperature before the trip is 25o C, determine the air temperature in the tire after the trip. (

1 answer:

Katarina [22]3 years ago

7 0

Answer:

$T_{f} = 312.348\,K\,(39.198\,K)$

Explanation:

Let assume that gas inside the automobile tire behaves as an ideal gas. Due to the absence of leakages, the number of moles remains constant during the trip. Air temperature can be found by using the following relation:

$\frac{P_{o}}{T_{o}}=\frac{P_{f}}{T_{f}}$

Final temperature is cleared with the expression:

$T_{f} = \frac{P_{f}}{P_{o}}\cdot T_{o}$

$T_{f} = \frac{220\,kPa}{210\,kPa}\cdot (298.15\,K)$

$T_{f} = 312.348\,K\,(39.198\,K)$

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yuradex [85]

Answer:

$v=3.564\ m.s^{-1}$

$\Delta v =2.16\ m.s^{-1}$

Explanation:

Given:

mass of John, $m_J=30\ kg$

mass of William, $m_W=30\ kg$

length of slide, $l=3\ m$

(A)

height between John and William, $h=1.8\ m$

<u>Using the equation of motion:</u>

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v_J = final velocity of John at the end of the slide

u_J = initial velocity of John at the top of the slide = 0

Now putting respective :

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$v_J=5.94\ m.s^{-1}$

<u>Now using the law of conservation of momentum at the bottom of the slide:</u>

<em>Sum of initial momentum of kids before & after collision must be equal.</em>

$m_J.v_J+m_w.v_w=(m_J+m_w).v$

where: v = velocity with which they move together after collision

$30\times 5.94+0=(30+20)v$

$v=3.564\ m.s^{-1}$ is the velocity with which they leave the slide.

(B)

frictional force due to mud, $f=105\ N$

<u>Now we find the force along the slide due to the body weight:</u>

$F=m_J.g.sin\theta$

$F=30\times 9.8\times \frac{1.8}{3}$

$F=176.4\ N$

<em><u>Hence the net force along the slide:</u></em>

$F_R=71.4\ N$

<em>Now the acceleration of John:</em>

$a_j=\frac{F_R}{m_J}$

$a_j=\frac{71.4}{30}$

$a_j=2.38\ m.s^{-2}$

<u>Now the new velocity:</u>

$v_J_n^2=u_J^2+2.(a_j).l$

$v_J_n^2=0^2+2\times 2.38\times 3$

$v_J_n=3.78\ m.s^{-1}$

Hence the new velocity is slower by

$\Delta v =(v_J-v_J_n)$

$\Delta v =5.94-3.78= 2.16\ m.s^{-1}$

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Sergeeva-Olga [200]

Answer:

0 m/s

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In this problem, a player runs the length of the 30-meter court and back. The player does this three times in 60 seconds.

As the player runs the court and returns to the original point. It would mean that the shortest path covered is 0.

Average velocity = displacement/time

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v = 0 m/s

Hence, the correct option is (1).

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Responder:

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Sphinxa [80]

Answer:

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