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3 years ago

13

Sam's bike tire contains 15 units of air particles and has a volume of 160mL. Under these conditions the pressure reads 13 psi.

The tire develops a leak. Now it contains 10 units of air particles and has contracted to a volume of 150mL. What would the tire pressure be now?

1 answer:

vichka [17]3 years ago

8 0

Use ideal gas equation, with T constant.

pV =nRT => pV / n = RT = constant

n = K* [units of particles]

pV / [units of particles] = constant

13 psi * 160 mL / 15 units = p * 150 mL / 10 units =>

=> p = [13psi*160mL/15units]*[10units/150mL] = 9.2 psi

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Answer:

The extension of the second wire is $e_2 = 0.0024 \ m = 2.4 mm$

Explanation:

From the question we are told that

The length of the wire is $L = 3 \ m$

The elongation of the wire is $e = 1.2mm = \frac{1.2}{1000} = 0.0012 m$

The tension is $F = 200 \ N$

The length of the second wire is $L_2 = 6 \ m$

Generally the Young's modulus(Y) of this material is

$Y = \frac{stress}{strain }$

Where $stress = \frac{F}{A}$

Where A is the area which is evaluated as

$A = \pi r^2$

and $strain = \frac{extention}{length} = \frac{e}{L}$

So

$Y = \frac{\frac{F}{\pi r^2 } }{ \frac{e}{L} }$

Since the wire are of the same material Young's modulus(Y) is constant

So we have

$\frac{F * L }{r^2 e} = \pi * Y = constant$

$F * L = constant * r^2 e$

Now the ration between the first and the second wire is

$\frac{F_1}{F_2} * \frac{L_1}{L_2} = \frac{r*2_1}{r^2} * \frac{e_1}{e_2}$

Since tension , radius are constant

We have

$\frac{L_1}{L_2} = \frac{e_1}{e_2}$

substituting values

$\frac{3}{6} = \frac{0.0012}{e_2}$

$0.5 e_2 = 0.0012$

$e_2 = \frac{ 0.0012 }{0.5}$

$e_2 = 0.0024 \ m = 2.4 mm$

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Answer:

a) 17.8 m/s

b) 28.3 m

Explanation:

Given:

angle A = 53.0°

sinA = 0.8

cosA = 0.6

width of the river,d = 40.0 m,

the far bank was 15.0 m lower than the top of the ramp h = 15.0 m,

The river itself was 100 m below the ramp H = 100 m,

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putting values h=15 m, v=0.8

$-15 = 0.8vt - 4.9t^2$ ............. (1)

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so v×t = d/0.6 = 40/0.6

plug it into (1) and get

$-15 = 0.8\times40/0.6 - 4.9t^2$

solving for t we get

t = 3.734 s

also, v = (40/0.6)/t = 40/(0.6×3.734) = 17.8 m/s

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t = 5.30 s

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