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MissTica
3 years ago
12

The two vectors and in fig. 3-28 have equal magnitudes of 10.0 m and the angles are 30° and 105°. find the (a) x and (b) y compo

nents of their vector sum , (c) the magnitude of , and (d) the angle makes with the positive direction of the x axis.
Physics
1 answer:
mylen [45]3 years ago
6 0

You can just use basic trigonometry to solve for the x & y components.

<span>vector a = 10cos(30) i + 10sin(30) j = <5sqrt(3), 5></span>

vector b is only slightly harder because the angle is relative to vector a, and not the positive x-axis. Anyway, this just makes vector b with an angle of 135deg to the positive x-axis.

<span>vector b = 10cos(135) i + 10sin(135) j = <-5sqrt(2), 5sqrt(2)></span>

So now we can do the questions:

r = a + b

r = <5sqrt(3)-5sqrt(2), 5+5sqrt(2)>

(a) 5sqrt(3)-5sqrt(2)

(b) 5+5sqrt(2)

(c)

|r| = sqrt( (5sqrt(3)-5sqrt(2))2 + (5+5sqrt(2))2 )

= 12.175

(d)

θ = tan-1 ( (5+5sqrt(2)) / (5sqrt(3)-5sqrt(2)) )

θ = 82.5deg

<span> </span>

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<u>Answer:</u> The volume and total pressure of the mixture is 0.876 L and 89.36 kPa respectively.

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

  • <u>For methane:</u>

Given mass of methane = 320 mg = 0.3 g     (Conversion factor:  1 g = 1000 mg)

Molar mass of methane = 16 g/mol

Putting values in equation 1, we get:

\text{Moles of methane}=\frac{0.3g}{16g/mol}=0.019mol

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Given mass of argon = 175 mg = 0.175 g

Molar mass of argon = 40 g/mol

Putting values in equation 1, we get:

\text{Moles of argon}=\frac{0.175g}{40g/mol}=0.0044mol

  • <u>For nitrogen:</u>

Given mass of nitrogen = 225 mg = 0.225 g

Molar mass of nitrogen = 28 g/mol

Putting values in equation 1, we get:

\text{Moles of nitrogen}=\frac{0.225g}{28g/mol}=0.0080mol

To calculate the volume of the mixture, we use the equation:

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We are given:

Partial pressure of argon = 12.52 kPa

Temperature = 300 K

R = Gas constant = 8.31\text{L kPa }mol^{-1}K^{-1}

n = number of moles of argon = 0.0044 moles

Putting values in equation 2, we get:

12.52kPa\times V=0.0044mol\times 8.31\text{L kPa }mol^{-1}K^{-1}\times 300K\\\\V=\frac{0.0044\times 8.31\times 300}{12.52}=0.876L

Now, calculating the total pressure of the mixture by using equation 2:

Total number of moles = [0.019 + 0.0044 + 0.0080] mol = 0.0314 mol

V= volume of the mixture = 0.876 L

Putting values in equation 2, we get:

P\times 0.876L=0.0314mol\times 8.31\text{L kPa }mol^{-1}K^{-1}\times 300K\\\\P=\frac{0.0341\times 8.31\times 300}{0.876}=89.36kPa

Hence, the volume and total pressure of the mixture is 0.876 L and 89.36 kPa respectively.

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