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3 years ago

7

The speeds of 22 particles are as follows: two at 5.30 cm/s, four at 1.40 cm/s, six at 7.14 cm/s, eight at 1.52 cm/s, two at 7.6

8 cm/s. What are (a) vavg, (b) vrms, and (c) vP?

1 answer:

Leona [35]3 years ago

8 0

Answer:

Average velocity is 3.93 cm/s

Root mean square velocity is 4.79 cm/s

Velocity peak to peak is 6.28 cm/s

Explanation:

Speed of 2 particles = 5.3 cm/s

Speed of 4 particles = 1.4 cm/s

Speed of 6 particles = 7.14 cm/s

Speed of 8 particles = 1.52 cm/s

Speed of 2 particles = 7.68 cm/s

$v_{avg}=\frac{2\times 5.3+4\times 1.4+6\times 7.14+8\times 1.52+2\times 7.68}{22}\\\Rightarrow v_{avg}=\frac{86.56}{22}\\\Rightarrow v_{avg}=3.93\ cm/s$

Average velocity is 3.93 cm/s

$v_{rms}=\sqrt{\frac{2\times 5.3^2+4\times 1.4^2+6\times 7.14^2+8\times 1.52^2+2\times 7.68^2}{22}}\\\Rightarrow v_{rms}=\sqrt{\frac{507.34}{22}}\\\Rightarrow v_{rms}=4.79\ cm/s$

Root mean square velocity is 4.79 cm/s

$v_p=7.68-1.4=6.28\ cm/s$

Velocity peak to peak is 6.28 cm/s

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A train at a constant 79.0 km/h moves east for 27.0 min, then in a direction 50.0° east of due north for 29.0 min, and then west

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Answer:

Magnitude of avg velocity, $|v_{avg}| = 18.9 km/h$

$\theta' = 56.85^{\circ}$

Given:

Constant speed of train, v = 79 km/h

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Angle, $\theta = 50^{\circ}$

Time taken in $50^{\circ}$ east of due North direction, t' = 29 min = $\frac{29}{60} h$

Time taken in west direction, t'' = 37 min = $\frac{27}{60} h$

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Displacement in $50^{\circ}$ east of due North for 29.0 min is given by:

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$\vec{s''} = - vt''hat{i} = - 79\frac{37}{60} = - 48.72\hat{i} km$

Now, the overall displacement,

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$\vec{s_{net}} = 35.5\hat{i} + 29.25\hat{i} + 24.54\hat{j} - 48.72\hat{i}$

$\vec{s_{net}} = 16.03\hat{i} + 24.54\hat{j} km$

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$v_{avg} = \frac{total displacement, \vec{s_{net}}}{total time, t}$

$v_{avg} = \frac{16.03\hat{i} + 24.54\hat{j}}{\frac{27 + 29 + 37}{60}}$

$v_{avg} = 10.34\hat{i} + 15.83\hat{j}) km/h$

Magnitude of avg velocity is given by:

$|v_{avg}| = \sqrt{(10.34)^{2} + (15.83)^{2}} = 18.9 km/h$

(b) angle can be calculated as:

$tan\theta' = \frac{15.83}{10.34}$

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