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2 years ago

5

Suppose a plot of inverse wavelength vs frequency has slope equal to 0.119, what is the speed of sound traveling in the tube to

2 decimal places?

1 answer:

strojnjashka [21]2 years ago

5 0

Answer:

8.40 m/s

Explanation:

Slope of the plot is 0.119

Slope of a plot is given by the change in y direction divided by the change in x direction

Here, the y axis represents inverse wavelength and the x axis represents frequency.

f = Frequency (Hz, assumed)

v = Phase velocity (m/s, assumed)

λ = Wavelength (m, assumed)

So, slope

$m=\frac{\frac{1}{\lambda}}{f}$

Now,

$\lambda=\frac{v}{f}\\\Rightarrow \lambda^{-1}=\frac{f}{v}$

$\\\Rightarrow m=\frac{\frac{f}{v}}{f}\\\Rightarrow 0.119=\frac{1}{v}\\\Rightarrow v=\frac{1}{0.119}\\\Rightarrow v=8.40\ m/s$

The speed of sound travelling in the tube is 8.40 m/s

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A Thomson's gazelle can run at very high speeds, but its acceleration is relatively modest. A reasonable model for the sprint of

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Answer:

a) 27.3 m/s

b) 3.78 s

c) 10.576 s

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

a)

$v=u+at\\\Rightarrow v=0+4.2\times 6.5\\\Rightarrow v=27.3\ m/s$

Top speed of the gazelle is 27.3 m/s

b)

$s=ut+\frac{1}{2}at^2\\\Rightarrow 30=0t+\frac{1}{2}\times 4.2\times t^2\\\Rightarrow t=\sqrt{\frac{30\times 2}{4.2}}\\\Rightarrow t=3.78\ s$

The gazelle would take 3.78 seconds to win a 30 m race.

c)

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=0\times 6.5+\frac{1}{2}\times 4.2\times 6.5^2\\\Rightarrow s=88.725\ m$

The gazelle would go 88.725 m with the acceleration after which there will be no acceleration.

Time = Distance / Speed

$\text{Time}=\frac{200-88.725}{27.3}=4.076\ s$

Total time taken by the gazelle to cover 200 m would be 6.5+4.076 = 10.576 seconds

4 0

2 years ago

A 3.50-meter length of wire with a cross-sectional area of 3.14 × 10-6 meter2 is at 20° Celsius. If the wire has a resistance of

maks197457 [2]

Answer:

$5.6\times 10^{-8}\ Ohm.m$

Explanation:

Resistivity is given by $\rho=\frac {AR}{L}$ where A is cross-sectional area, R is resistance, L is the length and $\rho$ is the reistivity. Substituting 0.0625 for R, 3.14 × 10-6 for A and 3.5 m for L then the resistivity is equivalent to

$\rho=\frac {3.14\times 10^{-6}\times 0.0625}{3.5}=5.60714285714285714285714285714285714285\times 10^{-8}\approx 5.6\times 10^{-8}\ Ohm.m$

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2 years ago

A skier is moving down a snowy hill with an acceleration of 0.40 m/s2. The angle of the slope is 5.0∘ to the horizontal. What is

kirill115 [55]

Answer:

1.25377 m/s²

Explanation:

m = Mass of person

g = Acceleration due to gravity = 9.81 m/s²

$\mu$ = Coefficient of friction

$\theta$ = Slope

From Newton's second law

$mgsin\theta-f=ma\\\Rightarrow mgsin\theta-\mu mgcos\theta=ma\\\Rightarrow \mu=\frac{gsin\theta-a}{gcos\theta}\\\Rightarrow \mu=\frac{9.81\times sin5-0.4}{9.81\times cos5}\\\Rightarrow \mu=0.04655$

Applying $\mu$ to the above equation and $\theta=10^{\circ}$

$mgsin\theta-\mu mgcos\theta=ma\\\Rightarrow a=gsin\theta-\mu gcos\theta\\\Rightarrow a=9.81\times sin10-0.04655\times 9.81\times cos10\\\Rightarrow a=1.25377\ m/s^2$

The acceleration of the same skier when she is moving down a hill is 1.25377 m/s²

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Gibbons, small Asian apes, move by brachiation, swinging below a handhold to move forward to the next handhold. A 8.6 kg gibbon

fiasKO [112]

Answer:

Explanation:

i dont knoe sorry

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2 years ago

A 150-N box is being pulled horizontally in a wagon accelerating uniformly at 3.00 m/s2. The box does not move relative to the w

Zepler [3.9K]

Answer:

Frictional force, F = 45.9 N

Explanation:

It is given that,

Weight of the box, W = 150 N

Acceleration, $a=3\ m/s^2$

The coefficient of static friction between the box and the wagon's surface is 0.6 and the coefficient of kinetic friction is 0.4.

It is mentioned that the box does not move relative to the wagon. The force of friction is equal to the applied force. Let a is the acceleration. So,

$m=\dfrac{W}{g}$

$m=\dfrac{150}{9.8}$

$m=15.3\ kg$

Frictional force is given by :

$F=ma$

$F=15.3\times 3$

F = 45.9 N

So, the friction force on this box is closest to 45.9 N. Hence, this is the required solution.

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2 years ago