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2 years ago

7

Four springs are stretched to the same distance from the equilibrium position. The spring constants are listed in the table. A 2

column table with 4 rows. The first column is labeled spring with entries W, X, Y, Z. The second column is labeled spring constant in newtons per meter with entries 24, 35, 22, 15. Which lists the springs based on the amount of elastic potential energy, from greatest to least?

1 answer:

RoseWind [281]2 years ago

4 0

Answer:

X,W,Y,Z

Explanation:

It is what it is. and i took the edge test

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A thin uniform rod (length = 1.2 m, mass = 2.0 kg) is pivoted about a horizontal, frictionless pin through one end of the rod. (

Anika [276]

Answer:

$a=9.8 rad/s^{2}$

Explanation:

Torque, $\tau$ is given by

$\tau=Fr$ where F is force and r is perpendicular distance

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Torque, $\tau$ can also be found by

$\tau=Ia$ where I is moment of inertia and a is angular acceleration

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Making a the subject, $a=\frac {Fr}{I}=\frac {mgr}{I}$ and already I is given as

$I=\frac {mL^{2}}{3} and r is 0.5Lcos\theta$ hence

$a=\frac {0.5mgLcos\theta}{1/3 mL^{2}}$

$a=\frac {3gcos\theta}{2L}$

Taking g as 9.81, $\theta$ is given as 37 and L is 1.2

$a=\frac {3*9.81cos37}{2*1.2}=9.7932679419$

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What type of clouds are associated with low pressure?

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Cumulus and cumulonimbus<span />

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As an airplane accelerates in level flight, what parameter must the pilot adjust to keep level and balance Lift = Weight?

Y_Kistochka [10]

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2 years ago

A child has an ear canal that is 1.3 cm long. Assume the speed of sound is v = 344 m/s.

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Answer:

The frequencies are $(f, f_1) = (6615.4 \ Hz , 19846.2\ Hz)$

Explanation:

From the question we are told that

The length of the ear canal is $l = 1.3 \ cm =\frac{1.3}{100} = 0.013 \ m$

The speed of sound is assumed to be $v_s = 344 \ m/s$

Now taking look at a typical ear canal we see that we assume it is a closed pipe

Now the fundamental harmonics for the pipe(ear canal) is mathematically represented as

$f = \frac{v_s}{4 * l }$

substituting values

$f = \frac{344}{4 * 0.013 }$

$f = 6615.4 \ Hz$

Also the the second harmonic for the pipe (ear canal) is mathematically represented as

$f_1 = \frac{3v_s}{4 * l}$

substituting values

$f_1 = \frac{3 * 344}{4 * 0.013}$

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Given that sound would be loudest in the pipe at the frequency, it implies that the child will have an increased audible sensitivity at this frequencies

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