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2 years ago

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Your ear is capable of differentiating sounds that arrive at the ear just 1.00 ms apart. What is the minimum distance (in centim

eters) between two speakers that produce sounds that arrive at noticeably different times on a day when the speed of sound is 337 m/s?

1 answer:

Andrej [43]2 years ago

5 0

Answer:Your ear is capable of differentiating sounds that arrive at the ear just 1.00 ms apart. What is the minimum distance between two speakers that produce sounds that arrive at noticeably different times on a day when the speed of sound is 340 m/s

Explanation:

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Anna [14]

Answer:(a)891.64 N

(b)0.7

Explanation:

Mass of crate $m=100 kg$

Crate slows down in $s=1.5 m$

initial speed $u=1.77 m/s$

inclination $\theta =30^{\circ}$

From Work-Energy Principle

Work done by all the Forces is equal to change in Kinetic Energy

$W_{friction}+W_{gravity}=\frac{1}{2}mv_i^2-\frac{1}{2}mv_f^2$

$W_{gravity}=mg(0-h)=mgs\sin \theta$

$W_{gravity}=-mgs\sin \theta$

$W_{gravity}=-100\times 9.8\times 1.5\sin 30=-735 N$

change in kinetic energy $=\frac{1}{2}\times 100\times 1.77^2=156.64 J$

$W_{friction}=156.64+735=891.645$

(b)Coefficient of sliding friction

$f_r\cdot s=W_{friciton}$

$891.645=f_r\times 1.5$

$f_r=594.43 N$

and $f_r=\mu mg\cos \theta$

$\mu 100\times 9.8\times \cos 30=594.43$

$\mu =0.7$

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A student bikes to school by traveling first dN = 1.00 miles north, then dW = 0.600 miles west, and finally dS = 0.200 miles sou

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Refer to the diagram shown below.

Define unit vectors along the x and y axes as respectively $\hat{i} \, and \, \hat{j}.$

Then the three successive displacements, written in component form, are respectively
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\vec{dW} = -0.6 \, \hat{i} \\ \vec{dS} = -0.2 \, \hat{j}$

The total displacement for the first leg of the trip is
$\vec{d} = \vec{dN} + \vec{dW} + \vec{dS} \\ \vec{d}= 1.0\hat{j}-0.6\hat{i}-0.2\hat{j} \\ \vec{d}=-0.6\hat{i}+0.8\hat{j}$

Answer:
$\vec{d} = -0.6\hat{i}+0.8\hat{j}$ or (-0.6, 0.8)

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T1 = ω1/α.............

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When laser light of wavelength 632.8 nm passes through a diffraction grating, the first bright spots occur at ± 17.8 ∘ from the

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Look on this website http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/sinslit.html

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Given:
Ca = 3Cb (1)
where
Ca = heat capacity of object A
Cb = heat capacity f object B

Also,
Ta = 2Tb (2)
where
Ta = initial temperature of object A
Tb = initial temperature of object B.

Let
Tf = final equilibrium temperature of both objects,
Ma = mass of object A,
Mb = mass of object B.

Assuming that all heat exchange occurs exclusively between the two objects, then energy balance requires that
Ma*Ca*(Ta - Tf) = Mb*Cb*(Tf - Tb) (3)

Substitute (1) and (2) into (3).
Ma*(3Cb)*(2Tb - Tf) = Mb*Cb*(Tf - Tb)
3(Ma/Mb)*(2Tb - Tf) = Tf - Tb

Define k = Ma/Mb, the ratio f the masses.
Then
3k(2Tb - Tf) = Tf - Tb
Tf(1+3k) = Tb(1+6k)
Tf = [(1+6k)/(1+3k)]*Tb

Answer:
$T_{f} =( \frac{1+6k}{1+3k} )T_{b}= \frac{1}{2}( \frac{1+6k}{1+3k})T_{a}$
where
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