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USPshnik [31]
3 years ago
5

A dolphin emits ultrasound at 100kHz and uses the timing of reflections to determine the position of objects in the water. What

is the wavelength of this ultrasound? Assume that temperature of water is 20C.
Physics
1 answer:
Bond [772]3 years ago
6 0
To answer the question above, 
First, let's determine the speed of sound under the water.
S
peed of sound in water at 20oC = 1482m/s 

<span>So λ = v/f = 1482m/s/(100kHz) I'm assuming your frequency is 100kHz </span>

<span>So λ would be = 0.0148m
I hope my answer helped you.</span>
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likoan [24]

The formula relating acceleration and angular velocity is:

a = ω^2 r

where a is acceleration, ω is angular velocity and r is radius

But the angular velocity ω is constant all throughout the disk therefore:

a1 / r1 = a2 / r2

 

So at points:

<span>r1 = 0.0130 m     -> a1 = 393 m/s^2</span>

<span>r2 = 0.0884 m     -> a2 = ?</span>

 

393 / 0.0130 = a2 / 0.0884

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6 0
3 years ago
Why are metals good conductors of both heat and electricity?
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They are both good conductors if both heat and electricity due to the sea of delocalized electrons that is floating around without getting bonded to an atom.

Such electrons can flow around freely to conduct heat and electricity.
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At what speed does a falling hailstone travel? Does the speed depend on the distance that the hailstone falls?
Anton [14]

Answer:The speed if hailstone dependly largely on its size. A hailstone with a diameter of 0.39 inches,falls wit a speed of 20mph while a hailstone with 3.1 inches in diameter falls at a speed of 110mph.

No speed does not depend on the distance that the hailstone falls.

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3 years ago
Which event demonstrates electromagnetic waves transferring energy?
Citrus2011 [14]

Answer:

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Explanation:

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4 0
3 years ago
Read 2 more answers
A small block of mass 20.0 grams is moving to the right on a horizontal frictionless surface with a speed of 0.68 m/s. The block
Usimov [2.4K]

Answer:

a) v'=-0.227\ m.s^{-1}

b) v=1.36\ m.s^{-1}

Explanation:

Given:

mass of the lighter block, m'=0.02\kg

velocity of the lighter block, u'=0.68\ m.s^{-1}

mass of the heavier block, m=0.04\ kg

velocity of the heavier block, u=0\ m.s^{-1}

a)

Using conservation of linear momentum:

m'.u'+m.u=m'.v'+m.v

where:

v'= final velocity of the lighter block

v= final velocity of the heavier block

m'.u'=m'.v'+m.v

m'(u'-v')=m.v ........................(1)

Since kinetic energy is conserved in elastic collision:

\frac{1}{2}m'.u'^2=\frac{1}{2}m'.v'^2+\frac{1}{2}m.v^2

m'(u'^2-v'^2)=m.v^2

m'(u'-v')(u'+v')= m.v^2

divide the above equation by eq. (1)

v=u'+v' .............................(2)

now we substitute the value of v from eq. (2) in eq. (1)

m'(u'-v')=m(u'+v')

\frac{m'+m}{m'-m} =\frac{u'}{v'}

\frac{0.02+0.04}{0.02-0.04} =\frac{0.68}{v'}

v'=-0.227\ m.s^{-1} (negative sign denotes that the direction is towards left)

b)

now we substitute the value of v' from eq. (2) in eq. (1)

m'(u'-v+u')=m.v

2m'.u'=(m-m')v

2\times 0.02\times 0.68=(0.04-0.02)\times v

v=1.36\ m.s^{-1}

6 0
3 years ago
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