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DENIUS [597]
2 years ago
6

If you wanted the pitch of a horn to drop relative to an observer, which way would you move the horn, relative to where that obs

erver is located?
A.Leave it in the same place and play the horn more softly


B.Away from the observer


C.Towards the observer
Physics
2 answers:
mel-nik [20]2 years ago
5 0

Answer: The correct option is B

Explanation:

In Doppler effect, there is an apparent change in the frequency when there is a relative motion between the source and the listener.

Pitch depends on the frequency.

When the source and the listener are moving away from each other then there is decrease in the frequency. In this case, the pitch will be lower.

In the given problem, to drop the pitch of the horn relative to an observer, you would move the horn away from the observer.

Therefore, the correct answer is "Away from the observer".

Ratling [72]2 years ago
4 0
(B) Away from the observer. 
The closer- the higher
The farther- the lower

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What instrument can be used to show the shape of a sound wave?
mafiozo [28]
"Oscilloscope"  can be used to show the shape of a sound wave

Hope this helps!
7 0
3 years ago
A wire of Nichrome (a nickel-chromium-iron alloy commonly used in heating elements) is 1.3 m long and 1.6 mm2 in cross-sectional
fredd [130]

Answer: 1.3 *10^6 Ω*m

Explanation: In order to explain this problem we have to use the following expression for the resistence:

R=L/(σ*A) where L and A are the length and teh area for the wire, respectively. σ is the conductivity of teh Nichrome.

Then, from mteh OHM law we have V=R*I so R=V/I=2/3.2=0.625 Ω

Finally we have:

σ=L/(R*A)=1.3/(0.625*1.6*10^-6)=1.3*10^6 Ω*m

5 0
3 years ago
The plates of a parallel-plate capacitor are 2.50 mm apart, and each carries a charge of magnitude 80.0 nC. The plates are in va
bonufazy [111]

Answer:

10000 V

0.00225988700565 m²

8\times 10^{-12}\ F

Explanation:

E = Electric field = 4\times 10^6\ V/m

d = Gap = 2.5 mm

Q = Charge = 80 nC

\epsilon_0 = Permittivity of free space = 8.85\times 10^{-12}\ F/m

Potential difference is given by

V=Ed\\\Rightarrow V=4\times 10^6\times 2.5\times 10^{-3}\\\Rightarrow V=10000\ V

The potential difference between the plates is 10000 V

Area is given by

A=\dfrac{Q}{\epsilon_0E}\\\Rightarrow A=\dfrac{80\times 10^{-9}}{8.85\times 10^{-12}\times 4\times 10^6}\\\Rightarrow A=0.00225988700565\ m^2

The area of the plate is 0.00225988700565 m²

Capacitance is given by

C=\dfrac{\epsilon_0A}{d}\\\Rightarrow C=\dfrac{8.85\times 10^{-12}\times 0.00225988700565}{2.5\times 10^{-3}}\\\Rightarrow C=8\times 10^{-12}\ F

The capacitance is 8\times 10^{-12}\ F

4 0
3 years ago
A lightweight vertical spring of force constant k has its lower end mounted on a table. You compress the spring by a distance d,
shusha [124]

Answer:

v=d\sqrt{\frac{k}{m}}

Explanation:

In order to solve this problem, we can do an analysis of the energies involved in the system. Basically the addition of the initial potential energy of the spring and the kinetic energy of the mass should be the same as the addition of the final potential energy of the spring and the kinetic energy of the block. So we get the following equation:

U_{0}+K_{0}=U_{f}+K_{f}

In this case, since the block is moving from rest, the initial kinetic energy is zero. When the block loses contact with the spring, the final potential energy of the spring will be zero, so the equation simplifies to:

U_{0}=K_{f}

The initial potential energy of the spring is given by the equation:

U_{0}=\frac{1}{2}kd^{2}

the Kinetic energy of the block is then given by the equation:

K_{f}=\frac{1}{2}mv_{f}^{2}

so we can now set them both equal to each other, so we get:

=\frac{1}{2}kd^{2}=\frac{1}{2}mv_{f}^{2}

This new equation can be simplified if we multiplied both sides of the equation by a 2, so we get:

kd^{2}=mv_{f}^{2}

so now we can solve this for the final velocity, so we get:

v=d\sqrt{\frac{k}{m}}

6 0
3 years ago
A string of length 100 cm is held fixed at both ends and vibrates in a standing wave pattern. The wavelengths of the constituent
azamat

The wavelengths of the constituent travelling waves CANNOT be 400 cm.

The given parameters:

  • <em>Length of the string, L = 100 cm</em>

<em />

The wavelengths of the constituent travelling waves is calculated as follows;

L = \frac{n \lambda}{2} \\\\n\lambda = 2L\\\\\lambda = \frac{2L}{n}

for first mode: n = 1

\lambda = \frac{2\times 100 \ cm}{1} \\\\\lambda = 200 \ cm

for second mode: n = 2

\lambda = \frac{2L}{2} = L = 100 \ cm

For the third mode: n = 3

\lambda = \frac{2L}{3} \\\\\lambda = \frac{2 \times 100}{3} = 67 \ cm

For fourth mode: n = 4

\lambda = \frac{2L}{4} \\\\\lambda = \frac{2 \times 100}{4} = 50  \ cm

Thus, we can conclude that, the wavelengths of the constituent travelling waves CANNOT be 400 cm.

The complete question is below:

A string of length 100 cm is held fixed at both ends and vibrates in a standing wave pattern. The wavelengths of the constituent travelling waves CANNOT be:

A. 400 cm

B. 200 cm

C. 100 cm

D. 67 cm

E. 50 cm

Learn more about wavelengths of travelling waves here: brainly.com/question/19249186

5 0
2 years ago
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