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zavuch27 [327]
3 years ago
15

What happens to the volume of sound as the amplitude gets larger?

Physics
2 answers:
tino4ka555 [31]3 years ago
7 0
The sound is perceived as louder if the amplifier increases, and softer if the amplitude decreases.
777dan777 [17]3 years ago
5 0
The sound gets louder :)
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ANSWERED: An object with a massive 5 kg is thrown with an initial velocity of 6 m/s. If the object has an initial height of 10 m
vodka [1.7K]

Explanation:

The velocity is not needed. (its needed to find kinetic energy but its irrelevant to the question)

Potential energy = Mgh

Mass=5kg, height=10meters, g=9.8

P.e=5×9.8×10

=5×98

=490 joules

Hope it helps

:)

4 0
3 years ago
A child blows a leaf from rest straight up in the air. The leaf has a constant upward acceleration of magnitude 1.0\,\dfrac{\tex
Novosadov [1.4K]

Answer:

Explanation:

Given

Acceleration a = 1.0m/s²

Displacement S = 1.0m

Required

Time t taken by the leaf to displace

Using the equation of motion

S = ut+1/2at²

Substitute

1.0 = 0+1/2(1)t²

1 = t²/2

Cross multiply

t² = 2

t = ±√2

t = 1.41secs

It takes the leaf to 1.41s to displace by 1m upward

6 0
3 years ago
The sound of frequency greater than 20,000 Hz is called ....
lapo4ka [179]

Answer:

ultra sound... follow me

3 0
4 years ago
Two loudspeakers are placed 1.8 m apart. They play tones of equal frequency. If you stand 3.0 m in front of the speakers, and ex
Ahat [919]

The question is incomplete. The complete question is :

Two loudspeakers are placed 1.8 m apart. They play tones of equal frequency. If you stand 3.0 m in front of the speakers, and exactly between them, you hear a minimum of intensity. As you walk parallel to the plane of the speakers, staying 3.0 m away, the sound intensity increases until reaching a maximum when you are directly in front of one of the speakers. The speed of sound in the room is 340 m/s.

What is the frequency of the sound?

Solution :

Given :

The distance between the two loud speakers, d = 1.8 \ m

The speaker are in phase and so the path difference is zero constructive interference occurs.

At the point D, the speakers are out of phase and so the path difference is $=\frac{\lambda}{2}$

Therefore,

$AD-BD = \frac{\lambda}{2}

$\sqrt{(1.8)^2+(3)^2-3} =\frac{\lambda}{2}$

$\lambda = 2 \times 0.4985$

$\lambda = 0.99714 \ m$

Thus the frequency is :

$f=\frac{v}{\lambda}$

$f=\frac{340}{0.99714}$

f=340.9744 Hz

3 0
3 years ago
The terminals of a 0.70 Vwatch battery are connected by a 80.0-m-long gold wire with a diameter of 0.200 mm What is the current
Molodets [167]

Answer:

I = 11.26 mA

Explanation:

given,

V = 0.7 V           length = 80 m

diameter   = 0.2 mm = 0.02 cm

radius = 0.01 × 10⁻² m

resistance = \dfarc{\rho l}{A}

ρ for gold wire = 2.44 × 10⁻⁸ ohm-m at 20 °C

A = cross sectional area = π r² = π (0.01 × 10⁻² )²

    = 31.4× 10⁻⁹ m²

resistance = \dfarc{2.44\times 10^{-8} \times 80}{31.4\times 10^{-9}}

R = 62.165 Ω

I = \dfrac{V}{R}

I = \dfrac{0.7}{62.165}

I = 11.26 mA

8 0
3 years ago
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