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Arada [10]
3 years ago
10

In general, as the energy of a sound wave increases, A. the sound gets softer and then louder. B. the sound gets louder. C. the

sound always remains the same. D. the sound gets softer.
Physics
2 answers:
agasfer [191]3 years ago
8 0
<h3><u>Answer;</u></h3>

B. the sound gets louder

<h3><u>Explanation;</u></h3>
  • <em><u>Sound waves are examples of mechanical longitudinal waves.</u></em> Mechanical waves because they require a material medium for transmission. Longitudinal waves because the vibration of particles in sound waves is parallel to the direction of wave motion.
  • <em><u>As the energy of a sound wave increases, then the sound becomes louder.</u></em> The higher the energy of a wave the higher the amplitude of sound waves. Therefore, <em><u>greater amplitude of sound waves means the waves possess higher energy and thus greater intensity and therefore, the sound will be louder.</u></em>
Vadim26 [7]3 years ago
5 0
The sound gets louder, like hitting a metal pole on a thin metal beam, the energy increases as the sound does.
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spayn [35]

Answer:

10

Explanation:

(r) = <10 cos 6t, 10 Sin 6t>

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The magnitude of vector r is given by

r = \sqrt{(10 Cos 6t)^{2}+(10 Sin 6t)^{2}}

r =10 \sqrt{(Cos^{2} 6t)+(Sin^{2} 6t)

r = 10        

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Suppose the coefficient of kinetic friction between mA and the plane in the figure(Figure 1) is μk = 0.15, and that mA=mB=2.7kg.
luda_lava [24]
A ) 
T = mB g + mB a
T + mA a - mA g sin 35° = (Mi) mA g cos 35°
------------------------------------------------------------
T = 2.7 · 9.81  + 2.7 a
T = 26.487 + 2.7 a
26.487 + 2.7 a + 2.7 a - 2.7 · 9.81 · 0.574 = 0.15 · 2.7 · 9.81 · 0.819
5.4 a + 26.487 - 15.2023 = 3.2539
5.4 a = 8.0296
a = 1.487 ≈ 1.5 m/s²
B )
T = 2,7 · 9.81 = 26.487 
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11.2835 = (Mi) · 21.69
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4 0
3 years ago
The quantity of charge Q in coulombs (C) that has passed through a point in a wire up to time t (measured in seconds) is given b
Mnenie [13.5K]

Explanation:

The quantity of charge Q in coulombs (C) that has passed through a point in a wire up to time t (measured in seconds) is given by :

Q=t^3-2t^2+4t+4

We need to find the current flowing. We know that the rate of change of electric charge is called electric current. It is given by :

I=\dfrac{dQ}{dt}\\\\I=\dfrac{d(t^3-2t^2+4t+4)}{dt}\\\\I=3t^2-4t+4

At t = 1 s,

Current,

I=3(1)^2-4(1)+4\\\\I=3\ A

So, the current at t = 1 s is 3 A.

For lowest current,

\dfrac{dI}{dt}=0\\\\\dfrac{d(3t^2-4t+4)}{dt}=0\\\\6t-4=0\\\\t=0.67\ s

Hence, this is the required solution.

7 0
3 years ago
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Naddika [18.5K]

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7 0
3 years ago
If a town was 90 miles away and you travel at 45 mph how long would it take to get there
notka56 [123]

Answer:

t = 2 hours

Explanation:

Given that,

Distance of the town, d = 90 miles

Speed, v = 45 mph

We need to find the time to get there. The speed of an object is given by :

v=\dfrac{d}{t}

Where

t is time

t=\dfrac{d}{v}\\\\t=\dfrac{90}{45}\\\\t=2\ h

So, the required time is 2 hours.

5 0
3 years ago
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