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Degger [83]
4 years ago
7

Normal conversation has a sound level of about 60 db. How many times more intense must a 100-hz sound be compared to a 1000-hz s

ound to be perceived as equal to 60 phons of loudness?
Physics
1 answer:
vampirchik [111]4 years ago
6 0

Answer:

5.65 times

Explanation:

60 db sound is equal to 60 phons sound when frequency is kept at 1000Hz.

But when the frequency of sound  is changed to 100 Hz , according to equal loudness curves , the loudness level on phon scale will be 35 phons.

A decrease of 10 phon on phon- scale makes sound 2 times less loud

Therefore a decrease of 25 phons will make loudness less intense by a factor equal to 2²°⁵ or 5.65 less intense . Therefore intensity at 100 Hz

must be increased 5.65 times so that its intensity matches intensity of 60 dB sound at  1000 Hz  frequency.

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While riding in a moving car, you toss an egg directly upward. Does the egg tend to land behind you, in front of you, or back in
bazaltina [42]

Answer:

Explanation:

The movement of a body can be analyzed using New's first law. In an inertial frame (without acceleration) every body is kept at rest or moving at constant speed until there is an external force that changes this state

Let's analyze these cases in the framework of this first law

a) If the vehicle is going at constant speed the two bodies (the egg and the hands) do not change movement so he had returned to the hands

b) If the vehicle accelerates the passenger goes faster, but the egg that is not subject to anything does not change the movement, so it falls behind the passenger

c) If the vehicle slows down, the passenger reduces its speed and the distance traveled in time, but the egg that is not attached follows its movement and falls in front of the passenger.

4 0
3 years ago
A 1.1-kg object is suspended from a vertical spring whose spring constant is 120 N/m. (a) Find the amount by which the spring is
andriy [413]

Answer:

e = 0.0898m

v = 2.07m/s

Explanation:

a) According to Hooke's law

F = ke

e is the extension

k is the spring constant

Since F = mg

mg = ke

e = mg/k

Substitute the given value

e = 1.1(9.8)/120

e = 10.78/120

e = 0.0898m

Hence it is stretched by 0.0898m from its unstrained length

2) Total Energy = PE+KE+Elastic potential

Total Energy = mgh +1/2mv²+1/2ke²

Substitute the given value

5.0= 1.1(9.8)(0.2)+1/2(1.1)v²+1/2(120)(0.0898)²

Solve for v

5.0 = 2.156+0.55v²+0.48338

5.0-2.156-0.48338= 0.55v²

2.36 =0.55v²

v² = 2.36/0.55

v² = 4.29

v ,= √4.29

v = 2.07m/s

Hence the required velocity is 9.28m/s

4 0
3 years ago
For a vehicle to negotiate a banked curve in poor weather conditions, where the force of friction f = 0, for a given velocity v
djverab [1.8K]

Answer:

R=240m

Explanation:

From the question we are told that:

Velocity v=25m/s

Force of friction f = 0

Angle \theta=15

Generally the equation for  Radius of curvature is mathematically given by

R=frac{v^2}{tan\theta *g}

R=frac{25^2}{tan 15 *9.81}

R=240m

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What are the units for mass and volume
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Those are the answers

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When solar radiation reaches the Earth it quickly dissipates as most of the radiation and UV rays are blocked by ozone layer, but more radiation and UV rays are able to get through because of global warming.
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