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melisa1 [442]
1 year ago
11

Traffic noise on Beethoven Boulevard has an intensity level of 80 dB; the traffic noise on Mozart Alley is only 60 dB. Part A Co

mpared to the sound intensity on Beethoven Boulevard, the sound intensity on Mozart Alley is Compared to the sound intensity on Beethoven Boulevard, the sound intensity on Mozart Alley is a. 25% lower.b. 100 times lower. c. 20 W/m2 lower. d. 20 times lower.
Physics
1 answer:
uranmaximum [27]1 year ago
8 0

Answer:

option (b)

Explanation:

Io = 10^-12 w/m^2

In Case I:

dB = 80 dB

The formula for the intensity of sound

dB=10log\left ( \frac{I}{I_{0}} \right )

80=10log\left ( \frac{I_{1}}{I_{0}} \right )

I_{1} = 10^{8}\times I_{0}

I_{1} = 10^{8}\times 10^{-12}=10^{-4}W/m^{2}   ... (1)

In Case II:

dB = 60 dB

The formula for the intensity of sound

dB=10log\left ( \frac{I_{2}}{I_{0}} \right )

60=10log\left ( \frac{I_{2}}{I_{0}} \right )

I_{2} = 10^{6}\times I_{0}

I_{2} = 10^{6}\times 10^{-12}=10^{-6}W/m^{2}   ... (2)

So, by equation 2 and 1 we get

I_{2}=\frac{I_{1}}{100}

Thus, the intensity of sound is 100 times lower.

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By 16.7% or 0.167 IPM

Explanation:

Substracting the final IPM (6.088) to the initial IPM (5.921) gives us the net difference, which is how much did it increase in IPM. Multiplying this number by 100 gives us the percentual increase in the feed rate.

4 0
1 year ago
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A 3.0-kg brick rests on a perfectly smooth ramp inclined at 34° above the horizontal. The brick is kept from sliding down the pl
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Answer:

d=0.137 m ⇒13.7 cm

Explanation:

Given data

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Solution

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F=29.4 N

As we also know that

Force parallel to the incline=FSinα

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F=16.44 N

d(distance)=F/Spring Constant

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4 0
1 year ago
3) 4 electrons are placed - one electron per corner - at the corners of a square of side 1 meter. One fixed proton is placed in
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1 year ago
You are exploring a distant planet. When your spaceship is in a circular orbit at a distance of 630 km above the planet's surfac
NemiM [27]

Answer:

The horizontal range of the projectile = 26.63 meters

Explanation:

Step 1: Data given

Distance above the planet's surface = 630 km = 630000

The ship's orbal speed = 4900 m/s

Radius of the planet = 4.48 *10^6 m

Initial speed of the projectile = 13.6 m/s

Angle = 30.8 °

Step 2: Calculate g

g= GM /R² = (v²*(R+h)) /(R²)

⇒ with v= the ship's orbal speed = 4900 m/S

⇒ with R = the radius of the planet = 4.48 *10^6 m

⇒ with h = the distance above the planet's surface = 630000 meter

g = (4900² * ( 4.48*10^6+ 630000)) / ((4.48*10^6)²)

g = 6.11 m/s²

<u>Step 3:</u> Describe the position of the projectile

Horizontal component: x(t) = v0*t *cos∅

Vertical component: y(t) = v0*t *sin∅ -1/2 gt² ( will be reduced to 0 in time )

⇒ with ∅ = 30.8 °

⇒ with v0 = 13.6 m/s

⇒ with t= v(sin∅)/g = 1.14 s

Horizontal range d = v0²/g *2sin∅cos∅  = v0²/g * sin2∅

Horizontal range d =(13.6²)/6.11 * sin(2*30.8)

Horizontal range d =26.63 m

The horizontal range of the projectile = 26.63 meters

6 0
1 year ago
Constants Periodic Table Suppose the top surface of the vessel in the figure (Figure 1) is subjected to an external gauge pressu
Gnom [1K]

Answer:

a)  v₁ = √ [2 (P₂-P₀) /ρ + 2 (y₂ -y₁)]

b) Water does not flow,

Explanation:

a) For this exercise we will use Bernoulli's equation, where index 1 is at the exit and index 2 on the surface of the water

            P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

This case does not indicate at the surface pressure is P₂, the pressure at the outlet is P₁ = P₀, the surface velocity is zero v₂ = 0

          P₀ + ½ ρ v₁² + ρ g y₁ = P₂ + 0 + ρ g y₂

           ½ ρ v₁² = P₂-P₀ + ρ (y₂ -y₁)

          v₁² = 2 (P₂-P₀) /ρ + 2 (y₂ -y₁)

          v₁ = √ [2 (P₂-P₀) /ρ + 2 (y₂ -y₁)]

b) Reduce the pressure to SI units

         P₂ = 0.86 atm (1.01 10⁵ Pa / 1 atm) = 0.8686 10⁵ Pa

         P₀ = 1.01 10⁵ Pa

         ρ = 1 10³ kg / m³

Let's calculate

         v₁ = √ [2 (0.8686 -1.01) 10⁵/10³ + 2 (2.6)]

         v₁ = √ [-0.2828 10² + 5.2] = Ra [-23]

Water does not flow, this is because the pressure of the inner part is less than atmospheric, so that the water flows the pressure P₂> = P₀

For example if the pressure P₂ = P₀

         v₁ = √ 5.2

          v₁ = 2.28 m / s

5 0
1 year ago
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