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umka21 [38]
3 years ago
14

Party hearing. As the number of people at a party increases, you must raise your voice for a listener to hear you against the ba

ckground noise of the other partygoers. However, once you reach the level of yelling, the only way you can be heard is if you move closer to your listener, into the listener’s "personal space." Model the situation by replacing you with an isotropic point source of fixed power P and replacing your listener with a point that absorbs part of your sound waves. These points are initially separated by ri = 1.20 m. If the background noise increases by Δβ = 5 dB, the sound level at your listener must also increase. What separation rf is then required?
Physics
1 answer:
dusya [7]3 years ago
6 0

Answer:

\frac{r_i}{1.77} m

Explanation:

Given that

At starting separated = 1.20m

And the increase in background noise by Δβ = 5 dB, due to which the level of sound also rises

Based on the above information, the separation rf that is needed is shown below:

As we know that

I_f = I_o \times 10^{\frac{\beta}{10} }\\\\I_f = I_i \times 10^{0.5}\\\\I_f = 3.16 \times  I_i\\\\I \alpha \frac{1}{r^2} \\\\\frac{r_i^2}{r_f^2} = 3.16\\\\r_f = \frac{\sqrt{r_i^2}}{{\sqrt3.16}} \\\\= \frac{r_i}{1.77} m

Hence, the separation r_f i.e. required is  \frac{r_i}{1.77} m

We simply applied the above equation so that the correct separation could come

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A)
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3 0
4 years ago
A circular test track for cars has a circumference of 4.7 km. A car travels around the track from the southernmost point to the
Alika [10]
Well, for the distance traveled, the car goes from the northernmost point to the southernmost point. So, it travels half of the circle's circumference = 4.7/2 = 2.35 km.

For the displacement, by going from the northernmost point to the southernmost point, the car basically just travels the diameter of circle.

So, using the formula: Circumference = 2πr = <span>πd

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7 0
3 years ago
8. During the last few kilometers of a marathon, runners play mental games to test each
o-na [289]

Answer:

The runner's average acceleration is 0.102 m/s²

Explanation:

The runner accelerates from 5 m/s to 5.2 m/s and covering 10 m

We need to find the runner's average acceleration

The given is:

→ Initial velocity 5 m/s

→ Final velocity 5.2 m/s

→ Distance 10 meters

→ Acceleration ?

We need a suitable rule for the given

→ v² = u² + 2 a s

where v is the final velocity, u is the initial velocity, a is the acceleration

and s is the distance

Substitute the values above in the rule

→ (5.2)² = (5)² + 2 a (10)

→ 27.04 = 25 + 20 a

Subtract 25 from both sides

→2.04 = 20 a

Divide both sides by 20

→ a = 0.102 m/s²

<em>The runner's average acceleration is 0.102 m/s²</em>

3 0
3 years ago
What is the mass in grams of 5.50 × 1014 lead (pb) atoms?
olchik [2.2K]
There will be no way that to happen aton is a positive charge
8 0
3 years ago
A 18-kg sled is being pulled along the horizontal snow-covered ground by a horizontal force of 30 N. Starting from rest, the sle
k0ka [10]

Answer:

Coefficient of kinetic friction = 0.146

Explanation:

Given:

Mass of sled (m) = 18 kg

Horizontal force (F) = 30 N

FInal speed (v) = 2 m/s

Distance (s) = 8.5 m

Find:

Coefficient of kinetic friction.

Computation:

Initial speed (u) = 0 m/s

v² - u² = 2as

2(8.5)a = 2² - 0²

a = 0.2352 m/s²

Nweton's law of :

F (net) = ma

30N - μf = 18 (0.2352)

30 - 4.2336 = μ(mg)

25.7664 =  μ(18)(9.8)

μ = 0.146

Coefficient of kinetic friction = 0.146

6 0
4 years ago
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