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tresset_1 [31]
3 years ago
7

At a distance of 8 m, the sound intensity of one speaker is 66 dB. If we were to place 3 speakers in a circle of radius 8 m, wha

t woud the sound intensity level be at the center of the circle
Physics
1 answer:
ludmilkaskok [199]3 years ago
3 0

Answer:

dβ = 70. 77 dβ

Explanation:

The intensity of sound in decibels is

         dβ = 10 log I/I₀

let's look for the intensity of this signal

         I / I₀ = 10 dβ/10

         I / I₀ = 3.981 10⁶

the threshold intensity of sound for humans is I₀ = 1 10⁻¹² W / m²

         I = 3.981 10 ⁶ 1 10⁻¹²

         I = 3,981 10⁻⁶ W / m²

It is indicated that 3 cornets are placed in the circle, for which total intensity is

        I_total - 3 I

        I_total = 3  3,981 10⁻⁶

        I_total = 11,943 10⁻⁶ W / m²

let's reduce to decibels

      dβ = 10 log (11,943 10⁻⁶/1 10⁻¹²)

      dβ = 10  7.077

      dβ = 70. 77 dβ

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mr_godi [17]

Ngan's mass on earth is 85kg.

Ngan has a weight on Mars = 14.5 N

Ngan’s weight on Earth = 833.0 N

Ngan’s mass on Earth = ?

<span>Fg,earth = mg(earth)</span>

<span>M = Fg,earth </span><span>/ g(earth)</span>

<span>M = 833.0 N / 9.8 m/s2</span>

<span>M = 85 kg</span>

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A water balloon is thrown horizontally at a speed of 2.00 m/s from the roof of a building that is 6.00 m above the ground. At th
Elis [28]

Answer:

  • <u>The water ballon that was thrown straight down at 2.00 m/s hits the ground first, 0.19 s before the other ballon.</u>

Explanation:

The motions of the two water ballons are ruled by the kinematic equations:

  • y=y_0+V_0t-gt^2/2

We are only interested in the vertical motion, so that equation is all what you need.

<u>1. Water ballon is thrown horizontally at sped 2.00 m/s.</u>

The time the ballon takes to hit the ground is independent of the horizontal speed.

Since 2.00 m/s is a horizontal speed, you take the initial vertical speed equal to 0.

Then:

y=y_0+V_0t-gt^2/2\\ \\ 0=6.00m-9.8\frac{m}{s^2} t^2/2\\ \\ t=\sqrt{2\times6.00m/9.8\frac{m}{s^2}}\\\\ t=1.11s

<u>2. Water ballon thrown straight down at 2.00 m/s</u>

Now the initial vertical speed is 2.00 m/s down. So, the equation is:

0=6.00m-2.00\frac{m}{s}t-9.8\frac{m}{s^2}t^2/2\\ \\ 4.9t^2+2t-6=0\\ \\ t=0.92s

To solve the equation you can use the quadratic formula.

t=\frac{-2+/-\sqrt{2^2-4(4.9)(-6)} }{2(4.9)}\\ \\ t=-1.33\\ \\ t=0.92

You get two times. One of the times is negative, thus it does not have physical meaning.

<u>3. Conclusion:</u>

The water ballon that was thrown straight down at 2.00 m/s hits the ground first by 1.11 s - 0.92s = 0.19 s.

5 0
3 years ago
A football quarterback runs 15.0 m straight down the playing field in 3.00 s. He is then hit and pushed 3.00 m straight backward
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Answer:

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b) v=+9.97m/s

Explanation:

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x_{1} =15m, t_{1}=3s

x_{2} =-3m, t_{1}=1.74s

x_{3} =29m, t_{3}=5.20s

From dynamics we know that the formula for average velocity is:

v=\frac{x_{2}-x_{1}  }{t_{2}-x_{1}  }

a) For the three intervals:

v_{1}=\frac{x_{2}-x_{1}  }{t_{2}-t_{1}  }=\frac{(-3-15)m}{(1.74-3)s}=14.29m/s

v_{2}=\frac{x_{3}-x_{2}  }{t_{3}-t_{2}  }=\frac{(29-(-3))m}{(5.20-1.74)s}=9.25m/s

v_{3}=\frac{x_{3}-x_{1}  }{t_{3}-t_{1}  }=\frac{(29-15)m}{(5.20-3)s}=6.36m/s

b) The average velocity for the entire motion can be calculate by the following formula:

v=\frac{v_{1}+v_{2}+v_{3}   }{n} =\frac{(14.29+9.25+6.36)m/s}{3}=+9.97m/s

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OlgaM077 [116]

Answer:

I think B is the answer of this question

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