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Anna [14]
3 years ago
14

A weather emergency siren is mounted on a tower, 105 m above the ground. On one hand, it would be a good idea to make the siren

very loud so that it will warn as many people as possible. On the other hand, safety regulations prohibit the siren from exceeding an intensity level of 101 dB for workers standing on the ground directly below the siren. Assuming that the sound is uniformly emitted, what is the maximum power that the siren can put out
Physics
1 answer:
Svetradugi [14.3K]3 years ago
4 0

Answer:

Explanation:

101 dB = 10.1 B.

Maximum intensity of sound allowed = 10.1 B

Intensity of sound in terms of W/m² can be found as follows

log (I / I₀) = 10.1

I / I₀ = 10¹⁰°¹

I = I₀ X 10¹⁰°¹

= 10⁻¹² X  10¹⁰°¹

= 10⁻¹°⁹ W/m²

105 m above the ground the this intensity will be 105² times

intensity at source point = 10⁻¹°⁹ x 105²

= 138.79 W/m²

energy of sound from source

= 4π times

= 4 x 3.14 x 138.79

= 1743.28W/m²

To calculate in terms of decibel :

log 1743.28 / 10⁻¹²

= log 1743.28 +12

= 15.24 B

= 152.4 dB .

152.4 dB .

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A mountain climber, in the process of crossing between two cliffs by a rope, pauses to rest. She weighs 514 N. As the drawing sh
n200080 [17]

Answer:

Part a)

T_L = 155.4 N

Part b)

T_R = 379 N

Explanation:

As we know that mountain climber is at rest so net force on it must be zero

So we will have force balance in X direction

T_L cos65 = T_R cos80

T_L = 0.41 T_R

now we will have force balance in Y direction

mg = T_L sin65 + T_Rsin80

514 = 0.906T_L + 0.985T_R

Part a)

so from above equations we have

514 = 0.906T_L + 0.985(\frac{T_L}{0.41})

514 = 3.3 T_L

T_L = 155.4 N

Part b)

Now for tension in right string we will have

T_R = \frac{T_L}{0.41}

T_R = 379 N

3 0
3 years ago
The graph represents the force applied on an 3.00kg crate while it moved 5.0m. A. How much total work is done on the crate? B. I
jeka57 [31]

a. We can calculate the amount of work by calculating the area under the graph.

first area (rectangular): 2.5 x 6 = 15

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b. the force was first applied = 6 N

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vf²=56

vf=7.5 m/s

8 0
2 years ago
An Atwood's machine consists of two different masses, both hanging vertically and connected by an ideal string which passes over
Tasya [4]

Answer:

V₁ = √ (gy / 3)

Explanation:

For this exercise we will use the concepts of mechanical energy, for which we define energy n the initial point and the point of average height and / 2

Starting point

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    Em₀ = m₁ g y₁ + m₂ g y₂

Let's place the reference system at the point where the mass m1 is

     y₁ = 0

    y₂ = y

    Em₀ = m₂ g y = 2 m₁ g y

End point, at height yf = y / 2

    E_{mf} = K₁ + U₁ + K₂ + U₂

    E_{mf} = ½ m₁ v₁² + ½ m₂ v₂² + m₁ g y_{f} + m₂ g y_{f}

Since the masses are joined by a rope, they must have the same speed

     E_{mf} = ½ (m₁ + m₂) v₁² + (m₁ + m₂) g y_{f}

   E_{mf}= ½ (m₁ + 2m₁) v₁² + (m₁ + 2m₁) g y_{f}

How energy is conserved

   Em₀ =  E_{mf}

   2 m₁ g y = ½ (m₁ + 2m₁) v₁² + (m₁ + 2m₁) g y_{f}

   2 m₁ g y = ½ (3m₁) v₁² + (3m₁) g y / 2

   3/2 v₁² = 2 g y -3/2 g y

   3/2 v₁² = ½ g y

   V₁ = √ (gy / 3)

5 0
2 years ago
What is the term for producing a current by moving a wire through a magnetic field?.
maria [59]
The answer is electromagnetic Induction.



I hope this answer will help you
6 0
2 years ago
Two trains collide in a messy train
Dvinal [7]

Answer:

b

Explanation:

8 0
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