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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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What is 16.558 m/s rounded to three significant figures?
In-s [12.5K]

Answer:

Option C. 16.6 m/s

Explanation:

To round this 16.558 m/s to 3sf, we need to count the number beginning from 1. When we get to the 3rd number( ie 5), we'll examine the fourth number(i.e 5)to see if it less than five or greater. If it less than five, then we'll discard it. But if it five or greater, we'll approximate it and add it to the 3rd number.

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16.558 m/s = 16.6m/s to 3sf

3 0
4 years ago
A circular disc of mass 20kg and radius 15cm is mounted in an horizontal cylindrical axle of radius
disa [49]

Using the concepts of energy, rotational Newton's second law and rotational kinematics we can find the kinematic energy of the system formed by the disk and the cylindrical axis

          KE = 0.23 J

given parameters

  • Disk radius R = 15 cm = 0.15 m
  • Cylinder radius r = 1.5 cm = 0.0015 m
  • Disk mass M = 20 kg
  • Time t = 1.2 s
  • Force F = 12 N

to find

  • Kinetic energy (KE)

This exercise must be solved in parts:

1st part. Endowment kinetic energy is the energy due to the circular motion of an object and is described by the equation

         KE = ½ I w²

Where KE is the kinetic energy, I the moment of inertia and w the angular velocity

The moment of inertia is a magnitude that measures the inertia for rotational movement, it is a scalar quantity, therefore it is additive. In this system it is composed of two bodies, the disk and the cylindrical axis, for which the total moment of inertia it is

         I_{ total} = I_{ disk} + I_{ cylinder}

the moments of inertia with respect to an axis passing through the center of mass are tabulated

disk          I_{disk} = ½ M R²

cylinder   I_{cylinder} = ½ m r²

where M and m are the masses of the disk and cylinder respectively, R and r their radii

         I_{total} = ½ (M R² + m r²) = ½ M R² ( 1 + \frac{m}{M} \ (\frac{r}{R})^2 )

         I_{total} = ½ M R² ( 1+ \frac{m}{20}  (\frac{0.015}{0.15} )^2 ) = \frac{1}{2} M R² (1 + 0.005 m)

As the shaft mass  is much lighter than the disk mass , the last term is very small, which is why we despise it.

         I_{total} = ½ M R²

2nd part. Let's use Newton's second law for endowment motion

        τ = I α

        α = \frac{\tau }{I_{total}}l

        τ = F R

        α = \frac{F \ R}{I_{total}}

With the rotational kinematics expressions, we assume that the system starts from rest (w₀ = 0)

        w = w₀ + α  t

where w is the angular velocity, alpha is the angular acceleration and t is the time

        w = 0 + \frac{\tau }{I_{total}} \ t

we substitute in the kinetic energy equation

        KE = ½ I_{total}  ( \frac{ \tau }{I_{total}} \ t )²

        KE = ½ \frac{ \tau^2 }{I_{total}} \ t^2

let's substitute

        KE = \frac{F^2 \ R^4}{M \ R^2 } \ t^2

        KE = F² R² t² / M

let's calculate

        KE = 12² 0.15² 1.2² / 20

        KE = 0.23 J

With the concepts of energy and rotational kinematics we can find the kinetic energy of the system is

       KE = 0.23 j

learn more about rotational kinetic energy here:

brainly.com/question/20261989

4 0
3 years ago
Of the three forces acting on the rock as it slides down the bowl, which (if any) are constant and which are not? explain.
choli [55]

Answer

Hi,

The forces are; weight (gravity), Normal/centripetal force and friction. Force due to gravity is constant where as friction and centripetal are not.

Explanation

Weight is constant, given by the force of gravity on the object. The centripetal force is a function of the angles occurring between the velocity vector and the weight vector that is at right angle with the perpendicular line drawn from the surface. Friction is a function of the centripetal force thus it also varies.

Hope this helps!

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4 years ago
QUESTION 1
Molodets [167]
Question 1: C Question 2: B, Hope this Helps!
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What will happen if you drop a golf ball, a baseball, and a bowling ball at the same instant from the top of a tall building
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Depending on the height of the building they can break due to impact on the floor.
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