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oee [108]
2 years ago
10

Suppose that a sound has initial intensity β1 measured in decibels. This sound now increases in intensity by a factor f. What is

the new level of sound β2?
Physics
1 answer:
topjm [15]2 years ago
4 0

Answer:

β2= β1+10*f

Explanation:

comparing β2 and β1, it is said that β2 is increased by a factor of f.

for each factor of f, there is a 10*f dB increase.

therefore if the β1 is increases by an intensity of factor f

the new intensity would be β1+ 10*f

You might be interested in
A uniform 1.0-N meter stick is suspended horizontally by vertical strings attached at each end. A 2.0-N weight is suspended from
fgiga [73]

Answer:

3.5 N

Explanation:

Let the 0-cm end be the moment point. We know that for the system to be balanced, the total moment about this point must be 0. Let's calculate the moment at each point, in order from 0 to 100cm

- Tension of the string attached at the 0cm end is 0 as moment arm is 0

- 2 N weight suspended from the 10 cm position: 2*10 = 20 Ncm clockwise

- 2 N weight suspended from the 50 cm position: 2*50 = 100 Ncm clockwise

- 1 N stick weight at its center of mass, which is 50 cm position, since the stick is uniform: 1*50 = 50 Ncm clockwise

- 3 N weight suspended from the 60 cm position: 3*60 = 180 Ncm clockwise

- Tension T (N) of the string attached at the 100-cm end: T*100 = 100T Ncm counter-clockwise.

Total Clockwise moment = 20 + 100 + 50 + 180 = 350Ncm

Total counter-clockwise moment = 100T

For this to balance, 100 T = 350

so T = 350 / 100 = 3.5 N

4 0
1 year ago
An engineer wants to design a circular racetrack of radius R such that cars of mass m can go around the track at speed without t
gtnhenbr [62]

1. tan \theta = \frac{v^2}{Rg}

For the first part, we just need to write the equation of the forces along two perpendicular directions.

We have actually only two forces acting on the car, if we want it to go around the track without friction:

- The weight of the car, mg, downward

- The normal reaction of the track on the car, N, which is perpendicular to the track itself (see free-body diagram attached)

By resolving the normal reaction along the horizontal and vertical direction, we find the following equations:

N cos \theta = mg (1)

N sin \theta = m \frac{v^2}{R} (2)

where in the second equation, the term m\frac{v^2}{R} represents the centripetal force, with v being the speed of the car and R the radius of the track.

Dividing eq.(2) by eq.(1), we get the  following expression:

tan \theta = \frac{v^2}{Rg}

2. F=\frac{m}{R}(w^2-v^2)

In this second situation, the cars moves around the track at a speed

w>v

This means that the centripetal force term

m\frac{v^2}{R}

is now larger than before, and therefore, the horizontal component of the normal reaction, N sin \theta, is no longer enough to keep the car in circular motion.

This means, therefore, that an additional radial force F is required to keep the car round the track in circular motion, and therefore the equation becomes

N sin \theta + F = m\frac{w^2}{R}

And re-arranging for F,

F=m\frac{w^2}{R}-N sin \theta (3)

But from eq.(2) in the previous part we know that

N sin \theta = m \frac{v^2}{R}

So, susbtituting into eq.(3),

F=m\frac{w^2}{R}-m\frac{v^2}{R}=\frac{m}{R}(w^2-v^2)

4 0
2 years ago
The 1.5-in.-diameter shaft AB is made of a grade of steel with a 42-ksi tensile yield stress. Using the maximum-shearing-stress
PolarNik [594]

Answer:

T = 0.03 Nm.

Explanation:

d = 1.5 in = 0.04 m

r = d/2 = 0.02 m

P = 56 kips = 56 x 6.89 = 386.11 MPa

σ = 42-ksi = 42 x 6.89 = 289.58 MPa

Torque = T =?

<u>Solution:</u>

σ = (P x r) / T

T = (P x r) / σ

T = (386.11 x 0.02) / 289.58

T = 0.03 Nm.

7 0
2 years ago
When the voltage and current have _____ polarities in a pure capacitive circuit, the capacitor is discharging and the energy is
tangare [24]

Answer:opposite

Explanation:for a capacitor to discharge (after charging) the polarities of the current and voltage have to be reversed

6 0
1 year ago
You are in a hot-air balloon that, relative to the ground, has a veloc- ity of 6.0 m/s in a direction due east. You see a hawk m
Ann [662]

Answer:

6.32 m/s 18.43° northeast

Explanation:

We express the velocity of hawk as:

v_{Hawk}=v_{balloon}+v_{HawkRelativetoBalloon}=6 x+2 y

We consider positive x towards east and positive y due north. So the magnitude is simply the square root of the square components:

|v_{hawk}|=\sqrt[]{6^2+2^2}=\sqrt{40}≈6.32 m/s

And the angle with respect to the east should be with:

arctan(\frac{2}{6} )=18.43 \°

8 0
2 years ago
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