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galina1969 [7]
3 years ago
8

When one person shouts at a football game, the sound intensity level at the center of the field is 58.4 dB. When all the people

shout together, the intensity level increases to 102 dB. Assuming that each person generates the same sound intensity at the center of the field, how many people are at the game?

Physics
2 answers:
elena-s [515]3 years ago
8 0

Explanation:

Below is an attachment containing the solution.

Tanzania [10]3 years ago
3 0

Answer:

The number of people at game are approximately 22909

Explanation:

Given data

When one person shout \beta _{1}=58.4dB

When n number of person shout together \beta _{n}=102dB

The sound intensity level during one person shout is given by:

\beta _{1}=10log(\frac{I_{1}}{I_{o}} )\\58.4=10log(\frac{I_{1}}{I_{o}} )\\5.84=log(\frac{I_{1}}{I_{o}} )\\\frac{I_{1}}{I_{o}} =10^{5.84}\\I_{1}=10^{5.84}*I_{o}

The sound intensity level during n number of person shout is given by:

\beta _{n}=10log(\frac{I_{n}}{I_{o}} )\\102=10log(\frac{I_{n}}{I_{o}} )\\10.2=log(\frac{I_{n}}{I_{o}} )\\\frac{I_{n}}{I_{o}}=10^{10.2}\\I_{n}=10^{10.2}*I_{o}

Since each person generates same sound intensity and hence total number  of persons can be determined as

 =\frac{I_{n}}{I_{1}}\\ =\frac{10^{10.2}I_{o}}{10^{5.84}I_{o}} \\=22909

Hence

The number of people at game are approximately 22909

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laila [671]
The answer is "friction and air resistance" gravity does some of the work by keeping the object from floating away, but friction and air resistance does the biggest part. Friction is how rough the ground it meaning on tile, dirt, grass, etc... that would slow down the object and air resistance is the gravity pushing on the object also making it stop. 

Hope this helps!
8 0
4 years ago
A 12-pack of Omni-Cola (mass 4.30 kg) is initially at rest ona horizontal floor. It is then pushed in a straight line for1.20 m
erastovalidia [21]

Answer:

a)  W = 46.8 J  and b)   v = 3.84 m/s

Explanation:

The energy work theorem states that the work done on the system is equal to the variation of the kinetic energy

    W = ΔK = k_{f} -K₀

a) work is the scalar product of force by distance

    W = F . d

Bold indicates vectors. In this case the dog applies a force in the direction of the displacement, so the angle between the force and the displacement is zero, therefore, the scalar product is reduced to the ordinary product.

    W = F d cos θ

    W = 39.0 1.20 cos 0

    W = 46.8 J

b) zero initial kinetic language because the package is stopped

    W -W_{fr} = k_{f} -K₀

    W - fr d= ½ m v² - 0

    W - μ N d = ½ m v

   on the horizontal surface using Newton's second law

     N-W = 0

     N = W = mg

 

     W - μ mg d = ½ m v

    v² = (W -μ mg d) 2/m  

    v = √(W -μ mg d) 2/m

    v = √[(46.8 -  0.30 4.30 9.8 1.20) 2/4.3 ]

    v = √(31.63 2/4.3)

    v = 3.84 m/s

8 0
3 years ago
particle of mass 59 g and charge 51 µC is released from rest when it is 32 cm from a second particle of charge −14 µC. Determine
AleksAgata [21]

Answer:

The initial acceleration of the 59g particle is 1062.7\frac{m}{s^{2}}

Explanation:

Newton's second laws relates acceleration (a), net force(F) and mass (m) in the next way:

F=ma (1)

We already know the mass of the particle so we should find the electric force on it to use on (1), the magnitude of the electric force between two charged objects by Columb's law is:

F=k\frac{\mid q_{1}q_{2}\mid}{r^{2}}

with q1 and q2 the charge of the particles, r the distance between them and k the constant k=9.0\times10^{9}\,\frac{Nm^{2}}{C^{2}}. So:

F=(9.0\times10^{9})\frac{\mid (51\times10^{-6})(-14\times10^{-6})\mid}{0.32^{2}}

F=62.7 N

Using that value on (1) and solving for a

a=\frac{F}{m}=\frac{62.7}{0.059}=1062.7\frac{m}{s^{2}}

4 0
4 years ago
I need help please I will give points
stira [4]

Answer:

-5 N of force

Explanation:

Hope it helps!

3 0
3 years ago
On the sonometer shown below, a horizontal cord of length 5 m has a mass of 1.45 g. When the cord was plucked the wave produced
Korolek [52]

Answer:

(a) T = 0.015 N

(b) M = 1.53 x 10⁻³ kg = 1.53 g

Explanation:

(a) T = 0.015 N

First, we will find the speed of waves:

v =f\lambda

where,

v = speed of wave = ?

f = frequency = 120 Hz

λ = wavelength = 6 cm = 0.06 m

Therefore,

v = (120 Hz)(0.06 m)

v = 7.2 m/s

Now, we will find the linear mass density of the coil:

\mu = \frac{m}{l}

where,

μ = linear mass density = ?

m = mass = 1.45 g = 1.45 x 10⁻³ kg

l = length = 5 m

Thereforre,

\mu = \frac{1.45\ x\ 10^{-3}\ kg}{5\ m}\\\\\mu = 2.9\ x\ 10^{-4}\ kg/m

Now, for the tension we use the formula:

v = \sqrt{\frac{T}{\mu}}\\\\7.2\ m/s = \sqrt{\frac{T}{2.9\ x\ 10^{-4}\ kg/m}}\\\\(51.84\ m^2/s^2)(2.9\ x\ 10^{-4}\ kg/m) = T

<u>T = 0.015 N</u>

<u></u>

(b)

The mass to be hung is:

T = Mg\\\\M = \frac{T}{g}\\\\M = \frac{0.015\ N}{9.8\ m/s^2}\\\\

<u>M = 1.53 x 10⁻³ kg = 1.53 g</u>

4 0
3 years ago
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