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Simora [160]
2 years ago
15

The sound level at 1.0 m from a certain talking person talking is 60 dB. You are surrounded by five such people, all 1.0 m from

you and all talking equally loud at the same time. The threshold of hearing is 1.0 × 10-12 W/m2. What sound level are you being exposed to? You can neglect any absorption, reflection, or interference of the sound. The threshold of hearing is 1.0 × 10-12 W/m2.
Physics
1 answer:
Hunter-Best [27]2 years ago
7 0

Answer:

66.98 db

Explanation:

We know that

L_T=L_S+10log(n)

L_T= Total signal level in db

n= number of sources

L_S= signal level from signal source.

L_T=60+10 log(5)

= 66.98 db

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A force of 150 N accelerates a 25 kg wooden chair across a wood floor at 4.3 m/s2 . How big is the frictional force on the block
solniwko [45]
We can first calculate the net force using the given information.

By Newton's second law, F(net) = ma:

F(net) = 25 * 4.3 = 107.5

We can now calculate the frictional force, f, which is working against the applied force, F(app) (this is why the net force is a bit lower):

f = F(net) - F(app) = 150 - 107.5 = 42.5 N

Now we can calculate the coefficient of friction, u, using the normal force, F(N):

f = uF(n) --> u = f/F(N)
u = 42.5/[25(9.8)]
u = 0.17
4 0
2 years ago
A passenger compartment of a rotating amusement park ride contains a bench on which a book of mass
Basile [38]

a) 120 s

b) v = 0.052R [m/s]

Explanation:

a)

The period of a revolution in a simple harmonic motion is the time taken for the object in motion to complete one cycle (in this case, the time taken to complete one revolution).

The graph of the problem is missing, find it in attachment.

To find the period of revolution of the book, we have to find the time between two consecutive points of the graph that have exactly the same shape, which correspond to two points in which the book is located at the same position.

The first point we take is t = 0, when the position of the book is x = 0.

Then, the next point with same shape is at t = 120 s, where the book returns at x = 0 m.

Therefore, the period is

T = 120 s - 0 s = 120 s

b)

The tangential speed of the book is given by the ratio between the distance covered during one revolution, which is the perimeter of the wheel, and the time taken, which is the period.

The perimeter of the wheel is:

L=2\pi R

where R is the radius of the wheel.

The period of revolution is:

T=120 s

Therefore, the tangential speed of the book is:

v=\frac{L}{T}=\frac{2\pi R}{120}=0.052R

8 0
2 years ago
When you urinate, you increase pressure in your bladder to produce the flow. For an elephant, gravity does the work. An elephant
weqwewe [10]

Answer:

a) v =  1.19 m / s , b)   P₁ = 0.922 10⁵ Pa

Explanation:

1) Let's use the fluid continuity equation

       Q = A v

The area of ​​a circle is

      A = π r2 = π d²/4

     

     v = Q / A = Q 4 / pi d²

     v = 0.006 4/π 0.08²

     v =  1.19 m / s

2) write Bernoulli's equation, where point 1 is the bladder and point 2 is the urine exit point

     P₁ + ½ rho v₁² + rho g y₁ = P₂ + ½ rho v₂² + rho g y₂

The exercise tell us

P₂ = 1.0013 105 Pa

v₁ = 0

y₁ = 1 m

y₂=0  

Rho (water) = 1000 kg / m³

      P₁ + rho y₁ = P₂ + ½ rho v₂²

      P₁ = P₂ + ½ rho v₂² - rho g y₁

      P₁ = 1.013 10⁵ + ½ 1000 (1.19)² - 1000 9.8 1

      P₁ = 1.013 10⁵ +708.5  - 9800

      P₁ =  92208.5Pa

      P₁ = 0.922 10⁵ Pa

8 0
2 years ago
If the rocket has an initial mass of 6300 kg and ejects gas at a relative velocity of magnitude 2000 m/s , how much gas must it
Rzqust [24]

Answer:

The amount of gas that is to be released in the first second in other to attain an acceleration of  27.0 m/s2  is

      \frac{\Delta m}{\Delta t}   = 83.92 \ Kg/s

Explanation:

From the question we are told that

   The mass of the rocket is m = 6300 kg

   The velocity at gas is being ejected is  u =  2000 m/s

    The initial acceleration desired is a =  27.0 \  m/s

   The time taken for  the gas to be ejected is  t = 1 s

Generally this desired acceleration is mathematically represented as

        a = \frac{u *  \frac{\Delta m}{\Delta t} }{M -\frac{\Delta m}{\Delta t}* t}

Here \frac{\Delta m}{\Delta  t }  is the rate at which gas is being ejected with respect to time

Substituting values

      27 = \frac{2000 *  \frac{\Delta m}{\Delta t} }{6300 -\frac{\Delta m}{\Delta t}* 1}

=>   170100 -27* \frac{\Delta m}{\Delta t} = 2000 *  \frac{\Delta m}{\Delta t}

=>   170100  = 2027 *  \frac{\Delta m}{\Delta t}

=>   \frac{\Delta m}{\Delta t}   = \frac{170100}{2027}

=>   \frac{\Delta m}{\Delta t}   = 83.92 \ Kg/s

     

3 0
2 years ago
A car experiences a centripetal acceleration of 4.4 m/s ^2 as ur rounds a corner with a speed of 15 m/s. What is the radius of t
damaskus [11]
The calculation of the centripetal acceleration of an object following a circular path is based on the equation,

                  a = v² / r

where a is the acceleration, v is the velocity, and r is the radius.

Substituting the known values from the given above,

             4.4 m/s² = (15 m/s)² / r

The value of r from the equation is 51.14 m.

Answer: 51.14 m
3 0
2 years ago
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