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rodikova [14]
3 years ago
11

Suppose a large housefly 3.0 m away from you makes sound with an intensity level of 40.0 dB. What would be the sound intensity l

evel if there 1000 identical flies at that same distance?
Physics
1 answer:
melisa1 [442]3 years ago
5 0

Answer:

Explanation:

Let the intensity of the noise be represented by I

Given that

40dB = 10 log 10 ⁡ ( I /I•) ........ 1

I• is the lowest or threshold intensity of sound made.

I represents the intensity of the sound/ noise

The intensity of noise of 1000flies will be

β = 10 log 10 ⁡(1000I/I•)

Open up the bracket

β = 10 log 10(1000)+ 10 log 10(I/I•)

10 log 10(10^3)+10 log 10(I/I•)

3×10(10 log 10) +10 log 10(I/I•)

Recall, 10 log 10 = 1

30×1 + 10 log 10(I/I•).........2

Put equation 1 into 2

β =30+40

= 70db

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An ac power generator produces 73.37 a (rms) at 4623 v. the voltage is stepped up to 105,033 v by an ideal transformer, and the
evablogger [386]

Step up transformer is a device which is used to step up the voltage which is input with some value.

This is based upon the principle of mutual inductance and in this the voltage input and voltage output is different because of number of turns.

Here if ideal transformer is given then power input and power output of the transformer must be same as there is no power loss in ideal transformer.

So we can write

i_pV_p = i_sV_s

here

i_p = 73.37 A

V_p = 4623 V

V_s = 105033 A

now using above equation we will have

73.37*4623 = 105033*i_s

solving above we will have

i_s = 3.23 A

7 0
3 years ago
A 50 Kg box sits at rest on a 30 degree ramp where the coef of static friction is 0.5773. If your push was directed at an angle
emmasim [6.3K]

<u>Given data:</u>

m= 50 Kg,

W= m×g = 50 × 9.81 = 490.5 N

ramp angle (α) = 30 degrees,

coefficient of friction (μs) = 0.5773,

Push at an angle (Θ) = 40 degrees,

Determine: Push to get box move up (P)=?

From the figure,

Resolving the forces along the plane

W sinα + μs.R = P cos Θ       --------------------- (i)

Resolving the forces perpendicular to the inclined plane

W cosα = R+Psin Θ  =>  R= W cosα - Psin Θ -------------- (ii)

Solving (i) and (ii) and keeping <em>μs = tan Φ, Φ = Θ </em>

<em>Pmin = W sin( α +Θ  )</em>

<em>          = W[ sin α.Cos Θ + cos α.sin Θ]</em>

<em>           = 490.5 [ (sin 30.cos40) + (cos30.sin 40)]</em>

<em>           = 460.9 N</em>

<em>Minimum push required to move the box up the ramp is 460.9 N</em>


7 0
3 years ago
A block weighting 400kg rests on a horizontal surface and support on top of it another block of weight 100kg placed on the top o
masha68 [24]

The horizontal force applied to the block is approximately 1,420.84 N

The known parameters;

The mass of the block, w₁ = 400 kg

The orientation of the surface on which the block rest, w₁ = Horizontal

The mass of the block placed on top of the 400 kg block, w₂ = 100 kg

The length of the string to which the block w₂ is attached, l = 6 m

The coefficient of friction between the surface, μ = 0.25

The state of the system of blocks and applied force = Equilibrium

Strategy;

Calculate the forces acting on the blocks and string

The weight of the block, W₁ = 400 kg × 9.81 m/s² = 3,924 N

The weight of the block, W₂ = 100 kg × 9.81 m/s² = 981 N

Let <em>T</em> represent the tension in the string

The upward force from the string = T × sin(θ)

sin(θ) = √(6² - 5²)/6

Therefore;

The upward force from the string = T×√(6² - 5²)/6

The frictional force = (W₂ - The upward force from the string) × μ

The frictional force, F_{f2} = (981 - T×√(6² - 5²)/6) × 0.25

The tension in the string, T = F_{f2} × cos(θ)

∴ T = (981 - T×√(6² - 5²)/6) × 0.25 × 5/6

Solving, we get;

T = \dfrac{5886}{\sqrt{6^2 - 5^2} + 28.8} \approx 183.27

Frictional \ force, F_{f2} = \left (981 -  \dfrac{5886}{\sqrt{6^2 - 5^2} + 28.8}  \times \dfrac{\sqrt{6^2 - 5^2} }{6} \times  0.25 \right) \approx 219.92

The frictional force on the block W₂, F_{f2} ≈ 219.92 N

Therefore;

The force acting the block w₁, due to w₂ F_{w2} = 219.92/0.25 ≈ 879.68

The total normal force acting on the ground, N = W₁ + \mathbf{F_{w2}}

The frictional force from the ground, \mathbf{F_{f1}} = N×μ + \mathbf{F_{f2}} = P

Where;

P = The horizontal force applied to the block

P = (W₁ + \mathbf{F_{w2}}) × μ + \mathbf{F_{f2}}

Therefore;

P = (3,924 + 879.68) × 0.25 + 219.92 ≈ 1,420.84

The horizontal force applied to the block, P ≈ 1,420.84 N

Learn more about friction force here;

brainly.com/question/18038995

3 0
3 years ago
In the diagram, q1= +8.0 C, q2= +3.5 C, and q3 = -2.5 C. q1 to q2 is 0.10 m, q2 to q3 is 0.15 m. What is the net force on q2? La
yulyashka [42]

Answer:

f(t) =  28,7 [N]

Explanation: IMPORTANT NOTE: IN PROBLEM STATEMENT CHARGES ARE IN C (COULOMBS) AND IN THE DIAGRAM IN μC. WE ASSUME CHARGES ARE IN μC.

The net force on +q₂  is the sum of the force of +q₁  on +q₂ ( is a repulsion force since charges of equal sign repel each other ) and the force of -q₃ on +q₂ ( is an attraction force, opposite sign charges attract each other)

The two forces have the same direction to the right of charge q₂, we have to add them

Then

f(t) = f₁₂ + f₃₂

f₁₂ = K * ( q₁*q₂ ) / (0,1)²

q₁  = + 8 μC     then   q₁ = 8*10⁻⁶ C

q₂ =  + 3,5 μC  then  q₂ = 3,5 *10⁻⁶ C

K = 9*10⁹  [ N*m² /C²]

f₁₂ = 9*10⁹ * 8*3,5*10⁻¹²/ 1*10⁻²   [ N*m² /C²]* C*C/m²

f₁₂ = 252*10⁻¹ [N]

f₁₂ = 25,2 [N]

f₃₂ =  9*10⁹*3,5*10⁻⁶*2,5*10⁻⁶ /(0,15)²

f₃₂ =  78,75*10⁻³/ 2,25*10⁻²

f₃₂ =  35 *10⁻¹

f₃₂ =  3,5 [N]

f(t) =  28,7 [N]

5 0
3 years ago
Read 2 more answers
How to convert from fahrenheit to celsius
kaheart [24]
The formula for Fahrenheit and Celsius conversion is 
T(°F)<span> = </span>T(°C)<span> × 1.8 + 32
where T is temperature in F or C ( Fahrenheit or Celsius whatever is the case)
</span>This means that keeping this FORMULA in mind we can add different values to it and  accordingly convert values from one to another.
Some examples of fahrenheit conversions to Celsius are :
32°F = 0°C  using F = (0 x 1.8) + 32


8 0
3 years ago
Read 2 more answers
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