Find the intensity III of the sound waves produced by four 60-WW speakers as heard by the driver. Assume that the driver is located 1.0 mm from each of the two front speakers and 1.5 mm from each of the two rear speakers.

Answers

Answer 1
Answer:

Given that,

Power = 60 W

Distance = 1.0 m

Distance between speakers = 1.5 m

We need to calculate the intensity

Using formula of intensity

I_(1)=(P)/(A)

I_(1)=(P)/(4\pi r^2)

Put the value into the formula

I_(1)=(60)/(4\pi*(1.0)^2)

I_(1)=4.77\ W/m^2

We need to calculate the intensity

Using formula of intensity

I_(2)=(P)/(A)

I_(2)=(P)/(4\pi r^2)

Put the value into the formula

I_(1)=(60)/(4\pi*(1.5)^2)

I_(1)=2.12\ W/m^2

We need to calculate the intensity of the sound waves produced by four speakers

Using formula for intensity

I'=(I_(1)+I_(2))*2

Put the value into the formula

I'=(4.77+2.12)*2

I'=13.78\ W/m^2

Hence, The intensity of the sound waves produced by four speakers is 13.78 W/m².


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What is the magnitude of a point charge that would create an electric field of 1.18 N/C at points 0.822 m away?

Answers

Answer:

q = 8.85 x 10⁻¹¹ C

Explanation:

given,

Electric field, E = 1.18 N/C

distance, r = 0.822 m

Charge magnitude = ?

using formula of electric field.

E = (kq)/(r^2)

k is the coulomb constant

q= (Er^2)/(k)

q= (1.18* 0.822^2)/(9* 10^9)

  q = 8.85 x 10⁻¹¹ C

The magnitude of charge is equal to q = 8.85 x 10⁻¹¹ C

At what distance from a long straight wire carrying acurrentof 5.0A is the magnitude of the magnetic field due to the
wireequal to the strength of the Earth's magnetic field of about
5.0 x10^-5 T?

Answers

Answer:

The distance is 2 cm

Solution:

According to the question:

Magnetic field of Earth, B_{E} = 5.0* 10^(- 5) T

Current, I = 5.0 A

We know that the formula of magnetic field is given by:

B = \farc{\mu_(o)I}{2\pi d}

where

d = distance from current carrying wire

Now,

d = (\mu_(o)I)/(2\pi B)

d = (4\pi* 10^(- 7)* 5.0)/(2\pi* 5.0* 10^(- 5))

d = 0.02 m 2 cm

A T-junction combines hot and cold water streams ( = 62.4 lbm/ft3 , cp = 1.0 Btu/lbm-R). The temperatures are measured to be T1 = 50 F, T2 = 120 F at the inlets and T3 = 80 F at the exit. The pipe diameters are d1 = d3 = 2" Sch 40 and d2 = 1¼" Sch 40. If the velocity at inlet 1 is 3 ft/s what is the mass flow rate at inlet 2? (3.27 kg/s)?

Answers

Answer:

m2=3.2722lbm/s

Explanation:

Hello!

To solve this problem follow the steps below

1. Find water densities and entlapies  in all states using thermodynamic tables.

note Through laboratory tests, thermodynamic tables were developed, which allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy, etc.)

through prior knowledge of two other properties, such as pressure and temperature.

D1=Density(Water;T=50;x=0)=62.41 lbm/ft^3

D2=Density(Water;T=120;x=0)=61.71 lbm/ft^3

D3=Density(Water;T=80;x=0)=62.21 lbm/ft^3

h1=Enthalpy(Water;T=50;x=0)=18.05 BTU/lbm

h2=Enthalpy(Water;T=120;x=0)=88  BTU/lbm

h3=Enthalpy(Water;T=80;x=0)=48.03 BTU/lbm

2. uses the continuity equation that states that the mass flow that enters a system is the same as the one that must exit

m1+m2=m3

3. uses the first law of thermodynamics that states that all the flow energy entering a system is the same that must come out

m1h1+m2h2=m3h3

18.05(m1)+88(m2)=48.03(m3)

divide both sides of the equation by 48.03

0.376(m1)+1.832(m2)=m3

4. Subtract the equations obtained in steps 3 and 4

m1            +      m2       =  m3

-

0.376m1   +  1.832(m2) =m3

--------------------------------------------

0.624m1-0.832m2=0

solving for m2

(0.624/0.832)m1=m2

0.75m1=m2

5. Mass flow is the product of density by velocity across the cross-sectional area

m1=(D1)(A)(v1)

internal Diameter for  2" Sch 40=2.067in=0.17225ft

A=(\pi )/(4) D^2=(\pi )/(4) (0.17225)^2=0.0233ft^2

m1=(62.41 lbm/ft^3)(0.0233ft^2)(3ft/S)=4.3629lbm/s

6.use the equation from step 4 to find the mass flow in 2

0.75m1=m2

0.75(4.3629)=m2

m2=3.2722lbm/s

Find the average speed of the electrons in a 1.0 cm diameter, copper power line, when it carries a current of 20 A.

Answers

Answer:

Average speed v_d=(i)/(neA)=(20)/(8.5* 10^(28)* 1.6* 10^(-19)* 7.8* 10^(-5))=1.87* 10^(-5)m/sec

Explanation:

We have given current through power i = 20 A

Diameter d = 1 cm = 0.01 m

So radius r = 0.005 m

So area A=\pi r^2=3.14* 0.005^2=7.8* 10^(-5)m^2

Charge on electron e = 1.6* 10^(-19)C

We know that current is given by i=neAv_d, here n is nuber density of free electron, e is charge on electron, A is area and v_d is average speed

We know that for copper n = 8.5* 10^(28)per\ m^3

So average speed v_d=(i)/(neA)=(20)/(8.5* 10^(28)* 1.6* 10^(-19)* 7.8* 10^(-5))=1.87* 10^(-5)m/sec

A round pipe of varying diameter carries petroleum from a wellhead to a refinery. At the wellhead, the pipe's diameter is 59.1 59.1 cm ( 0.591 0.591 m) and the flow speed of the petroleum is 11.9 11.9 m/s. At the refinery, the petroleum flows at 5.29 5.29 m/s. What is the volume flow rate of the petroleum along the pipe and what is the pipe's diameter at the refinery?

Answers

Answer:

The volume flow rate is 3.27m³/s

Diameter at the refinery is 88.64cm

Explanation:

Given

At the wellhead

Pipes diameter, d2 = 59.1cm = 0.591m

Flow speed of petroleum f2 = 11.9m/s

At the refinery,

Pipes diameter, d1 = ? Unknown

Flow speed of petroleum, f1 = 5.29m/s

Calculating the volume flow rate of petroleum along the pipe.

Volume flow rate = Flow rate * Area along the pipe

V = 11.9 * πd²/4

V = 11.9 * 22/7 * 0.591²/4

V = 3.265778m³/s

The volume flow rate is 3.27m³/s -------- Approximated

Since it's not stated if the flowrate is uniform throughout the pipe, we'll assume that flow rate is the same through out...

Using V1A1 = V2A2, where V1 & V2 Volume flow rate at both ends and area = Area of pipes at both ends

This gives;

V1A1 = V1A2

V1*πd1²/4 = V2 * πd2²/4 ----------- Divide through by π/4

So, we are left with

V1d1² = V2d2²

5.29 * d1²= 11.9 * 59.1²

d1² = 11.9 * 59.1²/5.29

d1² = 7857.172

d1 = √7857.172

d1 = 88.6406904305240618

d1 = 88.64cm --------------- Approximated

1. A sphere with a mass of 10 kg and radius of 0.5 m moves in free fall at sea level (where the air density is 1.22 kg/m3). If the object has a drag coefficient of 0.8, what is the object’s terminal velocity? What is the terminal velocity at an altitude of 5,000 m, where the air density is 0.736 kg/m3? Show all calculations in your answer.

Answers

The value of terminal velocity of object is 7.99 m/s and the value of terminal velocity at an altitude of 5000 m is 10.30 m/s.

Given data:

The mass of sphere is, m = 10 kg.

The radius of sphere is, r = 0.5 m.

The density of air is, \rho = 1.22 \;\rm kg/m^(3).

The drag coefficient of object is, C_(d)=0.8.

The altitude is, h = 5000 m.

The density of air at altitude is, \rho' =0.736 \;\rm kg/m^(3).

The mathematical expression for the terminal velocity of an object is,

v_(t)=\sqrt(2mg)/(\rho * A * C_(d))

here,

g is the gravitational acceleration.

A is the area of sphere.

Solving as,

v_(t)=\sqrt{(2 * 10 * 9.8)/(1.22 * (4 \pi r^(2)) * C_(d))}\n\n\nv_(t)=\sqrt{(2 * 10 * 9.8)/(1.22 * (4 \pi *  0.5^(2)) * 0.8)}\n\n\n\v_(t)=7.99 \;\rm m/s

Now, the terminal velocity at the altitude of 5000 m is given as,

v_(t)'=\sqrt(2mg)/(\rho' * A * C_(d))

Solving as,

v_(t)'=\sqrt{(2 * 10 * 9.8)/(0.736 * (4 \pi * 0.5^(2)) * 0.8)}\n\n\nv_(t)'=10.30 \;\rm m/s

Thus, we can conclude that the value of terminal velocity of object is 7.99 m/s and the value of terminal velocity at an altitude of 5000 m is 10.30 m/s.

Learn more about the terminal velocity here:

brainly.com/question/2654450

Answer:

The terminal velocity at sea level is 7.99 m/s

The terminal velocity at an altitude of 5000 m is 10.298 m/s

Explanation:

mass of sphere m  = 10 kg

radius of sphere r = 0.5 m

air density at sea level p = 1.22 kg/m^3

drag coefficient Cd = 0.8

terminal velocity = ?

Area of the sphere A = 4\pi r^(2) = 4 x 3.142 x 0.5^(2) = 3.142 m^2

terminal velocity is gotten from the relationship

Vt = \sqrt{(2mg)/(pACd) }

where g = acceleration due to gravity = 9.81 m/s^2

imputing values into the equation

Vt = \sqrt{(2*10*9.81)/(1.22*3.142*0.8) } = 7.99 m/s

If  at an altitude of  5000 m where air density = 0.736 kg/m^3, then we replace value of air density in the relationship as 0.736 kg/m^3

Vt = \sqrt{(2mg)/(pACd) }

Vt = \sqrt{(2*10*9.81)/(0.736*3.142*0.8) } = 10.298 m/s

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