PHYSICS COLLEGE

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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The ____ contains the highest concentration of ozone

Answers

I believe the term you are looking for is the ozone layer. This layer is in the highest region if the stratosphere.

A large aquarium has portholes of thin transparent plastic with a radius of curvature of 1.95 m and their convex sides facing into the water. A shark hovers in front of a porthole, sizing up the dinner prospects outside the tank.a) If one of the sharks teeth is exactly 46.5 cm from the plastic, how far from the plastic does it appear to be to observers outside the tank? (You can ignore refraction due to the plastic.)b) Does the shark appear to be right side up or upside down?c) If the tooth has an actual length of 5.00 cm, how long does it appear to the observers?

Answers

Answer:

Explanation:

For refraction through a curved surface , the formula is as follows

μ₂ / v - μ₁ / u = (μ₂ -μ₁ )/R , Here μ₂( air) = 1 , μ₁ ( water) = 4/3 , R = 1.95 m

u , object distance = - .465 m

1 / v + 1.333 / .465 = (1 -1.333 )/1.95

1 / v + 2.8667 = - .171

1 / v = - 2.8667 - .171 = - 3.0377

v = - .3292 m

= - 32.92 cm

image will be formed in water.

c ) magnification = μ₁v / μ₂u , μ₁ = 1.33 , μ₂ = 1 , u = 46.5 , v = 32.92 .

= (1.33 x 32.92) / (1 x 46.5)

= .94

size of image of teeth = .94 x 5

= 4.7 cm .

g Adjacent rows in the first part of the experiment are found to have potentials of 3.66 V and 4.22 V. If the distance between rows is found to be 0.4 cm, what is the magnitude of the electric field at the location between the rows

Answers

Answer:

$E=140V/m$

Explanation:

If the electric field is uniform, the electric field between two points at potentials $V_1$ and $V_2$ which are separated by a distance d will be given by the formula:

$E=(\Delta V)/(d)$

So in our case, we have $E=(4.22V-3.66V)/(0.004m)=140V/m$

A projectile is launched at some angle to the horizontal with some initial speed vi, and air resistance is negligible.(a) Is the projectile a freely falling body?Yes or No(b) What is its acceleration in the vertical direction? (Let up be the positive direction.)____? m/s2(c) What is its acceleration in the horizontal direction?

Answers

Answer:

A) No

B)-9,81 m/s^2

C)0 m/s^2

Explanation:

A free fallin object has only velocity on the vertical axis so any object that is moving in the Y and X axis has projetile motion not free falling, and when dealing with projectile motion the object is experiencing acceleration towards the ground of -9,81m/s^2 and in the Y axis, in the X axis there´s is only acceleration if the air is providing resistance, since it states that it isnot, then the accleration is 0.

What is the density, in Mg/m3, of a substance with a density of 0.14 lb/in3? (3 pts) What is the velocity, in m/sec, of a vehicle traveling 70 mi/hr?

Answers

Answer:

$276.74* 10^8Mg/m^3$

31.29 m/sec

Explanation:

We have given density of substance $0.14lb/in^3$

We have convert this into $Mg/m^3$

We know that 1 lb = 0.4535 kg. so 0.14 lb = 0.14×0.4535 = 0.06349 kg

We know that 1 kg = 1000 g ( 1000 gram )

So 0.06349 kg = 63.49 gram

And we know that 1 gram = 1000 milligram

So 63.49 gram $=63.49* 10^3\ Mg$

We know that $1 in^3=1.6387* 10^(-5)m^3$

So $0.14in^3=0.14* 1.6387* 10^(-5)=0.2294* 10^(-5)m^3$

So $0.14lb/in^3$ =\frac{63.49\times 10^3}{0.2249\times 10^{-5}}=276.74\times 10^8lb/m^3[/tex]

In second part we have to convert 70 mi/hr to m/sec

We know that 1 mi = 1609.34 meter

So 70 mi = 70×1609.34 = 112653.8 meter

1 hour = 3600 sec

So 70 mi/hr $=(70* 1609.34meter)/(3600sec)=31.29m/sec$

A rocket exhausts fuel with a velocity of 1500m/s, relative to the rocket. It starts from rest in outer space with fuel comprising 80 per cent of the total mass. When all the fuel has been exhausted its speed is:________