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kow [346]
3 years ago
14

The low-frequency speaker of a stereo set has a surface area of and produces 1W of acoustical power. What is the intensity at th

e speaker
Physics
1 answer:
AlekseyPX3 years ago
5 0

Answer:

      I =  \frac{1}{4\pi \ r^2}

we see the intensity decreases with the inverse of the distance squared

Explanation:

Intensity is defined as power per unit area,

           I = P / A

in this case we have that the sound is emitted in a spherical form therefore the area is

           A = 4 pi r2

therefore the intensity is

          I =  \frac{1}{4\pi \ r^2}

as we see the intensity decreases with the inverse of the distance squared

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Một đoạn mạch gồm R1=8 ôm;R2=12 ôm.biết R1 nối tiếp với đoạn mạch gồm R2 song song với R3; R3=1/2 R2 cường độ dòng điện chạy qua
LUCKY_DIMON [66]

Answer:

Explanation:

a.

b. R3= 1/2R2=6 ôm

R12= R1+R2=20 ôm

Rtđ=R12.R3/R12+R3= 4,62 ôm

c. U= I.Rtđ = 2.4,62=9,23 V

I1=I2= 9,23/20= 0,4615 A

I3= 2-0,4615=1,5385 A

7 0
2 years ago
PLEASE HELP!! DUE SOON!!!
max2010maxim [7]

Answer:

See attached file :)

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4 0
2 years ago
Read 2 more answers
What is the guage pressure of an object with a mass of 30 kg and 5ft height?
wariber [46]

Answer: D=pg

Explanation:

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8 0
3 years ago
You weigh 730N. What would you weigh if the earth were six times as massive as it is and its radius were four times its present
Ber [7]

The new weight is 273.8 N

Explanation:

The weight of an object on Earth is given by

W=mg

where

m is the mass of the object

g is the acceleration of gravity

The acceleration of gravity at the Earth's surface can be rewritten as

g=\frac{GM}{R^2}

where

G is the gravitational constant

M is the Earth's mass

R is the Earth's radius

Substituting into the previous equation,

W=\frac{GMm}{R^2}

We said that in normal conditions, the weight of the person is 730 N:

W=\frac{GMm}{R^2}=730 N

Later, we are said that:

  • The mass of the Earth increases by a  factor of 6, M'=6M
  • The radius of the Earth increases by a factor of 4, R'=4R

Substituting into the equation, we find the new weight of the person in these conditions:

W'=\frac{G(6M)m}{(4R)^2}=\frac{6}{16}(\frac{GMm}{R^2})=\frac{3}{8}W

So, the new weight is 3/8 of the original weight, therefore:

W'=\frac{3}{8}(730)=273.8 N

Learn more about gravity:

brainly.com/question/1724648

brainly.com/question/12785992

#LearnwithBrainly

3 0
3 years ago
How much does a 44 kg child weigh on Earth?​
Akimi4 [234]
Well if the child weighs 44 kg then 44 kg. Do you mean if they weigh 44 kg in space to earth?
4 0
3 years ago
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