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serg [7]
3 years ago
12

Find the intensity i of the sound waves produced by one 60-w speaker at a distance of 1.0 m.

Physics
1 answer:
Kipish [7]3 years ago
8 0

Answer:

the intensity of the sound waves produced by one 60-w speaker at a distance of 1.0 m is 60 w/m²

Explanation:

Intensity of sound , I, is given as;

I  = P/A

Where;

P is the power through an area = 60-w

A is the area = ?

A = 1.0m × 1.0m = 1.0 m²

I = 60-w / 1.0 m²

I = 60 w/m²

Therefore, the intensity of the sound waves produced by one 60-w speaker at a distance of 1.0 m is 60 w/m²

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Calculate the RMS voltage of the following waveforms with 10 V peak-to-peak:
Deffense [45]

Answer:

a) T=0.01s

b) T=0.001s

c) T=0.00001s

Explanation:

From the question we are told that:

Given Frequencies

a. 100 Hz,

b. 1 kHz,

c. 100 kHz.

Generally the equation for Waveform Period is mathematically given by

T=\frac{1}{f}

Therefore

a)

For

T=100 Hz

T=\frac{1}{100}

T=0.01s

b)

For

F=1kHz

T=\frac{1}{1000}

T=0.001s

c)

For

F=100kHz

T=\frac{1}{100*100}

T=0.00001s

6 0
3 years ago
Which of the following would increase the elastic force acting on that object
olga_2 [115]

Answer:

Which of the following would increase the elastic force acting on that object? Moving a spring to an unstretched position. Compressing a spring twice as much as its starting position.

Explanation:

4 0
3 years ago
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We can see Objects because of
Amiraneli [1.4K]

Answer:

I believe it's A.)

Explanation:

Although light comes into our atmosphere through refraction, it reaches our eyes only through reflection from objects. So when light rays reflect off an object and enter the eyes through the cornea you can then see that object.

Hope this helps you out : )

7 0
4 years ago
A student performs an activity to study how electric current flows in a circuit. The student constructs two different circuits,
natta225 [31]

Native_Americans_in_the_United_State0000000000000000000000000

Explanation:

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3 years ago
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A room has a wall with an R value of 18 F sq.ft. hr/BTU. The room is 15 feet long and 11 feet wide with walls that are 9 ft high
seropon [69]

Answer:

The heat transferred through the wall that day is  13728 BTUs

Explanation:

Here, we have the area of the wall given as

Area of wall = 2 × Length × Height + 2 × Width × Height

Length = 15 feet

Width = 11 Feet and

Height = 9 feet

Therefore, the area = 2×15×9 + 2×11×9 = 468 ft²

Temperature difference is given by

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Therefore the heat transferred through the wall that day (24 hours) at 18 sq.ft. hr/BTU is given by;

468 × 22 × 24/18 = 13728 = 13728 BTUs.

5 0
4 years ago
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