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erastova [34]
3 years ago
13

A wave with a period of 0.008 second has a frequency of

Physics
1 answer:
insens350 [35]3 years ago
7 0
A wave with a period of 0.008 second has a frequency of 125 Hz.
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An open organ pipe is 1.6m long. If the speed of sound is 343m/s, what are the pipes: a) fundamental , b) 1st overtone , & c
Yakvenalex [24]

Answer:

a) 107.1875 Hz

b) 214.375 Hz

c) 321.5625 Hz

Explanation:

L = length of the open organ pipe = 1.6 m

v = speed of sound = 343 m/s

f = fundamental frequency

fundamental frequency is given as

f = \frac{v}{2L}

inserting the values

f = \frac{343}{2(1.6)}

f = \frac{343}{2(1.6)}

f = 107.1875 Hz

b)

first overtone is given as

f' = 2f

f' = 2 (107.1875)

f' = 214.375 Hz

c)

first overtone is given as

f'' = 3f

f'' = 3 (107.1875)

f'' = 321.5625 Hz

3 0
3 years ago
Four blocks of weights are required using which any body whose weight is between 1kg and 40 kg can be weighed. Find the four wei
Vikentia [17]

Answer:

The weights are 1 kg, 3kg, 9kg and 27kg.

Explanation:

The weights are 1 kg, 3kg, 9kg and 27kg.

1+3+9+27= 40

27+9+3= 39

27+9+3-1=38

27+9+1=37

27+9=36

27+9-1=35

27+9+1-3=34

27+9-3=33

27+9-3-1=32

27+3+1=31

27+3=30

27+3-1=29

27+1=28

27

27-1=26

27+1-3=25

27-3=24

27-3-1=23

27+3+1-9=22

27+3-9=21

27+3-9-1=20

Like this all the weights from 1 to 40 kg can be made using 1,3,9 and 27 kg.

6 0
3 years ago
Two identical positive charges exert a repulsive force of 6.1 × 10−9 n when separated by a distance 3.6 × 10−10 m. calculate the
Natalka [10]
Im a jewish salesman wanting to know if you wanted to buy some jesus movies

3 0
3 years ago
I need help thanks I would appreciate your in-depth explanation thanks.
svlad2 [7]

Answer:

The coin is less dense than the water thefore it can float.

Explanation:

5 0
3 years ago
Read 2 more answers
A test charge of -1.4 x 10-7 coulombs experiences a force of 5.4 x 10-1 newtons. Calculate the magnitude of the electric field c
VARVARA [1.3K]

Answer:

3.86×10⁶ Newton/coulombs

Explaination:

Applying,

E = F/q....................... Equation 1

Where E = Electric Field, F  = Force, q = charge.

From the question,

Given: F = 5.4×10⁻¹ N, q = -1.4×10⁻⁷ coulombs

Substitute these values into equation 1

E = 5.4×10⁻¹/ -1.4×10⁻⁷

E = -3.86×10⁶ Newtons/coulombs

Hence the magnitude of the electric field created by the

negative test charge is 3.86×10⁶ Newton/coulombs

5 0
3 years ago
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