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musickatia [10]
4 years ago
13

Using two pieces of string stretched to the same tension, pythagoras discovered the ratio for creating the interval of a perfect

octave was ______________.
Physics
1 answer:
Katen [24]4 years ago
6 0
The answer to this question would be 1:2

When you making sound with a string, the frequency is reversed with the string length. If you use a string with 2x length the frequency will become half of the first string. For a perfect octave, it will need 1:2 ratio of frequency or 1:4 tension
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two-point charges are 10.0 cm apart and have charges of 2.0 uc and -2.0uc respectively What is the magnitude of the electrical f
Scorpion4ik [409]

Answer:

Electric field due to two charges is given as

E = 1.44 \times 10^7 N/C

Explanation:

As we know that two charges are opposite in nature

So the electric field at the mid point of two charges will add together

so the net field is given as

E = 2\frac{kq}{r^2}

now we have

q = 2\times 10^{-6} C

r = 5 cm = 0.05 m

now we have

E = 2\frac{(9\times 10^9)(2\times 10^{-6})}{0.05^2}

E = 1.44 \times 10^7 N/C

6 0
3 years ago
In a car, 75 percent of the chemical energy of gasoline is lost as thermal
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3 0
4 years ago
The sound intensity at a distance 2.00 m from a sound source is 5.00 Find the total sound energy emitted by the source in each s
notka56 [123]

Answer:

     P = 251, 3 W

Explanation:

The intensity is defined as the power emitted per unit area

           I = P / A

Since sound is distributed in all directions spherical shape, the area of ​​a sphere is

           A = 4π r²

let's clear the power and replace

         P = I A

         P = I (4π r²)

let's calculate

         P = 5.00 (4π 2²)

         P = 251, 3 W

6 0
3 years ago
(a) The Sun orbits the Milky Way galaxy once each 2.60 x 108y , with a roughly circular orbit averaging 3.00 x 104 light years i
PilotLPTM [1.2K]

Answer:

Part a)

a_c = 1.67 \times 10^{-10} m/s^2

Part b)

v = 2.18 \times 10^5 m/s

Explanation:

Time period of sun is given as

T = 2.60 \times 10^8 years

T = 2.60 \times 10^8 (365 \times 24 \times 3600) s

T = 8.2 \times 10^{15} s

Now the radius of the orbit of sun is given as

R = 3.00 \times 10^4 Ly

R = 3.00 \times 10^4 (3\times 10^8)(365 \times 24 \times 3600)m

R = 2.84 \times 10^20 m

Part a)

centripetal acceleration is given as

a_c = \omega^2 R

a_c = \frac{4\pi^2}{T^2} R

a_c = \frac{4\pi^2}{(8.2\times 10^{15})^2}(2.84 \times 10^{20})

a_c = 1.67 \times 10^{-10} m/s^2

Part b)

orbital speed is given as

v = \frac{2\pi R}{T}

v = \frac{2\pi (2.84 \times 10^{20})}{8.2 \times 10^{15}}

v = 2.18 \times 10^5 m/s

5 0
4 years ago
A sound wave is passing by a student 250 times every second. What is the frequency of the sound wave?
inysia [295]
My answer i believe is simply 250 Hz, because sounds or vibrations travel in 1 cycle/second, meaning the number of cycles, in your case 250, divided by the time,1 second, will ultimately be 250 Hertz.  For every Cycle/second it will equal 1 Hz, so 250/1 = 250Hz
7 0
3 years ago
Read 2 more answers
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