Charge q1 is distance s from the negative plate of a parallel-plate capacitor. Charge q2=q1/3 is distance 2s from the negative p
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Answer:
Decreased by a factor of 4.5
Explanation:
"We have Newton formula for attraction force between 2 objects with mass and a distance between them:
where is the gravitational constant on Earth.
are the masses of the object and Earth itself. and R distance between, or the Earth radius.
So when R is tripled and mass is doubled, we have the following ratio of the new gravity over the old ones:
Since and
So gravity would have been decreased by a factor of 4.5
When air is blown into the open pipe,
L =
where nis any integral number 1,2,3,4 etc. and λ is the wavelength of the oscillation
⇒λ=
Note here that n=1 is for fundamental, n=2 is first harmonic and so on..
⇒ third harmonic will be n=4
Given L=6m, n=4, solving for λ we get:
λ= =3m
Relationship of frequency(f), velocity of sound (c) and wavelength(λ) is:
c=f.λ Or f=
⇒f=
≈115 Hz
A. 90.1 m
The wavelength of a wave is given by:
where
v is the speed of the wave
f is its frequency
For the sound emitted by the whale, v = 1531 m/s and f = 17.0 Hz, so the wavelength is
B. 102 kHz
We can re-arrange the same equation used previously to solve for the frequency, f:
where for the dolphin:
v = 1531 m/s is the wave speed
is the wavelength
Substituting into the equation,
C. 13.6 m
Again, the wavelength is given by:
where
v = 340 m/s is the speed of sound in air
f = 25.0 Hz is the frequency of the whistle
Substituting into the equation,
D. 4.4-8.7 m
Using again the same formula, and using again the speed of sound in air (v=340 m/s), we have:
- Wavelength corresponding to the minimum frequency (f=39.0 Hz):
- Wavelength corresponding to the maximum frequency (f=78.0 Hz):
So the range of wavelength is 4.4-8.7 m.
E. 6.2 MHz
In order to have a sharp image, the wavelength of the ultrasound must be 1/4 of the size of the tumor, so
And since the speed of the sound wave is
v = 1550 m/s
The frequency will be