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Nuetrik [128]
3 years ago
11

Theoretically speaking when analyzing the frequency spectrum of dolphin vocalizations, which may go up to 150 khz, what is the l

owest sampling frequency necessary?
Physics
1 answer:
d1i1m1o1n [39]3 years ago
5 0
Near 40 hz to 20 hz youre welcome im not to sure on this answer because it was simply an experiment i did once
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1)Which conditions will result in the smallest change in momentum?. A)a large force over a long time period. B)a large force ove
fiasKO [112]
D, small force for short time, as it has the least influence on the velocity. 
C, momentum is velocity x mass, so if velocities are the same, then the object with less mass with have a lower momentum.
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3 years ago
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Calculate the potential energy of a 6kg bowling ball suspended 3m above the surface of the earth
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Answer:  176.4 Joules

Explanation: Potential energy of a body is defined as the energy possessed by virtue of position.

Potential energy =m\times g\times h

m = mass of the body = 6 kg

g = acceleration due to gravity = 9.8 ms^{-2}

h = height of body = 3 m

Potential energy =6 kg\times 9.8ms^{-2}\times 3m=176.4kgm^2s^{-2}

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8 0
3 years ago
A tuning fork produced four beats per second with a second, 270-Hz tuning fork. What are the two possible frequencies of the fir
EleoNora [17]

Answer:

frequency of the first tuning fork can be 274 Hz or 266 Hz

Explanation:

Given data :

we have number of beats per second = 4

thus, it can also be written as

Beat frequency of the tuning fork, v = 4 Hz

Frequency of the tuning fork, f = 270 Hz

now, the beat frequency is actually the difference of the consecutive frequency.

thus, we have

Frequency of the first tuning fork as = 270 Hz + 4 Hz = 274 Hz

or

Frequency of the first tuning fork as = 270 Hz - 4 Hz = 266 Hz

Therefore, the frequency of the first tuning fork can be 274 Hz or 266 Hz

7 0
3 years ago
A person pushes on the handle of a lawnmower with a force of 280. n. if the handle makes an angle of 40.0 degrees with the groun
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<span>Force F = 280 N Angle with the ground = 40 degrees Weight of the Lawnmower = 350 N Velocity is constant so Acceleration is 0 So Forward force Ff = F cos theta = 280 cos40 Frictional force with resists to back Fb = (u x Force from pressure) + vertical component of Force, where u is the coefficient of friction. Fb = (u x m x g) + (u x 280sin40) AS Ff = Fb => 280 cos40 = u x ((m x g) + 280sin40) u = 280 cos40 / ((350 x 9.81) + 280sin40) = 214.49 / () = 0.405 So the coefficient of friction u = 0.405</span>
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Ok so basically.. and then if you.. you get …
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