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2 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?

1 answer:

d1i1m1o1n [39]2 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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!!!!BRAINLIEST!!!!!

kykrilka [37]

Answer:

The jp2003parker guy is extremely wrong

So he says that the size wont matter and a physical change should occur, but how would the size change without having a physical change occur.

Explanation:

7 0

2 years ago

Read 2 more answers

How long does it take for the speed of light and sound to travel around the world

irina [24]

-- Light travels straight, not around in a circle. But if it did, it would cover
a distance equal to the length of the equator in about <em>0.13 second</em>.

-- At the speed of sound (in air at standard temperature and pressure),
it would take about <em>32.6 hours </em>to cover the same distance.

6 0

2 years ago

Match each example to its type of energy.

PtichkaEL [24]

Solar energy - A

nuclear energy - B

fossil fuel energy - C

wind energy - D

geothermal energy - E

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3 years ago

Use the Bohr model to address this question. When a hydrogen atom makes a transition from the 5 th energy level to the 2nd, coun

iris [78.8K]

Answer:

A. 2.82 eV

B. 439nm

C. 59.5 angstroms

Explanation:

A. To calculate the energy of the photon emitted you use the following formula:

$E_{n1,n2}=-13.4(\frac{1}{n_2^2}-\frac{1}{n_1^2})$ (1)

n1: final state = 5

n2: initial state = 2

Where the energy is electron volts. You replace the values of n1 and n2 in the equation (1):

$E_{5,2}=-13.6(\frac{1}{5^2}-\frac{1}{2^2})=2.82eV$

B. The energy of the emitted photon is given by the following formula:

$E=h\frac{c}{\lambda}$ (2)

h: Planck's constant = 6.62*10^{-34} kgm^2/s

c: speed of light = 3*10^8 m/s

λ: wavelength of the photon

You first convert the energy from eV to J:

$2.82eV*\frac{1J}{6.242*10^{18}eV}=4.517*10^{-19}J$

Next, you use the equation (2) and solve for λ:

$\lambda=\frac{hc}{E}=\frac{(6.62*10^{-34} kg m^2/s)(3*10^8m/s)}{4.517*10^{-19}J}=4.39*10^{-7}m=439*10^{-9}m=439nm$

C. The radius of the orbit is given by:

$r_n=n^2a_o$ (3)

where ao is the Bohr's radius = 2.380 Angstroms

You use the equation (3) with n=5:

$r_5=5^2(2.380)=59.5$

hence, the radius of the atom in its 5-th state is 59.5 anstrongs

8 0

2 years ago

A particle (charge = +0.8 mC) moving in a region where only electric forces act on it has a kinetic energy of 6.7 J at point A.

Maksim231197 [3]

Answer:

The kinetic energy of the particle as it moves through point B is 7.9 J.

Explanation:

The kinetic energy of the particle is:

$\Delta K = \Delta E_{p} = q\Delta V$

<u>Where</u>:

K: is the kinetic energy

$E_{p}$ : is the potential energy

q: is the particle's charge = 0.8 mC

ΔV: is the electric potential = 1.5 kV

$\Delta K = q \Delta V= 0.8 \cdot 10^{-3} C*1.5 \cdot 10^{3} V = 1.2 J$

Now, the kinetic energy of the particle as it moves through point B is:

$\Delta K = K_{f} - K_{i}$

$K_{f} = \Delta K + K_{i} = 1.2 J + 6.7 J = 7.9 J$

Therefore, the kinetic energy of the particle as it moves through point B is 7.9 J.

I hope it helps you!

8 0

3 years ago