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FromTheMoon [43]
2 years ago
11

Give Ideas to create hearing instruments to fix hearing problems in space, as you know sound waves can’t travel through empty sp

ace.
Physics
1 answer:
Andre45 [30]2 years ago
3 0
Hearing test provides an evaluation of the sensitivity of a person's sense of hearing and is most often performed by an audiologist using an audiometer. An audiometer is used to determine a person's hearing sensitivity at different frequencies.
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Can someone help me with this ?<br><br>I need proper explanation ~ ​
Pepsi [2]

Answer:

.

Explanation:

3 0
2 years ago
I attach a 4.1 kg block to a spring that obeys Hooke's law and supply 3.8 J of energy to stretch the spring. I release the block
borishaifa [10]

Answer:

The amplitude of the oscillation is 2.82 cm

Explanation:

Given;

mass of attached block, m = 4.1 kg

energy of the stretched spring, E = 3.8 J

period of oscillation, T = 0.13 s

First, determine the spring constant, k;

T = 2\pi \sqrt{\frac{m}{k} }

where;

T is the period oscillation

m is mass of the spring

k is the spring constant

T = 2\pi \sqrt{\frac{m}{k} } \\\\k = \frac{m*4\pi ^2}{T^2} \\\\k = \frac{4.1*4*(3.142^2)}{(0.13^2)} \\\\k = 9580.088 \ N/m\\\\

Now, determine the amplitude of oscillation, A;

E = \frac{1}{2} kA^2

where;

E is the energy of the spring

k is the spring constant

A is the amplitude of the oscillation

E = \frac{1}{2} kA^2\\\\2E = kA^2\\\\A^2 = \frac{2E}{k} \\\\A = \sqrt{\frac{2E}{k} } \\\\A =  \sqrt{\frac{2*3.8}{9580.088} }\\\\A = 0.0282 \ m\\\\A = 2.82 \ cm

Therefore, the amplitude of the oscillation is 2.82 cm

8 0
2 years ago
What is the wavelength, in nm, of the line in the hydrogen spectrum when one n value is 3 and the other n value is 6?
iren [92.7K]

Answer:

\lambda=1090nm

Explanation:

Rydberg formula is used to calculate the wavelengths of the spectral lines of many chemical elements. For the hydrogen, is defined as:

\frac{1}{\lambda}=R_H(\frac{1}{n_1^2}-\frac{1}{n_2^2})

Where R_H is the Rydberg constant for hydrogen and n_1, n_2 are the lower energy state and the higher energy state, respectively.

\frac{1}{\lambda}=1.10*10^{7}m^{-1}(\frac{1}{3^2}-\frac{1}{6^2})\\\frac{1}{\lambda}=9.17*10^{5}m^{-1}\\\lambda=\frac{1}{1.09*10^{6}m^{-1}}\\\lambda=1.09*10^{-6}m*\frac{10^{9}nm}{1m}\\\lambda=1090nm

4 0
3 years ago
A phosphodiester bond is used to:
bonufazy [111]

Answer:

A. Join glycerol to fatty acids

Explanation:

I majored in Physics.

4 0
2 years ago
Read 2 more answers
Dakdadakdadakdadakda
notsponge [240]

Answer: dakdadakdadakdadakda

Explanation:(sings) blah blah blah middle fingers in the air l-l-l-loser

4 0
2 years ago
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