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3 years ago

How do you determine the amplitude of a transverse wave

Chemistry

2 answers:

andrew11 [14]3 years ago

7 0

Answer:

The amplitude of a transverse wave is the difference between the height of crest which is the maximum point and the resting position the minimum point of one curve of a wave.

Please look at the attachment to see the figure to make the answer more clear.

Transverse waves are those that moves in the direction along x-axis

barxatty [35]3 years ago

4 0

To determine the amplitude of a transverse wave, measure the highest amount of disturbance from the equilibrium that the wave experiences.

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Chlorine–35 has 17 protons. How many protons and neutrons does the isotope chlorine–36 have? 19 protons and 17 neutrons 17 proto

pochemuha

Answer:

17 protons

19 neutrons

Explanation:

Chlorine will always have the same amount of protons, and that would be 17 protons.

The atomic mass will change according to how many neutrons are present.

Cl - 35 is comprised of 17 protons and 18 neutrons.

We want to find Cl - 36:

We simply add 1 neutron. 18 + 1 = 19 neutrons.

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3 years ago

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saw5 [17]

Answer: B) Wood Shavings

Reason/Explanation:
Notice what options A, C, and D have on common. Ice, salt and sugar. Those are all dissolvable. Insoluble means un-able to dissolve. It can't dissolve. Wood shavings can't dissolve in water. They will just float. Sugar, salt and ice will all dissolve. Wood shavings won't.

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3 years ago

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HELP When 400 J of heat is added to 5.6 g of olive oil at 23*C, the temperature increases to 87*C. What is the specific heat of

babunello [35]

Q=mc*temperature change

400=5.6/1000*c*(87-23)

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3 years ago

7. How many moles are in 5 x 10 2 atoms of gold?

goblinko [34]

Answer:

8.3028894e-22

Explanation:

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3 years ago

What volume of lead (of density 11.3 g/cm3 ) has the same mass as 395 cm3 of a piece of redwood (of density 0.38 g/cm3 )? Answer

Zepler [3.9K]

$d_{1}=\frac{m}{V_{1}}\\\\
V_{1}=395cm^{3}\\
d_{1}=0,38\frac{g}{cm^{3}} \ \ \ \ \Rightarrow \ \ \ \ m=395cm^{3}*0,38\frac{g}{cm^{3}}=150,1g\\\\\\
d_{2}=\frac{m}{V_{2}}\\\\
m=150,1g\\
d_{2}=11,3\frac{g}{cm^{3}} \ \ \ \ \Rightarrow \ \ \ \ V_{2}=\frac{150,1g}{11,3\frac{g}{cm^{3}}}\approx13,28cm^{3}$

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3 years ago