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insens350 [35]
4 years ago
13

What is the frequency of a clock waveform whose period is 750 microseconds?

Physics
1 answer:
stich3 [128]4 years ago
6 0
The frequency f of a wave is defined as the inverse of the period T:
f= \frac{1}{T}
The clock in the problem has a waveform with period T=750 ms=0.750 s. Therefore, its frequency is
f= \frac{1}{0.750 s}=1.33 Hz
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a) Two protons in a molecule are 3.40 10-10 m apart. Find the electric force exerted by one proton on the other. magnitude 1.99e
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Answer: a) F_{e}= 1.99 x 10⁻⁹N

              b) Electric force is 1.24 x 10³⁶ times larger than gravitational force

Explanation: <u>Electric</u> <u>Force</u> is an interaction between two charges at a distance. It is calculated as

F_{e}=k\frac{|q||Q|}{r^{2}}

k is electric constant that values 9 x 10⁹N.m²/C²

q and Q are the quantity of charges

r is distance between the charges

a) For the two protons, charge = 1.602 x 10⁻¹⁹C:

F_{e}=9.10^{9}\frac{(1.602.10^{-19})(1.602.10^{-19})}{(3.4.10^{-10})^{2}}

F_{e}= 1.99 x 10⁻⁹ N

Magnitude of electric force between 2 protons is 1.99 x 10⁻⁹ N and it is repulsive because they are both positive

<u>Gravitational</u> <u>Force</u> is the force of attraction that acts between all materials.

It is proportional to the object's masses and inversely proportional to their squared distance:

F_{g}=G\frac{mM}{r^{2}}

G is gravitational constant that values G = 6.67 x 10⁻¹¹ N.m²/kg²

b) For the two protons, m = 1.67 x 10⁻²⁷kg

F_{g}=6.67.10^{-11}\frac{(1.67.10^{-27})(1.67.10^{-27})}{(3.4.10^{-10})^{2}}

F_{g}= 1.61 x 10⁻⁴⁵N

Comparing electrical and gravitational forces:

\frac{F_{e}}{F_{g}} =\frac{1.99.10^{-9}}{1.61.10^{-45}}

\frac{F_{e}}{F_{g}}= 1.24 x 10³⁶

This shows that electrical force is <u>1.24 x 10³⁶</u> times greater than gravitational force.

4 0
3 years ago
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