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Readme [11.4K]
3 years ago
8

An example of an anaerobic exercise is what?

Physics
1 answer:
san4es73 [151]3 years ago
6 0

Answer:

1st: Theatre History

4th Quarter

Upcoming

Due today

Syllabus

Due Sunday1st: Theatre History

4th Quarter

Upcoming

Due today

Syllabus

Due Sunday1st: Theatre History

4th Quarter

Upcoming

Due today

Syllabus

Due Sunday

Explanation:

1st: Theatre History

4th Quarter

Upcoming

Due today

Syllabus

Due Sunday

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) Music. When a person sings, his or her vocal cords vibrate in a repetitive pattern that has the same frequency as the note tha
vaieri [72.5K]

(a) 0.0021 s, 2926.5 rad/s

The frequency of the B note is

f= 466 Hz

The time taken to make one complete cycle is equal to the period of the wave, which is the reciprocal of the frequency:

T=\frac{1}{f}=\frac{1}{466 Hz}=0.0021 s

The angular frequency instead is given by

\omega = 2\pi f

And substituting

f = 466 Hz

We find

\omega = 2\pi (466 Hz)=2926.5 rad/s

(b) 20 Hz, 125.6 rad/s

In this case, the period of the sound wave is

T = 50.0 ms = 0.050 s

So the frequency is equal to the reciprocal of the period:

f=\frac{1}{T}=\frac{1}{0.050 s}=20 Hz

While the angular frequency is given by:

\omega = 2\pi f = 2 \pi (20 Hz)=125.6 rad/s

(c) 4.30\cdot 10^{14} Hz, 7.48\cdot 1^{14} Hz, 2.33\cdot 10^{-15} s, 1.34\cdot 10^{-15}s

The minimum angular frequency of the light wave is

\omega_1 = 2.7\cdot 10^{15}rad/s

so the corresponding frequency is

f=\frac{\omega}{2 \pi}=\frac{2.7\cdot 10^{15}rad/s}{2\pi}=4.30\cdot 10^{14} Hz

and the period is the reciprocal of the frequency:

T=\frac{1}{f}=\frac{1}{4.30\cdot 10^{14}Hz}=2.33\cdot 10^{-15}s

The maximum angular frequency of the light wave is

\omega_2 = 4.7\cdot 10^{15}rad/s

so the corresponding frequency is

f=\frac{\omega}{2 \pi}=\frac{4.7\cdot 10^{15}rad/s}{2\pi}=7.48\cdot 10^{14} Hz

and the period is the reciprocal of the frequency:

T=\frac{1}{f}=\frac{1}{7.48\cdot 10^{14}Hz}=1.34\cdot 10^{-15}s

(d) 2.0\cdot 10^{-7}s, 3.14\cdot 10^{7} rad/s

In this case, the frequency is

f=5.0 MHz = 5.0 \cdot 10^6 Hz

So the period in this case is

T=\frac{1}{f}=\frac{1}{5.0\cdot 10^6  Hz}=2.0 \cdot 10^{-7} s

While the angular frequency is given by

\omega = 2\pi f=2 \pi (5.0\cdot 10^{6}Hz)=3.14\cdot 10^{7} rad/s

7 0
3 years ago
HELP Giving all the points I can. Physics virtual lab report: circuit design<br><br> Please help me.
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Answer:text me I can help

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Beams used in Heavy Timber construction are sometimes firecut. This is done to: a) allow air circulation at the beam’s end in an
ladessa [460]

Answer:

option C

Explanation:

The correct answer is option C

Fire cut of fireman cut is diagonal cut which is provided at the end of the beam to prevent the fall of masonry wall if a fire breaks out in the building.

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What 1columb=?<br> please help​
Juli2301 [7.4K]

Explanation:

it's a unit used to measure charge (C)

1C=1000millicoulombs

1millicoulomb=1000microcoulumbs

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The constants could be the day because the experiment is all on the same day or the person throwing the pumpkin because it will always be the same perosn
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