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2 years ago

If the second harmonic of a certain string is 42 Hz, what is the fundamental frequency of the string?

Physics

1 answer:

stich3 [128]2 years ago

5 0

I would think that you would have to do 42/2=21Hz, but I'm not sure...

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An earthquake emits both S-waves and P-waves which travel at different speeds through the Earth. A P-wave travels at 9 000 m/s a

Cerrena [4.2K]

Assuming constant speeds, the P-wave covers a distance d in time t such that

9000 m/s = d/(60 t)

while the S-wave covers the same distance after 1 more minute so that

5000 m/s = d/(60(t + 1))

Now,

d = 540,000 t

d = 300,000(t + 1) = 300,000 t + 300,000

Solve for t in the first equation and substitute it into the second equation, then solve for d :

t = d/540,000

d = 300,000/540,000 d + 300,000

4/9 d = 300,000

d = 675,000

So the earthquake center is 675,000 m away from the seismic station.

6 0

3 years ago

What is the sun's right ascension on the first day of winter?

Savatey [412]

From the point of view of the northern hemisphere:

The sun starts out the year at the Spring equinox, March 21,
Right Ascension 0, Declination 0.

3 months later, Summer solstice, June 21
RA = 6 hours, Dec. = +23.5°

3 months later, Fall equinox, September 21,
RA = 12 hours, Dec. = 0

3 months later, Winter solstice, December 21,
RA = 18 hours, Dec. = -23.5°

5 0

2 years ago

Read 2 more answers

A motorcycle, which has an initial linear speed of 8.0 m/s, decelerates to a speed of 2.2 m/s in 4.1 s. Each wheel has a radius

SVETLANKA909090 [29]

Answer:

Explanation:

Given

Initial linear speed $v_1=8 m/s$

initial angular velocity $\omega _1=\frac{v_1}{r}=\frac{8}{0.6}=13.33 rad/s$

Speed after 4.1 s is $v_2=2.2 m/s$

$\omega _2=\frac{2.2}{0.6}=3.66 rad/s$

using $\omega _2=\omega _1+\alpha t$

where $\alpha$ is angular acceleration

$3.66=13.33+\alpha \cdot 4.1$

$\alpha =-2.37 rad/s^2$ i.e. clockwise

(b)angular displacement

$\theta =\omega _1t+\frac{\alpha t^2}{2}$

$\theta =13.33\times 4.1-\frac{2.37\cdot 4.1^2}{2}$

$\theta =54.66-19.75$

$\theta =34.91 rad$

3 0

3 years ago

A solid cube of aluminum (density of 2.7 g/cm³) has a volume of 0.9 cm³. how many atoms are contained in the cube?

Reika [66]

Answer:

$0.542*10^{23}\ Aluminum\ Atoms$

Explanation:

$We\ are\ given:\\Density\ of\ aluminum=2.7\ g/cm^3\\Volume\ of\ aluminum-cube=0.9\ cm^3\\Hence,\\As\ we\ know\ that,\\Density=\frac{Mass}{Volume}\\Mass=Density*Volume\\Hence,\ here,\\Mass\ of\ the\ solid\ iron\ cube=2.7*0.9=2.43\ g\\Now,\\We\ also\ know\ that,\\Gram\ Atomic\ mass\ of\ Aluminum = 26.98 \approx 27\ g\\Hence,\\No.\ of\ particles=\frac{Mass}{GAM}*Avagadro's Constant\\Hence,\ here\\No.\ of\ Aluminum\ atoms=\frac{2.43}{27}*6.022*10^{23} \approx 0.542*10^{23}\ Aluminum\ Atoms$