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IRISSAK [1]
3 years ago
11

A device called oacillatwr is used to swnd waves along a streatched string to send waves along a streatched string. The string i

s 20 cm long, and four complete waves fit along its length when the oscillation vibrates 30 times per second the waves on the string.. What is its frequency and speed and wavelength
Physics
1 answer:
vitfil [10]3 years ago
3 0

Answer:

Frequency = 2m/s1m=2 waves/s, or 2 Hz

A wave is traveling at a speed of 2 m/s and has a frequency of 2 Hz.

Explanation:

wd Calculating Wave Frequency or Wavelength from Wave Speed

The equation for wave speed:

Frequency = SpeedWavelength or Wavelength = SpeedFrequency

Therefore, if you know the speed of a wave and either the wavelength or wave frequency, you can calculate the missing value.

For example, suppose that a wave is traveling at a speed of 2 meters per second and has a wavelength of 1 meter. Then the frequency of the wave is:

- Frequency = 2m/s1m=2 waves/s, or 2 Hz

A wave is traveling at a speed of 2 m/s and has a frequency of 2 Hz. What is its wavelength?

- Substitute these values into the equation for wavelength:

- Wavelength = 2m/s2waves/s=1 m

See more: https://www.ck12.org/physics/wave-speed/lesson/Wave-Speed-MS-PS/

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notsponge [240]

Answer:

180m

Explanation:

We can use the formula [ d = st ].

12 * 15 = 180m

Best of Luck!

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2 years ago
Atoms contain empty space true or false​
olga nikolaevna [1]
The answer would be false
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3 years ago
There are many ways to lose weight, but a sustainable plan generally involves behavior modification related to diet and exercise
Anna007 [38]

The answer is adequate nutrition, regular physical activity, and practical ways to reduce calories while retaining important nutrients.

Explanation:

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5 0
3 years ago
Find the potential inside and outside a uniformly charged solid sphere whose radius is R and whose total charge is q. Use infini
amid [387]

Answer:

Recall that the electric field outside  a uniformly charged solid sphere  is exactly the same as if the charge were all at a point in the centre of the  sphere:

E_{outside} =\frac{1}{4\pi(e_{0})}\frac{Q}{r^{2} } r^{'}

lnside the sphere, the electric field also acts like a point charge, but only for the proportion of the charge further inside than the point r:

E_{inside} =\frac{1}{4\pi(e_{0})}\frac{Q}{R^{2} } \frac{r}{R} r^{'}

To find the potential, we integrate the electric field on a path from infinity (where of course, we take the direct path so that we can write the it as a 1 D integral):

V(r>R)=\int\limits^r_\infty {\frac{1}{4\pi(e_{0)} }\frac{Q}{r^2}  } \, dr=\frac{q}{4\pi(e_{0)} } \frac{1}{r} \\V(r

=\frac{q}{4\pi e_{0} } [\frac{1}{R} -\frac{r^{2}-R^{2}  }{2R^{3} } ]

∴NOTE: Graph is attached

8 0
3 years ago
2 pts
const2013 [10]

Answer:

I think the answer 1

Explanation:

im probably wrong too i dont know

5 0
3 years ago
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