Is there more potentioal energy at the top of bottom of a ramp?

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

vitfil [10]

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/

3 0

3 years ago

Suppose you want to move a big rock in your yard. It is about waist high and 4 feet long, has a volume of 1.2 m3 and a density o

beks73 [17]

Answer:

6613.87 lbs

Explanation:

1.2 m³ = 1200000 cm³

Mass = Density * Volume

M = (2.5 g/cm³) * 1200000 cm³ = 3000000 g

1 lb = 453.592 g

3000000 g * (1 lbs / 453.592 g) = 6613.87 lbs

3 0

3 years ago

Two sound waves (wave X and wave Y) are moving through a medium at the same speed. If wave X has a greater frequency than wave Y

Naya [18.7K]

Answer:

Wave X has a shorter wavelength.

Explanation:

It is given that, two sound waves (wave X and wave Y) are moving through a medium at the same speed.

Speed of a wave is given by :

$v=\nu\times \lambda$

Where,

$\nu$ is frequency

$\lambda$ is wavelength

The relation between the frequency and the wavelength is inverse. Wave X has greater frequency than Y, then wave X has a shorter wavelength.

Hence, the correct option is (B) "has a shorter wavelength".

8 0

3 years ago

GIVING BRAINLIEST PLEASE HELP!!

Marrrta [24]

The answer is D hope it helps

8 0

3 years ago

Returning once again to our table top example of a horizontal mass on a low-friction surface with m = 0.254 kg and k = 10.0 N/m

Julli [10]

Explanation:

Given that,

Mass = 0.254 kg

Spring constant [tex[\omega_{0}= 10.0\ N/m[/tex]

Force = 0.5 N

y = 0.628

We need to calculate the A and d

Using formula of A and d

$A=\dfrac{\dfrac{F_{0}}{m}}{\sqrt{(\omega_{0}^2-\omega^{2})^2+y^2\omega^2}}$ .....(I)

$tan d=\dfrac{y\omega}{(\omega^2-\omega^2)}$ ....(II)

Put the value of $\omega=0.628\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-0.628)^2+0.628^2\times0.628^2}}$

$A=0.0198$

From equation (II)

$tan d=\dfrac{0.628\times0.628}{((10.0^2-0.628)^2)}$

$d=0.0023$

Put the value of $\omega=3.14\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-3.14)^2+0.628^2\times3.14^2}}$

$A=0.0203$

From equation (II)

$tan d=\dfrac{0.628\times3.14}{((10.0^2-3.14)^2)}$

$d=0.0120$

Put the value of $\omega=6.28\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-6.28)^2+0.628^2\times6.28^2}}$

$A=0.0209$

From equation (II)

$tan d=\dfrac{0.628\times6.28}{((10.0^2-6.28)^2)}$

$d=0.0257$

Put the value of $\omega=9.42\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-9.42)^2+0.628^2\times9.42^2}}$

$A=0.0217$

From equation (II)

$tan d=\dfrac{0.628\times9.42}{((10.0^2-9.42)^2)}$

$d=0.0413$

Hence, This is the required solution.

5 0

3 years ago