All categories

2 years ago

2. Apply Mathematics If the amplitude of the 6 PM wave increases to 0.6 m, how many times greater would the energy become?

Physics

1 answer:

RoseWind [281]2 years ago

7 0

Answer:

8 and 093 over 0.6 then 303

Explanation:

cross over the numbers and get the answer

You might be interested in

Red laser light from a He-Ne laser (λ = 632.8 nm) creates a second-order fringe at 53.2° after passing through the grating. What

Svetlanka [38]

Explanation:

It is given that,

Wavelength of red laser light, $\lambda=632.8\ nm=632.8\times 10^{-9}\ m$

The second order fringe is formed at an angle of, $\theta=53.2^{\circ}$

For diffraction grating,

$d\ sin\theta=n\lambda$

$d=\dfrac{n\lambda}{sin\theta}$ , n = 2

$d=\dfrac{2\times 632.8\times 10^{-9}}{sin(53.2)}$

$d=1.58\times 10^{-6}\ m$

The wavelength λ of light that creates a first-order fringe at 22 is given by :

$\lambda=d\ sin\theta$

$\lambda=1.58\times 10^{-6}\ sin(22)$

$\lambda=5.91\times 10^{-7}\ m$

$\lambda=591\ nm$

Hence, this is the required solution.

6 0

3 years ago

The distance between Earth and Mars is 225 million km. When converted using the conversion factor 1 AU = 1.5 × 108 km, the dista

noname [10]

To convert km to AU, we divide 225,000,000 km by the factor of 1.5 x 10^8 = 150,000,000 km. This gives us 225,000,000 / 150,000,000 = 1.5 AU. Therefore, the distance between Earth and Mars in AU is 1.5 AU.
The AU is not equivalent to a light-year. A light-year is equivalent to around 9.5 x 10^12 kilometers.

4 0

3 years ago

The action that has the ability to change the motion of an object is

Troyanec [42]

The answer to this question is force

3 0

2 years ago

Two cars, one in front of the other, are traveling down the highway at 25 m/s. the car behind sounds its horn, which has a frequ

netineya [11]

You are given two cars, one in front of the other, that are traveling down the highway at 25 m/s. You are also given a frequency of 500 Hz of the car travelling behind it. You are asked what is the frequency heard by the driver of the lead car. This problem can be solved using the Doppler effect

sound frequency heard by the lead car = [(speed of sound + lead car velocity)/( speed of sound + behind car velocity)] * (sound of frequency of the behind car)
sound frequency heard by the lead car = [(340 m/s + 25 m/s)/(340 m/s - 25 m/s)] * (500 Hz)
sound frequency heard by the lead car = 579 Hz

7 0

3 years ago

if an object is being acted on by two forces a push and a pull of 6N what is the net force of the object

Paul [167]

The concepts of force, mass, and weight play critical roles. Newton's Laws of. Motion ... the person stops pushing? ... F net =10 N 2 N. = 8 N (to the right) a = F net m. = 8 N. 5 kg. =1.6 m s. 2 ... Two equal forces act on an object in the directions shown. If these ... Two connected carts being accelerated by a force F applied by.

6 0

2 years ago