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Afina-wow [57]
3 years ago
12

THE RIGHT ANSWER WILL RECEIVE A BRAINLESS AND POINTS AND THANKS!!! THE RIGHT ANSWER WILL RECEIVE A BRAINLESS AND POINTS AND THAN

KS!!! What is the fundamental frequency of a mandolin string that is 42.0 cm long when the speed of waves of the string is 329 m/s? with working outs
Physics
2 answers:
Iteru [2.4K]3 years ago
7 0

Explanation:

이것이 정답입니다!!! (this is the correct answer)

#폴로 (주)따라 (follow me)

#이 대답을 마음(heart this answer)

Karo-lina-s [1.5K]3 years ago
6 0

Answer:

450km because this night founder

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"Force is applied to an object and the object is moved over a distance in the same direction of the applied force" is the defini
DaniilM [7]
I believe this is gravitational force

8 0
3 years ago
Read 2 more answers
A vertical spring has a spring constant of 2900 N/m. The spring is compressed 80 cm and a 8 kg spider is placed on the spring. T
Serga [27]

Answer:

a)  k_{e} = 928 J , b)U = -62.7 J , c) K = 0 , d) Y = 11.0367 m,  e)  v = 15.23 m / s  

Explanation:

To solve this exercise we will use the concepts of mechanical energy.

a) The elastic potential energy is

      k_{e} = ½ k x²

      k_{e} = ½ 2900 0.80²

      k_{e} = 928 J

b) place the origin at the point of the uncompressed spring, the spider's potential energy

     U = m h and

     U = 8 9.8 (-0.80)

     U = -62.7 J

c) Before releasing the spring the spider is still, so its true speed and therefore the kinetic energy also

      K = ½ m v²

      K = 0

d) write the energy at two points, maximum compression and maximum height

     Em₀ = ke = ½ m x²

     E_{mf} = mg y

     Emo = E_{mf}

     ½ k x² = m g y

     y = ½ k x² / m g

     y = ½ 2900 0.8² / (8 9.8)

     y = 11.8367 m

As zero was placed for the spring without stretching the height from that reference is

     Y = y- 0.80

     Y = 11.8367 -0.80

     Y = 11.0367 m

Bonus

Energy for maximum compression and uncompressed spring

     Emo = ½ k x² = 928 J

     E_{mf}= ½ m v²

     Emo = E_{mf}

     Emo = ½ m v²

      v =√ 2Emo / m

     v = √ (2 928/8)

     v = 15.23 m / s

8 0
3 years ago
How long does it take (in minutes) for light to reach venus from the sun, a distance of 1.152 × 108 km?
7nadin3 [17]
Using the precise speed of light in a vacuum (299,792,458 \ \frac{m}{s}), and your given distance of 1.152 * 10^{8} km, we can convert and cancel units to find the answer. The distance in m, using \frac{1000 \ m}{1 \ km}, is 1.152 * 10^{11} m. Next, for the speed of light, we convert from s to min, using \frac{1 \ min}{60 \ s}, so we divide the speed of light by 60. Finally, dividing the distance between the Sun and Venus by the speed of light in km per min, we find that it is 6.405 min.

7 0
3 years ago
Which of the following air conditions would be least likely to have precipitation?
Lynna [10]
The best and most correct answer among the choices provided by the question is the first choice "warm, dry air"


In meteorology, precipitation<span> is any product of the condensation of atmospheric water vapor that falls under gravity. The main forms of </span>precipitation<span> include drizzle, rain, sleet, snow, graupel and hail.</span>

I hope my answer has come to your help. God bless and have a nice day ahead!
7 0
3 years ago
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"A thin film with an index of refraction of 1.50 is placed in one of the beams of a Michelson interferometer. If this causes a s
Tamiku [17]

Answer:

The film thickness is 4.32 * 10^-6 m

Explanation:

Here in this question, we are interested in calculating the thickness of the film.

Mathematically;

The number of fringes shifted when we insert a film of refractive index n and thickness L in the Michelson Interferometer is given as;

ΔN = (2L/λ) (n-1)

where λ is the wavelength of the light used

Let’s make L the subject of the formula

(λ * ΔN)/2(n-1) = L

From the question ΔN = 8 , λ = 540 nm, n = 1.5

Plugging these values, we have

L = ((540 * 10^-9 * 8)/2(1.5-1) = (4320 * 10^-9)/1 = 4.32 * 10^-6 m

6 0
3 years ago
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