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klemol [59]
3 years ago
13

What is the period of a wave with a frequency of 100 Hz and a wavelength of 2.0 m

Physics
2 answers:
spin [16.1K]3 years ago
5 0

The answer for the following answer is answered below.

  • <u><em>Therefore the time period of the wave is 0.01 seconds.</em></u>
  • <u><em>Therefore the option for the answer is "B".</em></u>

Explanation:

Frequency (f):

The number of  waves that pass a fixed place in a given amount of time.

The SI unit of frequency is Hertz (Hz)

Time period (T):

The time taken for one complete cycle of vibration to pass a given point.

The SI unit of time period is seconds (s)

Given:

frequency (f) = 100 Hz

wavelength (λ) = 2.0 m

To calculate:

Time period (T)

We know;

According to the formula;

<u>f =</u>\frac{1}{T}<u></u>

Where,

f represents the frequency

T represents the time period

from the formula;

  T = \frac{1}{f}

 T = \frac{1}{100}

  T = 0.01 seconds

<u><em>Therefore the time period of the wave is 0.01 seconds.</em></u>

pychu [463]3 years ago
4 0

Answer:

0.01

Explanation:

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bekas [8.4K]

Answer:

The work done on the canister by the 5.0 N force during this time is

54.06 Joules.

Explanation:

Let the initial kinetic energy of the canister be

KE₁ = \frac{1}{2} mv_1^{2} = \frac{1}{2} *3*3.6^{2} = 19.44 J in the x direction

Let the the final kinetic energy of the canister be

KE₂ = \frac{1}{2} mv_2^{2} = \frac{1}{2} *3*7.0^{2} = 73.5 J in the y direction

Therefore from the Newton's first law of motion, the effect of the force is the change of momentum and the difference in energy between the initial and the final

= 73.5 J - 19.44 J = 54.06 J

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3 years ago
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denis23 [38]
Answer

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3 years ago
How many moles are in 73.4 grams of Phosphorus?
mylen [45]
23.2


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3 years ago
What is the velocity of a wave with a wavelength of 3.7 m and a frequency of 22.5 Hz?
ExtremeBDS [4]

Answer: 343 m/s. The sound wave has a frequency of. 436 Hz. What is the period of the wave? T = = 436 Hz. = 2.29x10-3 s. C. What is the wave's wavelength? To halve

Explanation:

8 0
2 years ago
The force exerted by the wind on the sails of a sailboat is Fsail = 330 N north. The water exerts a force of Fkeel = 210 N east.
Elena L [17]

Answer:

The magnitude of the acceleration is a_r = 1.50 \ m/s^2

The direction is  \theta =  32.5 6^o north of  east

Explanation:

From the question we are told that

   The force exerted by the wind is  F_{sail} =  (330 ) \ N \ north

   The force exerted by water is  F_{keel} =  (210  ) \ N \ east

      The mass of the boat(+ crew) is  m_b  =  260  \ kg

Now Force is mathematically represented as

      F =  ma

Now the acceleration towards the north is mathematically represented as

      a_n  =  \frac{F_{sail}}{m_b}

substituting values

       a_n  =  \frac{330 }{260}

      a_n  =  1.269 \ m/s^2

Now the acceleration towards the east is mathematically represented as

       a_e = \frac{F_{keel}}{m_b }

substituting values

      a_e = \frac{210}{260}

      a_e =0.808 \ m/s^2

The resultant acceleration is  

      a_r =  \sqrt{a_e^2 + a_n^2}

substituting values

     a_r =  \sqrt{(0.808)^2 + (1.269)^2}

      a_r = 1.50 \ m/s^2

The direction with reference from the north is evaluated as

Apply SOHCAHTOA

        tan \theta =  \frac{a_e}{a_n}

       \theta = tan ^{-1} [\frac{a_e}{a_n } ]

substituting values

     \theta = tan ^{-1} [\frac{0.808}{1.269 } ]

    \theta = tan ^{-1} [0.636 ]

   \theta =  32.5 6^o

     

   

       

5 0
3 years ago
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