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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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An indirect measurement of the speed of molecules is temperature

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A small rocket with mass 20.0 kg is moving in free fall toward the earth. Air resistance can be neglected. When the rocket is 80
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Answer:

= 308.5 N

Explanation:

acceleration  of rocket for safe landing

v_{y} ^{2} = v_{0} ^{2} +2ay

a = \frac{v_{y}^{2} - v_{0}^{2}  }{2y}

v_{0} = initial velocity

v_{y} =final Velocity

m = mass of rocket

\frac{0^{2} - 30^{2}  }{2\times80}

-5.625 m/s^{2}

Upward force

F - mg = ma

F = ma+ mg

F = m(a+g)

m= mass

a = 5.625 m/s^{2}

g = 9.8m/s^{2}

= 20(5.625 m/s^{2}+ 9.8m/s^{2})

= 308.5 N

7 0
3 years ago
A river has a steady speed 0.500 m/s. A student swims upstream a distance of1.00 km and swims back to the starting point. a) If
Ivahew [28]

Answer:

a) 33.6 min

b) 13.9 min

c)  Intuitively, it takes longer to complete the trip when there is current because, the swimmer spends much more time swimming at the net low speed (0.7 m/s) than the time he spends swimming at higher net speed (1.7 m/s).


Explanation:


The problem deals with relative velocities.

  • Call Vr the speed of the river, which is equal to 0.500 m/s
  • Call Vs the speed of the student in still water, which is equal to 1.20 m/s
  • You know that when the student swims upstream, Vr and Vs are opposed and the net speed will be Vs - Vr
  • And when the student swims downstream, Vr adds to Vs and the net speed will be Vs + Vr.

Now, you can state the equations for each section:

  • distance = speed × time
  • upstream: distance = (Vs - Vr) × t₁ = 1,000 m
  • downstream: distance = (Vs + Vr) × t₂ = 1,000 m

Part a). To state the time, you substitute the known values of Vr and Vs and clear for the time in each equation:

  • (Vs - Vr) × t₁ = 1,000 m
  • (1.20 m/s - 0.500 m/s) t₁ = 1,000 m⇒ t₁ = 1,000 m / 0.70 m/s ≈ 1429 s
  • (1.20 m/s + 0.500 m/s) t₂ = 1,000 m ⇒ t₂ = 1,000 m / 1.7 m/s ≈ 588 s
  • total time = t₁ + t₂ = 1429s + 588s =  2,017s
  • Convert to minutes: 2,0147 s ₓ 1 min / 60s ≈ 33.6 min

Part b) In this part you assume that the complete trip is made at the velocity Vs = 1.20 m/s


  • time = distance / speed = 1,000 m / 1.20 m/s ≈ 833 s ≈ 13.9 min

Part c) Intuitively, it takes longer to complete the trip when there is current because the swimmer spends more time swimming at the net speed of 0.7 m/s than the time than he spends swimming at the net speed of 1.7 m/s.

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Answer:

c

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Answer:

Explanation:

The magnitude of the acceleration in the x direction is always zero: TRUE.

At the apex, in the y direction the velocity is zero and the accelration is positive. TRUE.

At the apex, in the x direction, the velocity is zero and the acceleration is zero. FALSE. The accelration is zero, but the velocity is the same it had when it was shot.

The magnitude of the velocity in the y direction is always constant. FALSE, it's subject to gravity and it's velocity varies as v=v_0 - 9.81ms^{-2} t

At the apex, in the Y direction, the velocity is negative and the acceleration is zero. FALSE. Velocity is zero, Acceleration is 9.81 ms^{-2}, towards the negative y axis

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