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amm1812
3 years ago
10

A slinky is stretched across a classroom and moved up and down at a frequency of 2 hz. if the corresponding wave velocity is 4.2

m s , determine the wavelength of the slinky wave.
Physics
1 answer:
marusya05 [52]3 years ago
4 0
Velocity of a wave is the amount of wavelength multiplied by its frequency. Moreover, derivations would be;wavelength?
Hence wavelength would be velocity divided by the frequency.Result is 2.1 mThank you for your question. Please don't hesitate to ask in Brainly your queries. 
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Suppose you are on a cart that is moving at a constant speed v toward the left on a frictionless track. If you throw a massive b
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The correct answer is A
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2 years ago
Gyroscopic wander can be divided in to two categories, one is DRIFT, what is the other?
Tju [1.3M]

Gyroscopic wander can be divided into two categories and these are:

  1. Drift
  2. Topple

<h3>What is gyroscopic wander?</h3>

Gyroscopic wander can be defined as a movement of the spin axis (axis of rotation) away from a specific fixed direction.

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4 0
1 year ago
A sound is produced at one extremity of a metallic pipe of 1000 m of length. A person being at
prohojiy [21]

The sound was repeated because of the phenomenon of echo. The speed of sound in the metal is 800 m/s.

Echo results from the reflection of sound waves. The reason why a sound may be heard twice owes to the phenomenon of reflection which leads to echo.

To determine the speed of sound in the metal;

Length of metal = 1000 m

Time taken between the two sounds = 2.5 s

Using the formula;

V = 2d/t

V = 2(1000)/2.5

V = 800 m/s

6 0
2 years ago
What is the orbital velocity in km/s and period in hours of a ring particle at the outer edge of Saturn's A ring
mote1985 [20]

Answer:

The orbital velocity v  =  16.4 \ km/s

The period is  T =  14.8 \ hours

Explanation:

Generally centripetal force acting ring particle is equal to the gravitational  force between the ring particle and the planet , this is mathematically represented as

       \frac{GM_s  *  m }{r^2 }  = m w^2 r

=>    w = \sqrt{ \frac{GM}{r^3} }

Here G is the gravitational constant with value  G = 6.67*10^{-11}

        M_s  is the mass of with value  M_s  =5.683*10^{26} \  kg

        r is the is distance from the center of the  to the  outer edge of the  A ring

i.e r = R  + D  

Here R  is the radius of the planet   with value  R  = 60300 \ km

         D  is the distance from the  equator to the outer edge of the  A ring with value  D = 80000 \  kg

So  

       r =80000 + 60300

=>    r =140300 \ km  = 1.4*10^{8} \  m

So

    =>    w = \sqrt{ \frac{ 6.67*10^{-11}*  5.683*10^{26}}{[1.4*10^{8}]^3} }

    =>    w =  1.175*10^{-4} \ rad/s

Generally the orbital velocity is mathematically represented as

       v  = w * r

=>     v  = 1.175*10^{-4}   * 1.4*10^{8}

=>     v  = 1.64*10^{4} \  m /s =  16.4 \ km/s

Generally the period is mathematically represented as

     T   =  \frac{2 \pi }{w }

=> T   =  \frac{2 *  3.142  }{ 1.175 *10^{-4} }

=> T   = 53473 \ second = 14.8 \ hours

Answer:

The orbital velocity v  =  16.4 \ km/s

The period is  T =  14.8 \ hours

Explanation:

Generally centripetal force acting ring particle is equal to the gravitational  force between the ring particle and the , this is mathematically represented as

       \frac{GM_s  *  m }{r^2 }  = m w^2 r

3 0
3 years ago
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