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Mekhanik [1.2K]
3 years ago
15

A rope has one end tied to a vertical support. You hold the other end so that the rope is horizontal. If you move the end of the

rope back and forth with a frequency of 4 Hz, the transverse wave you produce has a wavelength of 0.5 m. What is the speed of the wave in the rope?
Physics
2 answers:
Aliun [14]3 years ago
8 0

Answer:

2 m/s.

Explanation:

The relationship between frequency, wavelength and speed of a wave is given as:

v = λ × f

where,

v is the speed of the wave

λ is the wavelength

f is the frequency

Given:

f = 4 Hz

λ = 0.5 m

v = (0.5 × 4)

= 2 m/s.

son4ous [18]3 years ago
4 0

Answer:

2 m/s.

Explanation:

The relationship between frequency, wavelength and speed of a wave is given as:

v = λ × f

where,

v is the speed of the wave

λ is the wavelength

f is the frequency

Given:

f = 4 Hz

λ = 0.5 m

v = (0.5 × 4)

= 2 m/s.

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Answer:

force becomes one - ninth

Explanation:

According to Coulomb's law in electrostatics, two charges can exert a force of attraction or repulsion on each other which is directly proportional to the product of two charges and inversely proportional to the square of distance between them.

Here both the charges remains same but the distance is variable.

So, we can say that

F \alpha \frac{1}{d^{2}}    .... (1)

Where d be the distance between the tow charges

As the distance between two charges increases by factor of three, let the new force be F'.

F' \alpha \frac{1}{9d^{2}}   .... (2)

Divide equation (2) by equation (1), we get

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2 years ago
A satellite orbits a planet of unknown mass in a circular orbit of radius 2.3 x 104 km. The gravitational force on the satellite
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Answer:

The  kinetic energy is KE  =  7.59  *10^{10} \  J

Explanation:

From the question we are told that

       The  radius of the orbit is  r =  2.3 *10^{4} \ km  = 2.3  *10^{7} \ m

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The kinetic energy of the satellite is mathematically represented as

       KE  =  \frac{1}{2} * mv^2

where v is the speed of the satellite which is mathematically represented as

     v  = \sqrt{\frac{G  M}{r^2} }

=>  v^2  =  \frac{GM }{r}

substituting this into the equation

      KE  =  \frac{ 1}{2} *\frac{GMm}{r}

Now the gravitational force of the planet is mathematically represented as

      F_g  = \frac{GMm}{r^2}

Where M is the mass of the planet and  m is the mass of the satellite

 Now looking at the formula for KE we see that we can represent it as

     KE  =  \frac{ 1}{2} *[\frac{GMm}{r^2}] * r

=>    KE  =  \frac{ 1}{2} *F_g * r

substituting values

       KE  =  \frac{ 1}{2} *6600 * 2.3*10^{7}

         KE  =  7.59  *10^{10} \  J

 

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