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2 years ago

14

Please help if you have the time

1 answer:

Alex777 [14]2 years ago

6 0

We know, length of segment wave is half the wavelength .

Let, wavelength of wave is $\lambda$ .

So, length of segment will be $\dfrac{\lambda}{2}$ .

Now, it is given that the string vibrates in four segments.

So,

$L = 4\times \dfrac{\lambda}{2}\\\\L = 2\lambda\\\\\lambda = 1\ m$

Speed can be given by :

$speed = wavelength \times frequency\\\\speed = 1\times 120\ m/s\\\\speed = 120\ m/s$

Therefore, the wave speed in the string is 120 m/s.

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3 years ago

Read 2 more answers

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Answer:

$V_{ft}= 317 cm/s$

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a) Using the law of the conservation of the linear momentum:

$P_i = P_f$

Where:

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$P_f = M_cV_{fc} + M_tV_{ft}$

Now:

$M_cV_{ic} + M_tV_{it} = M_cV_{fc} + M_tV_{ft}$

Where $M_c$ is the mass of the car, $V_{ic}$ is the initial velocity of the car, $M_t$ is the mass of train, $V_{fc}$ is the final velocity of the car and $V_{ft}$ is the final velocity of the train.

Replacing data:

$(1.1 kg)(4.95 m/s) + (3.55 kg)(2.2 m/s) = (1.1 kg)(1.8 m/s) + (3.55 kg)V_{ft}$

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$V_{ft}= 3.17 m/s$

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b) The kinetic energy K is calculated as:

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$K_i = \frac{1}{2}M_cV_{ic}^2+\frac{1}{2}M_tV_{it}^2$

$K_i = \frac{1}{2}(1.1)(4.95)^2+\frac{1}{2}(3.55)(2.2)^2$

$K_i = 22.06 J$

And the final K is:

$K_f = \frac{1}{2}M_cV_{fc}^2+\frac{1}{2}M_tV_{ft}^2$

$K_f = \frac{1}{2}(1.1)(1.8)^2+\frac{1}{2}(3.55)(3.17)^2$

$K_f = \frac{1}{2}(1.1)(1.8)^2+\frac{1}{2}(3.55)(3.17)^2$

$K_f = 19.61 J$

Finally, the change in the total kinetic energy is:

ΔK = Kf - Ki = 22.06 - 19.61 = 2.45 J

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