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3 years ago

13

What relationship exists between the amplitude of a wave and amount of disturbance in the water?

1 answer:

guapka [62]3 years ago

6 0

A wave is a result of the disturbance in the equilibrium state. There are two types of wave, transverse and longitudinal. Transverse wave affects amplitude while longitudinal wave affects the frequency of the wave. As for the transverse wave, the magnitude of the perpendicular disturbance of the wave is directly proportional to the amplitude of the wave. The higher the transverse disturbance the higher the amplitude.

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A 0.5 kg block slides down a frictionless semicircular track from a height, h, and collides with a second 1.5 kg block at the bo

Ede4ka [16]

Answer:

$h' = 0.062\cdot h$

Explanation:

The speed of the 0.5 kg block just before the collision is found by the Principle of Energy Conservation:

$U_{g} = K$

$m\cdot g \cdot h = \frac{1}{2}\cdot m \cdot v^{2}$

$v = \sqrt{2\cdot g \cdot h}$

$v = \sqrt{2\cdot (9.807\,\frac{m}{s^{2}} )\cdot h}$

$v \approx 4.429\cdot \sqrt{h}$

Knowing that collision is inelastic, the speed just after the collision is determined with the help of the Principle of Momentum Conservation:

$(0.5\,kg) \cdot (4.429\cdot \sqrt{h}) = (2\,kg)\cdot v$

$v = 1.107\cdot \sqrt{h}$

Lastly, the height reached by the two blocks is:

$K = U_{g}$

$\frac{1}{2}\cdot m \cdot v^{2} = m\cdot g \cdot h'$

$h' = \frac{v^{2}}{2\cdot g}$

$h' = \frac{(1.107\cdot \sqrt{h})^{2}}{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)}$

$h' = 0.062\cdot h$

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What is the wavelength of a wave with f = 0.9 Hz and v = 154.9 cm/s

Nataly_w [17]

Answer:

The wavelength of a wave is 1.72 m.

Explanation:

Given,

Frequency, f = 0.9 Hz

Speed of a wave, v = 154.9 cm/s = 1.549 m/s

The speed of a wave is given by :

$v=f\lambda$

$\lambda=\dfrac{v}{f}\\\\\lambda=\dfrac{1.549\ m/s}{0.9\ Hz}\\\\\lambda=1.72\ m$

So, the wavelength of a wave is 1.72 m. Hence, this is the required solution.

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4 years ago