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3 years ago

Capillary waves travel what than long waves

Physics

2 answers:

7nadin3 [17]3 years ago

6 0

Faster than. Hope this helps!!!

SOVA2 [1]3 years ago

4 0

Answer:

Capillary waves travel faster than long waves.

Explanation

Capillary waves are also known as ripples these waves has wavelength of the order of about 1.5 cm or less that it. In capillary waves surface tension of water act as a restoring force. This restoring force is very less in comparison to the air which below on the surface of water due to which these waves are produced. Hence these waves travel faster that long waves.

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In a very busy off-campus eatery one chef sends a 201 g broccoli-tomato-pickle-onion-mushroom pizza sliding down the counter fro

Murrr4er [49]

Answer:

The correct answer is:

(A) to the left

(B) at speed -0.8725 m/s

Explanation:

The given values are:

Plate 1:

Mass,

m₁ = 201 g

Velocity,

v₁ = +1.79 m/s

Plate 2:

Mass,

m₁ = 335 g

Velocity,

v₁ = -2.47 m/s

According to the conservation of momentum, we get

⇒ $m_1v_1+m_2v_2=(m_1+m_2)v$

then,

⇒ $v=\frac{m_1v_1+m_2v_2}{m_1+m_2}$

On substituting the values, we get

⇒ $=\frac{201\times 1.79+335\times (-2.47)}{201+335}$

⇒ $=\frac{359.79-827.45}{536}$

⇒ $=\frac{-467.66}{536}$

⇒ $=-0.8725$ (to the left)

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bagirrra123 [75]

Answer:

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Answer:

Explanation:

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3 years ago

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A 1.2 kg block is held at rest against the spring with a force constant k= 730 N/m. Initially, the spring is compressed a distan

Nesterboy [21]

Answer:

Compression distance: $d \approx 0.102\,m$

Explanation:

According to this statement, we know that system is non-conservative due to the rough patch. By Principle of Energy Conservation and Work-Energy Theorem, we have the following expression that represents the system having a translational kinetic energy ( $K$ ), in joules, at the expense of elastic potential energy ( $U$ ), in joules, and overcoming work losses due to friction ( $W_{l}$ ), in joules:

$K + W_{l} = U$ (1)

By definitions of translational kinetic and elastic potential energies and work losses due to friction, we expand the equation described above:

$\frac{1}{2}\cdot m \cdot v^{2} +\mu\cdot m\cdot g \cdot s = \frac{1}{2} \cdot k \cdot d^{2}$ (2)

Where:

$m$ - Mass of the block, in kilograms.

$v$ - Final velocity of the block, in meters per second.

$\mu$ - KInetic coefficient of friction, no unit.

$g$ - Gravitational acceleration, in meters per square second.

$s$ - Width of the rough patch, in meters.

$k$ - Spring constant, in newtons per meter.

$d$ - Compression distance, in meters.

If we know that $m = 1.2\,kg$ , $v = 2.3\,\frac{m}{s}$ , $\mu = 0.44$ , $g = 9.807\,\frac{m}{s^{2}}$ , $s = 0.05\,m$ and $k = 730\,\frac{N}{m}$ , then the compression distance of the spring is: