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3 years ago

Which of the following best describes transverse waves?

Physics

2 answers:

gladu [14]3 years ago

7 0

Answer:

Displacement of particles is back and forth, perpendicular to the direction of wave motion.

den301095 [7]3 years ago

5 0

Best Answer: perpendicular to the direction of wave motion

think of you and a friend holding different ends of a rope. you shake your end side to side and the wave travels down to your friend's end. if you pay attention to a given segment of the rope, it moves, but it moves to the right and left while the wave travels forward. in a transverse wave, the constituents of the wave move perpendicular to the direction of motion of the wave itself.

a longitudinal wave, on the other hand, is the opposite. the constituents of the wave alternate moving in the same or opposite direction as the wave moves. this one is more like if you and your friend were holding onto a slinky. you give your end a push towards your friend, and you can see the wave move towards the other end.

hope that helps :D

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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

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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:

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

$m\cdot v^{2} + 2\cdot m\cdot g \cdot s = k\cdot d^{2}$

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

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ThIs is the same type of problem
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