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goldfiish [28.3K]
3 years ago
7

PLZ HELP

Physics
1 answer:
marissa [1.9K]3 years ago
7 0
We just keep meeting huh? I believe the wave to be longitudinal, because the energy is going left to right.
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The vertical displacement of the wave is measured from the ?
sergey [27]
The whole question is talking about the amplitude of a wave
that's transverse and wiggling vertically.

Equilibrium to the crest . . . that's the amplitude.

Crest to trough . . . that's double the amplitude.

Trough to trough . . . How did that get in here ?  Yes, that's
                               the wavelength, but it has nothing to do
                               with vertical displacement.

Frequency . . . that's how many complete waves pass a mark
                       on the ground every second.  Doesn't belong here.

Notice that this has to be a transverse wave.  If it's a longitudinal wave,
like sound or a slinky, then it may not have any displacement at all
across the direction it's moving.

It also has to be a vertically 'polarized' wave.  If it's wiggling across
the direction it's traveling BUT it's wiggling side-to-side, then it has
no vertical displacement.  It still has an amplitude, but the amplitude
is all horizontal.
6 0
3 years ago
If a 5000-kg is moving at a speed of 43 m/s, what is its momentum?
Eduardwww [97]

Answer:

215000kgm/s

Explanation:

Given parameters:

Mass of the moving body  = 5000kg

Velocity  = 43m/s

Unknown:

Momentum  = ?

Solution:

The momentum of a body is the amount of motion a body possess.

 It is mathematically expressed as:

  Momentum  = mass x velocity

 Now:

  Momentum  = 5000 x 43  = 215000kgm/s

4 0
3 years ago
In moving out of a dormitory at the end of the semester, a student does 1.20 x 104 J of work. In the process, his internal energ
Zielflug [23.3K]

Answer:

(a) W=1.20×10⁴J

(b) U= -5.46×10⁴J

(c) Q= -4.26×10⁴J

Explanation:

Given that student does 1.20×10⁴J work

(a) W=1.20×10⁴J

Work done by student,so positive sign

During the process, his internal energy decreases by 5.46×10⁴J.

(b) U= -5.46×10⁴J.

As the Energy decreases therefore negative sign

For (c) Q

We know the formula

Q=W+U\\Q=1.2*10^{4}+(-5.46*10^{4} )\\ Q=-4.26*10^{4}J

8 0
3 years ago
Ocean currents near the equator are primarily: eastward westward north and south clockwise
sladkih [1.3K]
Equatorial currents are primarily westward. This is because the dominant current in the northern hemisphere has a clockwise direction, while the southern hemisphere has a counterclockwise direction. When these two currents meet at the equator, a common westward current exists.
3 0
3 years ago
For a star with a parallax angle of 1/2 of an at arcsecond, what will be its distance in parsec?
avanturin [10]

(1 parsec) is the distance at which an object has a parallax of 1 arcsecond. The distance is about 3.26 light years.

Another way to understand it is:  The distance from which the Earth's orbit appears 1 arcsecond across.

For a parallax angle of 1/2 arcsecond, the distance is <em>2 parsecs </em>(about 6.52 light years).

1 arcsecond is 1/3600 of a degree, 0.00028 degree.  

8 0
3 years ago
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