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REY [17]
3 years ago
10

What's the difference between a wavelength and an an amplitude?

Physics
2 answers:
erastova [34]3 years ago
7 0
A wave is propagated by a point or a particle in a medium that vibrates or oscillates between its mean position and a maximum displacement position.

<u>Amplitude is the maximum displacement from the mean position.</u>

Due to the vibration of the particle, neighbouring particles get disturbed and they also oscillate but some small time (phase) behind.

Along the direction of propagation of the wave, the wave travels a wavelength distance in the same time that is taken by a particle to complete one oscillation.
It is the<u> distance between two successive crests (</u>maximum displacement positions - When a snap shot of wave is taken at one point of time.

valkas [14]3 years ago
3 0
Wavelength - the distance from one wave crest or trough to another wave crest or trough. Amplitude - the distance from the median point or "middle" of the wave straight up to a crest (a maximum) or straight down to a trough (or minimum), which is the peak amplitude; or the distance from a trough straight up to a crest, or a crest straight down to a trough, called peak-to-peak amplitude.
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How many significant digits are measurement 0.00210 mg?
jek_recluse [69]
There are 3 significant figures, if that answers the question.
8 0
3 years ago
Read 2 more answers
A 0.311 kg tennis racket moving 30.3 m/s east makes an elastic collision with a 0.0570 kg ball moving 19.2 m/s west. Find the ve
Troyanec [42]

The velocity of tennis racket after collision is 14.96m/s

<u>Explanation:</u>

Given-

Mass, m = 0.311kg

u1 = 30.3m/s

m2 = 0.057kg

u2 = 19.2m/s

Since m2 is moving in opposite direction, u2 = -19.2m/s

Velocity of m1 after collision  = ?

Let the velocity of m1 after collision be v

After collision the momentum is conserved.

Therefore,

m1u1 - m2u2 = m1v1 + m2v2

v1 = (\frac{m1-m2}{m1+m2})u1 + (\frac{2m2}{m1+m2})u2

v1 = (\frac{0.311-0.057}{0.311+0.057})30.3 + (\frac{2 X 0.057}{0.311 + 0.057}) X-19.2\\\\v1 = (\frac{0.254}{0.368} )30.3 + (\frac{0.114}{0.368}) X -19.2\\ \\v1 = 20.91 - 5.95\\\\v1 = 14.96

Therefore, the velocity of tennis racket after collision is 14.96m/s

7 0
3 years ago
An example of a folkway violation would be driving 70 mph in a 50-mph zone. Group of answer choices True False
shutvik [7]

Answer:

Yes.

Explanation:

That is also a law violation.

3 0
3 years ago
A point source of light is located at the bottom of a steel tank, and an opaque circular card of radius is placed horizontally o
pochemuha
Start by using the addition as a sign and use multiplying
3 0
3 years ago
Consider the two moving boxcars in Example 5. Car 1 has a mass of m1 = 65000 kg and a velocity of v01 = +0.80 m/s. Car 2 has a m
Amiraneli [1.4K]

Answer:

1.034m/s

Explanation:

We define the two moments to develop the problem. The first before the collision will be determined by the center of velocity mass, while the second by the momentum preservation. Our values are given by,

m_1 = 65000kg\\v_1 = 0.8m/s\\m_2 = 92000kg\\v_2 = 1.2m/s

<em>Part A)</em> We apply the center of mass for velocity in this case, the equation is given by,

V_{cm} = \frac{m_1v_1+m_2v_2}{m_1+m_2}

Substituting,

V_{cm} = \frac{(65000*0.8)+(92000*1.2)}{92000+65000}

V_{cm} = 1.034m/s

Part B)

For the Part B we need to apply conserving momentum equation, this formula is given by,

m_1v_1+m_2v_2 = (m_1+m_2)v_f

Where here v_f is the velocity after the collision.

v_f = \frac{m_1v_1+m_2v_2}{m_1+m_2}

v_f = \frac{(65000*0.8)+(92000*1.2)}{92000+65000}

v_f = 1.034m/s

8 0
3 years ago
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