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3 years ago

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Describe the difference between a transverse wave and longitudinal wave. Include the parts that you would find on a transverse w

ave and longitudinal wave. Give an example of each.
PLEASE HELP! ITS A WRITTEN RESPONSE!

2 answers:

irina1246 [14]3 years ago

7 0

Answer:

A transverse wave is a moving wave in which the current is perpendicular to the direction of the wave or path of propagation. A longitudinal wave are waves in which the displacement of the median is in the direction of the propagation.

Explanation:

hi

spayn [35]3 years ago

6 0

A transverse wave is a moving wave in which the current is perpendicular to the direction of the wave or path of propagation. A longitudinal wave are waves in which the displacement of the median is in the direction of the propagation.
Example:
Transverse- pond ripple
Longitudinal- crest and troff

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10. How far does a transverse pulse travel in 1.23 ms on a string with a density of 5.47 × 10−3 kg/m under tension of 47.8 ?????

KATRIN_1 [288]

Answer: Tension = 47.8N, Δx = 11.5× $10^{-6}$ m.

Tension = 95.6N, Δx = 15.4× $10^{-5}$ m

Explanation: A speed of wave on a string under a tension force can be calculated as:

$|v| = \sqrt{\frac{F_{T}}{\mu} }$

$F_{T}$ is tension force (N)

μ is linear density (kg/m)

Determining velocity:

$|v| = \sqrt{\frac{47.8}{5.47.10^{-3}} }$

$|v| = \sqrt{0.00874 }$

$|v| =$ 0.0935 m/s

The displacement a pulse traveled in 1.23ms:

$\Delta x = |v|.t$

$\Delta x = 9.35.10^{-2}*1.23.10^{-3}$

Δx = 11.5× $10^{-6}$

With tension of 47.8N, a pulse will travel Δx = 11.5× $10^{-6}$ m.

Doubling Tension:

$|v| = \sqrt{\frac{2*47.8}{5.47.10^{-3}} }$

$|v| = \sqrt{2.0.00874 }$

$|v| = \sqrt{0.01568}$

|v| = 0.1252 m/s

Displacement for same time:

$\Delta x = |v|.t$

$\Delta x = 12.52.10^{-2}*1.23.10^{-3}$

$\Delta x =$ 15.4× $10^{-5}$

With doubled tension, it travels $\Delta x =$ 15.4× $10^{-5}$ m

4 0

3 years ago

This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive an

nalin [4]

Answer:

The maximum volume is 1417.87 $inch^3$

Explanation:

<u>Optimization Using Derivatives</u>

We have a 24x30 inch piece of metal and we need to make a rectangular box by cutting a square from each corner of the piece and bending up the sides. The width of the piece is 24 inches and its length is 30 inches

When we cut a square of each corner of side x, the base of the box (after bending up the sides) will be (24-2x) and (30-2x), width and length respectively. The volume of the box is

$V=(24-2x)(30-2x)x$

Operating

$V=4x^3-108x^2+720x$

To find the maximum value of V, we compute the first derivative and equate it to zero

$V'=12x^2-216x+720=0$

Simplifying by 12

$x^2-18x+60=0$

Completing squares

$x^2-18x+81-81+60=0$

$(x-9)^2=21$

We have two values for x

$x=9+\sqrt{21}=13.58\ inch$

$x=9-\sqrt{21}=4.42\ inch$

The first value is not feasible because it will produce a negative width (24-2(13.58))=-6.16

We'll keep only the solution

$x=4.42\ inch$

The width is

$w=(24-2(4.42))=15.16\ inch$

The length is

$l=(30-2(4.42))=21.16\ inch$

And the height

$x=4.42\ inch$

The maximum volume is

$V=(15.16)(21.16)(4.42)=1417.87\ inch^3$

4 0

3 years ago

Bohr found experimental evidence for his atomic model by studying _____.

nevsk [136]

For the answer to te question above, it is a. The Electron Cloud.
For additional information, The Electron Cloud is an informal term in physics. It is used to describe where electrons are when they go around the nucleus of an atom.
I hope my answer helped you.

6 0

3 years ago

Read 2 more answers

The SI unit of refractive index is ___<br>a)km/s. b)m/s. c)sec. d)no unit

cupoosta [38]

Answer:

d)no unit

Explanation:

refractive index is a unit less quantity.

6 0

3 years ago

Why are all the noncurrent-carrying metal parts bonded together in a wiring system?

Yuliya22 [10]

Explanation:

This process is called Bonding and it is done to provide low resistance path to ground.

"Bonding — a low impedance direction achieved by permanently connecting all non-current-carrying metal parts to maintain electrical stability and to be able to conduct any current that is likely to be put on it safely.

6 0

3 years ago