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leva [86]
3 years ago
8

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!
Physics
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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La velocidad de un tren se reduce uniformemente de 12m/s a 5m/s. Sabiendo que durante ese tiempo recorre una distancia de 100 m.
MA_775_DIABLO [31]

Responder:

Explicación:

Dados los siguientes datos

Valor inicial u = 12 m / s

velocidad final v = 5 m / s

Distancia S = 100 m

Necesario

aceleración del tren.

Usando la ecuación de movimiento

v² = u² + 2as

2as = v²-u²

a = v²-u² / 2s

Sustituyendo los valores dados para obtener la aceleración que tenemos;

a = 5²-12² / 2 (100)

a = 25-144 / 200

a = -119/200

a = -0,595 m / s²

Por tanto, la aceleración del tren durante este período es de -0,595 m / s²

b) Si el tren viaja a una parada desde 5 m / s, su velocidad final será cero y su velocidad inicial u será 5 m / s

Para obtener la distancia durante este período, sustituiremos u = 5 m / s, v = 0 m / sy a = -1,19 m / s² en la ecuación de movimiento anterior;

v² = u² + 2as

0² = 5²-2 (0.595) s

0 = 25-1,19 s

1,19 s = 25

s = 25 / 1,19

Responder:

Explicación:

Dados los siguientes datos

Valor inicial u = 12 m / s

velocidad final v = 5 m / s

Distancia S = 100 m

Necesario

aceleración del tren.

Usando la ecuación de movimiento

v² = u² + 2as

2as = v²-u²

a = v²-u² / 2s

Sustituyendo los valores dados para obtener la aceleración que tenemos;

a = 5²-12² / 2 (100)

a = 25-144 / 200

a = -119/200

a = -0,595 m / s²

Por tanto, la aceleración del tren durante este período es de -0,595 m / s²

b) Si el tren viaja a una parada desde 5 m / s, su velocidad final será cero y su velocidad inicial u será 5 m / s

Para obtener la distancia durante este período, sustituiremos u = 5 m / s, v = 0 m / sy a = -1,19 m / s² en la ecuación de movimiento anterior;

v² = u² + 2as

0² = 5²-2 (0.595) s

0 = 25-1,19 s

1,19 s = 25

s = 25 / 1,19

s = 21,0 m

Por lo tanto, la distancia que recorre hasta detenerse asumiendo la misma aceleración es 21.0 m

6 0
2 years ago
In any system, the _____ in the system remains constant, while the PE or KE may change.
vladimir1956 [14]

Answer:

The sum of PE and KE remains constant

Explanation:

6 0
3 years ago
Diffraction supports the:<br><br> A. wave theory of light.<br><br> B. particle theory of light.
Artemon [7]
The answer is a hope its helps you
8 0
2 years ago
Read 2 more answers
Lasers are classified according to the eye-damage danger they pose. Class 2 lasers, including many laser pointers, produce visib
Alexus [3.1K]

Answer:

<em>a) 318.2 W/m^2</em>

<em>b) 2.5 x 10^-4 J</em>

<em>c) 1.55 x 10^-8 v/m</em>

<em></em>

Explanation:

Power of laser P = 1 mW = 1 x 10^-3 W

exposure time t = 250 ms = 250 x 10^-3 s

If beam diameter = 2 mm = 2 x 10^-3 m

then

cross-sectional area of beam A = \pi d^{2} /4 = (3.142 x (2*10^{-3} )^{2})/4

A = 3.142 x 10^-6 m^2

a) Intensity I = P/A

where P is the power of the laser

A is the cros-sectional area of the beam

I = ( 1 x 10^-3)/(3.142 x 10^-6) = <em>318.2 W/m^2</em>

<em></em>

b) Total energy delivered E = Pt

where P is the power of the beam

t is the exposure time

E = 1 x 10^-3 x 250 x 10^-3 = <em>2.5 x 10^-4 J</em>

<em></em>

c) The peak electric field is given as

E = \sqrt{2I/ce_{0} }

where I is the intensity of the beam

E is the electric field

c is the speed of light = 3 x 10^8 m/s

e_{0} = 8.85 x 10^9 m kg s^-2 A^-2

E = \sqrt{2*318.2/3*10^8*8.85*10^9}  = <em>1.55 x 10^-8 v/m</em>

6 0
3 years ago
In a college homecoming competition, eighteen students lift a sports car. While holding the car off the ground, each student exe
Nata [24]

Answer:

Explanation:

Given

Each student exert a force of F=400 N

Let mass of car be m

there are 18 students who lifts the car

Total force by 18 students F=18\times 400=7200 N

therefore weight of car W=7200

mass of car m=\frac{W}{g}

m=\frac{7200}{9.8}=734.69 kg

(b)7200 N \approx 1618.624\ Pound-force

734.69 kg\approx 1619.71 Pounds                  

6 0
3 years ago
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