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3 years ago

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A wave is sent through a medium. The energy travels from left to right, and the direction of the wave travels from left to right

. Identify the type of wave and explain your reasoning.

2 answers:

evablogger [386]3 years ago

7 0

Answer:

Longitudinal Waves

Explanation:

As we know that longitudinal waves are those in which medium molecules oscillates parallel to the direction of propagation of wave or propagation of energy.

So here we know that when wave is sent into the medium then the energy will travels from left to right and also the propagation is given from left to right also.

So here we will say that the wave here must be longitudinal wave.

While if the medium particles will oscillates perpendicular to the propagation of energy or propagation of waves, then it is known as transverse type of wave.

So here correct answer will be

LONGITUDINAL WAVE

Art [367]3 years ago

5 0

It's a longitudinal wave surely because information passes parallel to the direction of travel or something like that?

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Early black-and-white television sets used an electron beam to draw a picture on the screen. The electrons in the beam were acce

lutik1710 [3]

Answer:

speed of electrons = 3.25 × $10^{7}$ m/s

acceleration in term g is 3.9 × $10^{17}$ g.

radius of circular orbit is 2.76 × $10^{-4}$ m

Explanation:

given data

voltage = 3 kV

magnetic field = 0.66 T

solution

law of conservation of energy

PE = KE

qV = 0.5 × m × v²

v = $\sqrt{\frac{2qV}{m}}$

v = $\sqrt{\frac{2\times 1.6 \times 10^{-19}\times 3}{9.1\times 10^{-31}}$

v = 3.25 × $10^{7}$ m/s

and

magnetic force on particle movie in magnetic field

F = Bqv

ma = Bqv

a = $\frac{Bqv}{m}$

a = $\frac{0.67\times 1.6\times 10^{-19}\times 3.25\times 10^7}{9.1\times 10^{-31}}$

a = 3.82 × $10^{18}$ m/s²

and acceleration in term g

a = $\frac{3.82\times 10^{18}}{9.81}$

a = 3.9 × $10^{17}$ g

acceleration in term g is 3.9 × $10^{17}$ g.

and

electron moving in circular orbit has centripetal force

F = $\frac{mv^2}{r}$

Bqv = $\frac{mv^2}{r}$

r = $\frac{mv}{Bq}$

r = $\frac{9.1\times 10^{-31}\times 3.25\times 10^7}{0.67\times 1.6\times 10^{-19}}$

r = 2.76 × $10^{-4}$ m

radius of circular orbit is 2.76 × $10^{-4}$ m

8 0

3 years ago

Light of wavelength 560 nm passes through a slit of width 0. 170 mm. (a) the width of the central maximum on a screen is 8. 00 m

faltersainse [42]

The distance between slit and the screen is 1.214m.

To find the answer, we have to know about the width of the central maximum.

<h3>How to find the distance between slit and the screen?</h3>

It is given that, wavelength 560 nm passes through a slit of width 0. 170 mm, and the width of the central maximum on a screen is 8. 00 mm.

We have the expression for slit width w as,

$w=\frac{2*wavelength*d}{a}$

where, d is the distance between slit and the screen, and a is the slit width.

Thus, distance between slit and the screen is,

$d=\frac{w*a}{2*wavelength} =\frac{8*10^{-3}*0.17*10^{-3}}{560*10^{-9}*2} \\\\d=1.214m$

Thus, we can conclude that, the distance between slit and the screen is 1.214m.

Learn more about the width of the central maximum here:

brainly.com/question/13088191

#SPJ4

3 0

1 year ago

Helppppppp pleaseeee

SVETLANKA909090 [29]

Answer: (Sorry, but I don't know how to calculate mass)

1. 15 N

2. 0.4921 $\frac{ft}{s^2}$ (feet per second squared)

4. 150 N

5. 8.202 feet per second squared

3 0

2 years ago

an athlete whirls an 8.71 kg hammer tied to the end of a 1.5 m chain in a simple horizontal circle where you should ignore any v

-BARSIC- [3]

Answer:

T = 692.42 N

Explanation:

Given that,

Mass of hammer, m = 8.71 kg

Length of the chain to which an athlete whirls the hammer, r = 1.5 m

The angular sped of the hammer, $\omega=1.16\ rev/s=7.28\ rad/s$

We need to find the tension in the chain. The tension acting in the chain is balanced by the required centripetal force. It is given by the formula as follows :

$F=m\omega^2r\\\\=8.71\times (7.28)^2\times 1.5\\\\=692.42\ N$

So, the tension in the chain is 692.42 N.

5 0

2 years ago

I will Give Brainliest To whoever actually answers A 500 kg satellite experiences a gravitational force of 3000 N, while moving

snow_lady [41]

Answer:

$9.7\times 10^{-4}\ rad/s$

Explanation:

Given:

$m=500 kg\\F=3000 N$

$Radius of earth , R=6371 \times 10^3\ m\\Angular speed =\omega\\We\ know\ that\ \\F= m\times \omega^{2} \times R\\\omega^{2}=\frac{F}{m*R} \\\\=\frac{3000}{500*6371 \times 10^3\ m}$

= $\frac{6}{6371 \times 10^3\ m}$

= $9.7\times 10^{-4}\ rad/s$

7 0

3 years ago