All categories

3 years ago

Transverse and longitudinal waves both

Physics

2 answers:

e-lub [12.9K]3 years ago

8 0

Transverse and longitudinal waves both transfer energy through a medium.

Answer: Option B

<u>Explanation: </u>

Waves are a medium of transfer of energy via medium. Generally waves are classified as longitudinal and transverse waves. In longitudinal waves, the medium transfer energy by displacing the media along the wave's direction.

But in transverse wave energy propagation occurs by the medium travelling perpendicular to the wave's direction. But both longitudinal and transverse waves transfer energy through medium and thus these waves need a medium to propagate. So, both waves transfer the energy through the medium.

UkoKoshka [18]3 years ago

5 0

Check this attachment

You might be interested in

The electric field in a region of space increases from 0 to 2150 N/C in 5.00 s. What is the magnitude of the induced magnetic fi

Feliz [49]

To solve this problem we will use the Ampere-Maxwell law, which describes the magnetic fields that result from a transmitter wire or loop in electromagnetic surveys. According to Ampere-Maxwell law:

$\oint \vec{B}\vec{dl} = \mu_0 \epsilon_0 \frac{d\Phi_E}{dt}$

Where,

B= Magnetic Field

l = length

$\mu_0$ = Vacuum permeability

$\epsilon_0$ = Vacuum permittivity

Since the change in length (dl) by which the magnetic field moves is equivalent to the perimeter of the circumference and that the electric flow is the rate of change of the electric field by the area, we have to

$B(2\pi r) = \mu_0 \epsilon_0 \frac{d(EA)}{dt}$

Recall that the speed of light is equivalent to

$c^2 = \frac{1}{\mu_0 \epsilon_0}$

Then replacing,

$B(2\pi r) = \frac{1}{C^2} (\pi r^2) \frac{d(E)}{dt}$

$B = \frac{r}{2C^2} \frac{dE}{dt}$

Our values are given as

$dE = 2150N/C$

$dt = 5s$

$C = 3*10^8m/s$

$D = 0.440m \rightarrow r = 0.220m$

Replacing we have,

$B = \frac{r}{2C^2} \frac{dE}{dt}$

$B = \frac{0.220}{2(3*10^8)^2} \frac{2150}{5}$

$B =5.25*10^{-16}T$

Therefore the magnetic field around this circular area is $B =5.25*10^{-16}T$

3 0

3 years ago

An object is dropped from a height of 75.0 m above ground level. (a) Determine the distance traveled during the first second. (b

lys-0071 [83]

Answer:

a)Distance traveled during the first second = 4.905 m.

b)Final velocity at which the object hits the ground = 38.36 m/s

c)Distance traveled during the last second of motion before hitting the ground = 33.45 m

Explanation:

a) We have equation of motion

S = ut + 0.5at²

Here u = 0, and a = g

S = 0.5gt²

Distance traveled during the first second ( t =1 )

S = 0.5 x 9.81 x 1² = 4.905 m

Distance traveled during the first second = 4.905 m.

b) We have equation of motion

v² = u² + 2as

Here u = 0, s= 75 m and a = g

v² = 0² + 2 x g x 75 = 150 x 9.81

v = 38.36 m/s

Final velocity at which the object hits the ground = 38.36 m/s

c) We have S = 0.5gt²

75 = 0.5 x 9.81 x t²

t = 3.91 s

We need to find distance traveled last second

That is

S = 0.5 x 9.81 x 3.91² - 0.5 x 9.81 x 2.91² = 33.45 m

Distance traveled during the last second of motion before hitting the ground = 33.45 m

3 0

3 years ago

Calculate, for the judge, how fast you were going in miles per hour when you ran the red light because it appeared Doppler-shift

sammy [17]

Answer:

The doppler effect equation is:

$f' = \frac{v +v0}{v - vs}*f$

In the equation we have frequencies, but then we have the wavelengths of the lights, remember the relation:

v = f*λ

then:

f = v/λ

and v is the speed of light, then:

f = c/λ

where:

f' is the observed frequency, in this case, is equal to f = (3*10^17nm/s)/550 nm

f is the real frequency, in this case, is (3*10^17nm/s)/650 nm

vs is the speed of the source, in this case, the source is not moving, then vs = 0 m/s.

v is the speed of the wave, in this case, is equal to the speed of light, v = 3*10^8 m/s

v0 is your speed, this is what we want to find.

Replacing those quantities in the equation, we get:

(3*10^17nm/s)/550 = (3*10^8 m/s + v0)/(3*10^8 m/s)*(3*10^17nm/s)/650 nm

(650nm)/(550nm) = (3*10^8 m/s + v0)/(3*10^8 m/s)

1.182*(3*10^8 m/s) = (3*10^8 m/s + v0)

1.182*(3*10^8 m/s) - (3*10^8 m/s) = v0 = 54,600,000 m/s

So your speed was 54,600,000 m/s, which is a lot.

6 0

3 years ago

Electricity is the flow of electrons. The questions relate to how electricity is quantified. Electrons are charged particles. Th

xxTIMURxx [149]

The amount of charge that passes per unit time is called <em>electric current</em> .

Current has dimensions of [Charge] / [Time] .

It's measured and described in units of ' Ampere ' .

1 Ampere means 1 Coulomb of charge passing a point every second.

7 0

3 years ago

16. Two capacitors have an equivalent

Gennadij [26K]

Answer:

C1 + C2 = 30 parallel connection

C1 * C2 / (C1 + C2) = 7.2 series connection

C1 * C2 = 7.2 * (C1 + C2) = 216

C2 + 216 / C2 = 30 using first equation

C2^2 + 216 = 30 C2

C2^2 - 30 C2 + 216 = 0

C2 = 12 or 18 solving the quadratic

Then C1 = 18 or 12

5 0

3 years ago