1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
dimulka [17.4K]
3 years ago
15

1. An EM wave representing 650 nm laser light is traveling in the +zˆ direction. At one point in space and time, the electric fi

eld is in the +yˆ direction (i.s., the polarization of the wave is along the yˆ direction). Assume that the amplitude of the electric field is 20.0 V/m.
(a) At this same location in space/time, in what direction is the magnetic field pointing?
(b) What is the magnitude of the magnetic field?
(c) What is the magnitude of the average power per unit area of this wave?
Physics
1 answer:
ivann1987 [24]3 years ago
5 0

Answer:

Explanation:

a ) Direction of the magnetic field will be in positive x direction.

The direction of the vector E X B gives the direction of motion of wave.

b ) Magnitude of magnetic field is given by the relation

E₀ / B₀ = c , c is velocity of light

B₀ = E₀ / c

= 20 / (3 x 10⁸)

= 6.67 x 10⁻⁸ T

c ) Average power flowing per unit area by this wave is called Poynting vector

c ε₀E₀² , ε₀ = 8.85X10⁻¹²

= 3 X 10⁸ X 8.85 X 10⁻¹² X 20²

= 1.062 W m⁻²

You might be interested in
NEED HELP ASAP Which of the following chemical equations shows the Law of Conservation of Mass? (1 point)
natulia [17]

Answer:

OK + MgBR arrow KBR + MG

Explanation:

I know nothing about this topic, but if it has to be balanced I am pretty certain that's the only balanced equation

4 0
1 year ago
A uniformly charged ring of radius 10.0 cm has a total charge of 50.0 μC Find the electric field on the axis of the ring at 30.0
Grace [21]

Answer: 4.27 *10^6 N/C

Explanation: In order to calculate the electric field along the axis of charged ring we have to use the following expression:

E=k*x/(a^2+x^2)^3/2    where a is the ring radius and x the distance to the point measured from the center of the ring.

Replacing the data we have:

E= (9* 10^9* 0.3* 50 * 10^-6)/(0.1^2+0.3^2)^3/2

then

E=4.27 * 10^6 N/C

8 0
3 years ago
A uniform rod is hung at one end and is partially submerged in water. If the density of the rod is 5/9 that of water, find the f
VashaNatasha [74]

Answer:

\frac{y}{L} = 0.66

Hence, the fraction of the length of the rod above water = \frac{y}{L} = 0.66

and fraction of the length of the rod submerged in water = 1 - \frac{y}{L} = 1 - 0.66 = 0.34  

Explanation:

Data given:

Density of the rod = 5/9 of the density of the water.

Let's denote density of Water with w

And density of rod with r

So,

r = 5/9 x w

Required:

Fraction of the length of the rod above water.

Let's denote total length of the rod with L

and length of the rod above with = y

Let's denote the density of rod = r

And density of water = w

So, the required is:

Fraction of the length of the rod above water = y/L

y/L = ?

In order to find this, we first need to find out the all type of forces acting upon the rod.

We know that, a body will come to equilibrium if the net torque acting upon a body is zero.

As, we know

F = ma

Density = m/v

m = Density x volume

Volume = Area x length = X ( L-y)

So, let's say X is the area of the cross section of the rod, so the forces acting upon it are:

F = mg

F = (Density x volume) x g

g = gravitational acceleration

F1 = X(L-y) x w x g (Force on the length of the rod submerged in water)

where,

X (L-y) = volume

w = density of water.

Another force acting upon it is:

F = mg

F2 =  X x L x r x g

Now, the torques acting upon the body:

T1 + T2 = 0

F1 ( y + (\frac{L-y}{2}) ) g sinФ - F2 x (\frac{L}{2}) x gsinФ = 0

plug in the  equations of F1 and F2 into the above equation and after simplification, we get:

(L^{2} - y^{2} ) . w = L^{2} . r

where, w is the density of water and r is the density of rod.

As we know that,

r = 5/9 x w

So,

(L^{2} - y^{2} ) . w = L^{2} . 5/9 x w

Hence,

(L^{2} - y^{2} ) = \frac{5L^{2} }{9}

\frac{L^{2} - y^{2}  }{L^{2} } = \frac{5}{9}

Taking L^{2} common and solving for \frac{y}{L}, we will get

\frac{y}{L} = 0.66

Hence, the fraction of the length of the rod above water = \frac{y}{L} = 0.66

and fraction of the length of the rod submerged in water = 1 - \frac{y}{L} = 1 - 0.66 = 0.34

8 0
3 years ago
Calculate the speed of a car that travels 355km in 15 hours , a good answer please
Ludmilka [50]

Answer:

23.67 km / hr

Explanation:

car travels 355 km (d)

duration = 15 hrs  (t)

average speed formula = v = d / t

v = 355 km / 15 hr

v = 23.67 km / hr

8 0
3 years ago
Read 2 more answers
Two particles are moving along the x axis. Particle 1 has a mass m₁ and a velocity v₁ = +4.7 m/s. Particle 2 has a mass m₂ and a
nirvana33 [79]

Answer:

m₁ / m₂ = 1.3

Explanation:

We can work this problem with the moment, the system is formed by the two particles

The moment is conserved, to simulate the system the particles initially move with a moment and suppose a shock where the particular that, without speed, this determines that if you center, you should be stationary, which creates a moment equal to zero

    p₀o = m₁ v₁ + m₂ v₂

    pf = 0

    m₁ v₁ + m₂ v₂ = 0

    m₁ / m₂ = -v₂ / v₁

    m₁ / m₂=  - (-6.2) / 4.7

     m₁ / m₂ = 1.3

Another way to solve this exercise is to use the mass center relationship

    Xcm = 1/M    (m₁ x₁ + m₂ x₂)

We derive from time

   Vcm = 1/M   (m₁ v₁ + m₂v₂)

As they say the velocity of the center of zero masses

    0 = 1/M   (m₁ v₁ + m₂v₂)

   m₁ v₁ + m₂v₂ = 0

    m₁ / m₂ = -v₂ / v₁

   m₁ / m₂ = 1.3

4 0
3 years ago
Other questions:
  • When a mortar shell is fired with an initial
    8·1 answer
  • 14. the sun's layer is called the photosphere . A. true . . b. false . . 15. meteoroids can form when comets break apart , creat
    12·1 answer
  • A bear spies some honey and takes off from rest, accelerating at a rate of 2.0 m/s^2. If the honey is 16m away, how fast will hi
    15·1 answer
  • QuesLUI 2 UI IU
    9·1 answer
  • An electron of mass 9.11*10-31kg has an initial speed of 3.00*105m/s. It travels in a straight line and its speed increases to 7
    9·1 answer
  • 1. Two wires - A and B - with circular cross-sections have identical lengths and are made of the same material. Yet, wire A has
    15·1 answer
  • Where would you encounter each portion of the spectrum in your daily life
    15·2 answers
  • A flat-plate solar collector is used to heat water by having water flow through tubes attached at the back of the thin solar abs
    13·1 answer
  • An air track car with a mass of 6 kg and velocity of 4 m/s to the right collides with a 3 kg car moving to the left with a veloc
    12·1 answer
  • A wire that is 1.0 m long with a mass of 90 g is under a tension of 710 N. When a transverse wave travels on the wire, its wavel
    15·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!