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soldi70 [24.7K]
3 years ago
11

g Light is a traveling wave! The oscillations are oscillations of electric fields. The electric fields oscillate in the y-direct

ion while the light travels in the x direction, so we can create a wave equation Ey(x,t); Ey is the y-component of the electric field, x is position in meters, and t is time in seconds. Consider light with a wave length of 550 nanometers, a wave speed of 3.00 x 108 m/s , and an amplitude of 10 V/m (volts/meters). Construct the traveling wave equation for this light if at t = 0 seconds, the light is at its maximum positive value and is traveling in the negative x-direction. Your equation, Ey(x,t) , should only have the variables x and t (i.e. calculate the values of the relevant traveling wave quantities to three sig figures and use those in the equation you are constructing, do not use the symbols in your final answer).
Physics
1 answer:
Ahat [919]3 years ago
3 0

Answer:

  E_{y}  (x,t)= 10 cos (1.15 10⁻¹¹ x - 3.427 10¹⁵ t)

Explanation:

The general equation of a traveling wave on the x-axis is

        E_{y}(x,t) = E₀ cos (kx -wt)

the amplitude of the wave is E₀ = 10 V / m

the wave number is

         k = 2π /λ

indicate the value of   λ = 550 nm = 550 10⁻⁹ m

we substitute

         k = 2π / 550 10⁻⁹

         k = 1.15 10⁻¹¹ m⁻¹

angular velocity is related to frequency

          w = 2πf

           

the speed of light is related to the wavelength and frequency

           c = λ f

           f = c /λ

we substitute

           w = 2\pi  \frac{c}\lambda }

           w = 2π  3 10⁸/550 10⁻⁹

           w = 3.427 10¹⁵ rad / s

we substitute in the electric field equation

           E_{y}  (x,t)= 10 cos (1.15 10⁻¹¹ x - 3.427 10¹⁵ t)

Let's test the value of this wave for t = 0 at the point x = 0, the wave is worth

           Ey (0,0) = 10

which corresponds to its maximum positive value, therefore this is the expression of the traveling wave

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A given wave has a wavelength of 1.4 m and a frequency of 2.0 Hz. How fast
stiks02 [169]

Answer:

The wave speed is 2.8 m.

Explanation:

Wavelength = 1.4 m

frequency, f = 2 Hz

the wave speed is given by

wave speed = wavelength x frequency

wave speed = 1.4 x 2 = 2.8 m

option (D) is correct.  

3 0
3 years ago
Refer to the first diagram. What is the weight of the person hanging on the end of the seesaw in Newtons?
irina1246 [14]

Due to equilibrium of moments:

1) The weight of the person hanging on the left is 250 N

2) The 400 N person is 3 m from the fulcrum

3) The weight of the board is 200 N

Explanation:

1)

To solve the problem, we use the principle of equilibrium of moments.

In fact, for the seesaw to be in equilibrium, the total clockwise moment must be equal to the total anticlockwise moment.

The moment of a force is defined as:

M=Fd

where

F is the magnitude of the force

d is the perpendicular distance of the force from the fulcrum

In the first diagram:

- The clockwise moment is due to the person on the right is

M_c = W_2 d_2

where W_2 = 500 N is the weight of the person and d_2 = 2 m is its distance from the fulcrum

- The anticlockwise moment due to the person hanging on the left is

M_a = W_1 d_1

where W_1 is his weight and d_1 = 4 m is the distance from the fulcrum

Since the seesaw is in equilibrium,

M_c = M_a

So we can find the weight of the person on the left:

W_1 d_1 = W_2 d_2\\W_1 = \frac{W_2 d_2}{d_1}=\frac{(500)(2)}{4}=250 N

2)

Again, for the seesaw to be in equilibrium, the total clockwise moment must be equal to the total anticlockwise moment.

- The clockwise moment due to the person on the right is

M_c = W_2 d_2

where W_2 = 400 N is the weight of the person and d_2 is its distance from the fulcrum

- The anticlockwise moment due to the person on the left is

M_a = W_1 d_1

where W_1 = 300 N is his weight and d_1 = 4 m is the distance from the fulcrum.

Since the seesaw is in equilibrium,

M_c = M_a

So we can find the distance of the person on the right:

W_1 d_1 = W_2 d_2\\d_2 = \frac{W_1 d_1}{W_2}=\frac{(300)(4)}{400}=3 m

3)

As before, for the seesaw to be in equilibrium, the total clockwise moment must be equal to the total anticlockwise moment.

- The clockwise moment around the fulcrum this time is due to the weight of the seesaw:

M_c = W_2 d_2

where W_2 is the weight of the seesaw and d_2 = 3 m is the distance of its centre of mass from the fulcrum

- The anticlockwise moment due to the person on the left is

M_a = W_1 d_1

where W_1 = 600 N is his weight and d_1 = 1 m is the distance from the fulcrum

Since the seesaw is in equilibrium,

M_c = M_a

So we can find the weight of the seesaw:

W_1 d_1 = W_2 d_2\\W_2 =\frac{W_1 d_1}{d_2}= \frac{(600)(1)}{3}=200 N

#LearnwithBrainly

8 0
3 years ago
A ball is dropped from the top of a tall building. As the ball falls, the upward force of air rsistance becomes equal to the dow
iren [92.7K]
Fall at a constant speed

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3 years ago
A force of 40 N accelerates a 5 block at 6 m/s ^ 2 along a horizontal surface a. What would the block's acceleration be if the s
Bas_tet [7]

Answer:

a = 8 m/s^2, Ffriction = 10 N, μk = 0.205

Explanation:

a. Force = Mass*Acceleration,

(since you didn't add the units..."5 block"....for the mass, I will assume it to be in kg, per SI units)

40 N = 5 kg*acceleration,

a = 40/5 = 8 m/s^2

b. As you know newtons second law (F=m*a) is actually in the form Fnet = m*a. Which means that if the friction force comes into play, it would be Fapplied - Ffriction = m*a.

Fapplied - Ffriction = m*a,

40 - Ffriction = 5*6,

40 - Ffriction = 30,

Ffriction = 40 - 30 = 10 N

c. The coefficient of kinetic friction is calculated by the formula "Ffriction = μk*Fnormal".

10 = μk*Fnormal (Fnormal = m*g = 5*9.8)

10 = μk*49,

μk=10/49 ≈ 0.205

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3 years ago
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sergiy2304 [10]
You performed 0 work for the fact that work means the distance of movement made on an object not the amount of force it is exposed to. 0 work because it didn't move
7 0
3 years ago
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