When the frequency of an electromagnetic wave increases, its energy A. increases.
B. decreases.
C. stays the same.
D. It depends on the exact type of electromagnetic wave.

Answers

Answer 1

Answer:

At the time when the frequency of an electromagnetic wave increases, its energy increased.

The following information should be considered:

When the electromagnetic wave increases so the energy should be increased.
Due to all this, the speed, wavelength all should be decreased.

Therefore we can conclude that At the time when the frequency of an electromagnetic wave increases, its energy increased.

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Answer 2

Answer: When the frequency of an electromagnetic wave increases, its energy also must increase. As this occurs, its speed, wavelength, and amplitude all decreases. So, the correct choice would be A. Increases.

I hope I was able to satisfyingly answer your question. :)

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Two narrow slits separated by 0.30 mm are illuminated with light of wavelength 496 nm. (a) How far are the first three bright fringes from the center of the pattern if observed on the screen 130 cm distant? (b) How far are the first three dark fringes from the center of the pattern?

Answers

Answer:

Explanation:

d = separation of the slits = 0.30 mm = 0.30 x 10⁻³ m

λ = wavelength of the light = 496 nm = 496 x 10⁻⁹ m

n = order of the bright fringe

D = screen distance = 130 cm = 1.30 m

$x_(n)$ = Position of nth bright fringe

Position of nth bright fringe is given as

$x_(n) =( n D \lambda )/(d)$

For n = 1

$x_(1) =( (1) (1.30)(496* 10^(-9)))/(0.30* 10^(-3))$

$x_(1) = 2.15* 10^(-3)m$

For n = 2

$x_(2) =( (2) (1.30)(496* 10^(-9)))/(0.30* 10^(-3))$

$x_(2) = 4.30* 10^(-3)m$

For n = 3

$x_(2) =( (2) (1.30)(496* 10^(-9)))/(0.30* 10^(-3))$

$x_(2) = 6.45* 10^(-3)m$

Position of nth dark fringe is given as

$y_(n) =( (2n+1) D \lambda )/(2d)$

For n = 1

$y_(1) =( (2(1)+1) (1.30)(496* 10^(-9)))/(2(0.30* 10^(-3)))$

$y_(1) = 3.22* 10^(-3)m$

For n = 2

$y_(2) =( (2(2)+1) (1.30)(496* 10^(-9)))/(2(0.30* 10^(-3)))$

$y_(2) = 5.4* 10^(-3)m$

For n = 3

$y_(3) =( (2(3)+1) (1.30)(496* 10^(-9)))/(2(0.30* 10^(-3)))$

$x_(3) = 7.5* 10^(-3)m$

When dots are further apart on a ticker-tape diagram, it indicates an object is moving

Answers

At a higher velocity.

Hope this helps!

Answer:

At a higher velocity.

Two particles are traveling through space. At time t the first particle is at the point (−1 + t, 4 − t, −1 + 2t) and the second particle is at (−7 + 2t, −6 + 2t, −1 + t). (a) (5 Points) Do the paths of the two particles cross? If so, where?

Answers

Answer:

Yes, the paths of the two particles cross.

Location of path intersection = ( 1 , 2 , 3)

Explanation:

In order to find the point of intersection, we need to set both locations equal to one another. It should be noted however, that the time for each particle can vary as we are finding the point where the paths meet, not the point where the particles meet themselves.

So, we can name the time of the first particle $T_F$ , and the time of the second particle $T_S$ .

Setting the locations equal, we get the following equations to solve for $T_F$ and $T_S$ :

$(-1 + T_F) = (-7 + 2T_S)$ Equation 1

$(4 - T_F) = (-6 + 2T_S)$ Equation 2

$(-1 + 2T_F) = (-1 + T_S)$ Equation 3

Solving these three equations simultaneously we get:

$T_F =$ 2 seconds

$T_S =$ 4 seconds

Since, we have an answer for when the trajectories cross, we know for a fact that they indeed do cross.

The point of crossing can be found by using the value of $T_F$ or $T_S$ in the location matrices. Doing this for the first particle we get:

Location of path intersection = ( -1 + 2 , 4 - 2 , -1 + 2(2) )

Location of path intersection = ( 1 , 2 , 3)

The concentration of Biochemical Oxygen Demand (BOD) in a river just downstream of a wastewater treatment plant’s effluent pipe is 75 mg/L. If the BOD is destroyed through a first-order reaction with a rate constant equal to 0.05/day, what is the BOD concentration 50 km downstream? The velocity of the river is 15 km/day.

Answers

Answer:

The BOD concentration 50 km downstream when the velocity of the river is 15 km/day is 63.5 mg/L

Explanation:

Let the initial concentration of the BOD = C₀

Concentration of BOD at any time or point = C

dC/dt = - KC

∫ dC/C = -k ∫ dt

Integrating the left hand side from C₀ to C and the right hand side from 0 to t

In (C/C₀) = -kt + b (b = constant of integration)

At t = 0, C = C₀

In 1 = 0 + b

b = 0

In (C/C₀) = - kt

(C/C₀) = e⁻ᵏᵗ

C = C₀ e⁻ᵏᵗ

C₀ = 75 mg/L

k = 0.05 /day

C = 75 e⁻⁰•⁰⁵ᵗ

So, we need the BOD concentration 50 km downstream when the velocity of the river is 15 km/day

We calculate how many days it takes the river to reach 50 km downstream

Velocity = (displacement/time)

15 = 50/t

t = 50/15 = 3.3333 days

So, we need the C that corresponds to t = 3.3333 days

C = 75 e⁻⁰•⁰⁵ᵗ

0.05 t = 0.05 × 3.333 = 0.167

C = 75 e⁻⁰•¹⁶⁷

C = 63.5 mg/L

Final answer:

The BOD concentration 50 km downstream from the wastewater treatment plant is approximately 15.865 mg/L.

Explanation:

To calculate the BOD concentration 50 km downstream, we need to consider the rate of dilution due to the flow of the river and the first-order reaction that destroys BOD. The concentration of BOD downstream can be calculated using the equation C2 = C1 * exp(-k * d/v), where C1 is the initial concentration, k is the rate constant, d is the distance, and v is the velocity of the river.

Plugging in the given values, we have C2 = 75 * exp(-0.05 * 50/15), which gives us a BOD concentration of approximately 15.865 mg/L 50 km downstream from the wastewater treatment plant.

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How long is a bus route across a small town

Answers

Answer:

4 kilometers

Explanation:

Final answer:

The length of a bus route in a small town can greatly depend on the specifics of the town and the route. It can range from a couple miles in a very small town to 20 miles or more for larger towns.

Explanation:

The length of a bus route across a small town can vary greatly depending on the size of the town and the specifics of the bus route. In a very small town, the bus route might only be a mile or two long. For larger towns, it could easily be 10-20 miles, or more. If you know the specifics of the route (streets it travels along, the number of stops, etc.), you could use a tool like Go_gle Maps to calculate an approximate distance.

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An ideal step-down transformer is needed to reduce a primary voltage of 120 V to 6.0 V. What must be the ratio of the number of turns in the secondary to the number of turns in the primary