A certain part of the electromagnetic spectrum ranges from 200 nm to 400 nm. What is the lowest frequency associated with this portion of the spectrum?

Answers

Answer 1
Answer:

Answer:

the lowest frequency is 7.5* 10^(14) Hz

Explanation:

In the question it is given that wavelength(L) in the range of 200μm to 400μm.

let ν be frequency of wave v velocity = 3\times 10^8

velocity v= Lν

therefore ν=(v)/(L)

frequency ν be lopwest when L will be heighest

ν(lowest)=(3* 10^8)/(400* 10^-9)

ν=7.5* 10^(14) Hz


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(a) Is the velocity of car A greater than, less than, or the same as thevelocity of car B?
(b) Is the initial position of car A greater than, less than, or equal to the
initial position of car B?
(c) In the time period from t = 0 tot = 1 s, is car A ahead of car B,
behind car B, or at the same position as car B?

Answers

a. ) Is the velocity of car A  less than the velocity of car B b. the initial position of car A greater than the initial position of car B  c. ahead In the time period from t = 0 tot = 1 s, is car A ahead of car B?.

what is velocity ?

Velocity is the parameter which is different from speed,  can be defined as the rate at which the position of the object is changed with respect to time, it is basically speeding the object in a specific direction in a specific rate.

Velocity is a  vector quantity which shows both magnitude  and direction  and The SI unit of velocity is meter per second (ms-1). If there is a change in magnitude or the direction of velocity of a body, then it is said to be accelerating.

Finding the final velocity is simple but few calculations and basic conceptual knowledge are needed.

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Answer:

a. less than, b. greater than, c. ahead

Explanation:

An airplane is traveling 835 km/h in a direction 41.5 ∘ west of north. Find the components of the velocity vector in the northerly and westerly directions. How far north and how far west has the plane traveled after 2.20 h ?

Answers

I assume the graph is looking like in the picture bellow.

North component:
cos(41.5) * 835 = 625.37 km/h

West component of speed:
sin(41.5) * 835 = 553.29 km/h

After 2.2 hours plane will fly:
2.2*625.37 = 1375.81 km north
2.2*553.29 = 1217.23 km  west

Final answer:

To find the components of the velocity vector, you can use trigonometry. The north component is calculated using the sine function and the west component is calculated using the cosine function. After 2.20 hours, the distance traveled north and west can be found by multiplying the velocity components by the time.

Explanation:

To find the components of the velocity vector in the northerly and westerly directions, we can use trigonometry. The velocity vector is 835 km/h and is traveling in a direction 41.5° west of north. To find the north component, we can use the sine function: North component = velocity * sin(angle). To find the west component, we can use the cosine function: West component = velocity * cos(angle).

After 2.20 hours, we can find the distance traveled north and west by multiplying the velocity components by the time: Distance north = North component * time and Distance west = West component * time.

Let's calculate the values:

  1. North component = 835 km/h * sin(41.5°)
  2. West component = 835 km/h * cos(41.5°)
  3. Distance north = North component * 2.20 h
  4. Distance west = West component * 2.20 h

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Two parallel wires carry currents in the same direction. If the currents in the wires are 1A and 4A and the wires are 5 m apart. Calculate the force between the 2 wires. Is it attractive or repulsive? Calculate the magnetic field midway between the wires.

Answers

Answer:

1.6* 10^(-7) N

2.4* 10^(-7) N

Explanation:

i_(1) = 1 A

i_(2) = 4 A

r = distance between the two wire = 5 m

F = Force per unit length acting between the two wires

Force per unit length acting between the two wires is given as

F = (\mu _(o))/(4\pi )(2i_(1)i_(2))/(r)

F = (10^(-7))(2(1)(4))/(5)

F = 1.6* 10^(-7) N

r'} = distance of each wire from the midpoint = 2.5 m

Magnetic field midway between the two wires is given as

B = (\mu _(o))/(4\pi ) \left \left ( (2i_(2))/(r') \right - (2i_(1))/(r') \right \right ))

B = (10^(-7)) \left \left ( (2(4))/(2.5) \right - (2(1))/(2.5) \right \right ))

B = 2.4* 10^(-7)

Net force is the sum of all the forces acting on an object. If a spring balance pulls on a body with a force of 10 N, and friction acts on the body in the opposite direction with a force of 1 N, the net force would be 9 N in the direction of the spring balance (10 N – 1 N = 9 N).What is the net force acting on the object when the spring balance pulls the rope with a force of 25 N and friction acts on the body with a force of 20N?

Answers

Answer:

5N

Explanation:

(25 N - 20 N = 5 N)

A woman who weighs 500 N stands on an 8.0-m-long board that weighs 100 N. The board is supported at each end. The support force at the right end is 3 times the support force at the left end. How far from the right end is the woman standing?

Answers

Answer:

The woman's distance from the right end is 1.6m = (8-6.4)m.

The principles of moments about a point or axis running through a point and summation of forces have been used to calculate the required variable.

Principle of moments: the sun of clockwise moments must be equal to the sun of anticlockwise moments.

Also the sun of upward forces must be equal to the sun of downward forces.

Theses are the conditions for static equilibrium.

Explanation:

The step by step solution can be found in the attachment below.

Thank you for reading this solution and I hope it is helpful to you.

A uniform electric field is parallel to the x axis. In what direction can a charge be displaced in this field without any external work being done on the charge?

Answers

Answer:

θ=π/2

Explanation:

The definition of work is W = → F ⋅ → d = q E c o s θ d W=F→⋅d→=qEcosθd. So if no work is done, the displacement must be in the direction perpendicular to the force ie c o s θ = 0 → θ = π / 2 cosθ=0→θ=π/2

Final answer:

A charged particle can be displaced without any external work done on it in a uniform electric field when its movement is perpendicular to the direction of the electric field.

Explanation:

In a uniform electric field, the electric force is the same in every direction. Therefore, if a charge were to be displaced perpendicular to the original direction of the electric field (i.e., in the y or z direction), it would not encounter any extra electric forces. This means there would be no external work being done on the charge. When a charge is moved perpendicular to an electric field, the field does not affect it, and hence, no work is done by the field.

In other words, a charge can be displaced in this field without any external work being done on it when it is moved in a direction perpendicular to the uniform electric field, either in y-axis or z-axis, assuming the electric field is constant in the x-axis direction.

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