1. What is the frequency of light waves with wavelength of 5 x 10-⁷ m? ​

Answers

Answer 1
Answer:

Taking into account the definition of wavelength, frecuency and propagation speed, the frequency of light waves with wavelength of 5×10⁻⁷ m is 6×10¹⁴ Hz.

Definition of wavelength

First of all, wavelength is the minimum distance between two successive points on the wave that are in the same state of vibration. It is expressed in units of length (m).

Definition of frequency

On the other side, frequency is the number of vibrations that occur in a unit of time. Its unit is s⁻¹ or hertz (Hz).

Definition of propagation speed

Finally, the propagation speed is the speed with which the wave propagates in the medium, that is, it is the magnitude that measures the speed at which the wave disturbance propagates along its displacement.

The propagation speed relate the wavelength (λ) and the frequency (f) inversely proportional using the following equation:

v = f× λ

All electromagnetic waves propagate in a vacuum at a constant speed of 3×10⁸ m/s, the speed of light.

Frequency of light waves with wavelength of 5×10⁻⁷ m

In this case, you know:

  • v= 3×10⁸ m/s
  • f= ?
  • λ= 5×10⁻⁷ m

Replacing in the definition of propagation speed:

3×10⁸ m/s = f× 5×10⁻⁷ m

Solving:

3×10⁸ m/s ÷ 5×10⁻⁷ m= f

f= 6×10¹⁴ Hz

In summary, the frequency of light waves with wavelength of 5×10⁻⁷ m is 6×10¹⁴ Hz.

Learn more about wavelength, frecuency and propagation speed:

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Answer 2
Answer:

Answer:

Speed of light =m/s

wavelength = m

frequency = ?

we have

Speed = frequency × wavelength

3* 10^8 = frequency × 5 * 10^(-7)

 Frequency = (3*10^8)/(5*10^(-7))=6*10^(14)hz


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A parallel-plate capacitor is charged and then disconnected from the battery. By what factor does the stored energy change when the plate separation is then doubled?

Answers

Answer:

U/U₀ = 2

(factor of 2 i.e U = 2U₀)

Therefore, the energy stored in the capacitor is doubled when the plate separation is doubled while the capacitor has been disconnected

Explanation:

Energy stored in a capacitor can be expressed as;

U = 0.5CV^2 = Q^2/2C

And

C = ε₀ A/d

Where

C = capacitance

V = potential difference

Q = charge

A = Area of plates

d = distance between plates

So

U = Q^2/2C = dQ^2/2ε₀ A

The initial energy of the capacitor at d = d₀ is

U₀ = Q^2/2C = d₀Q^2/2ε₀ A ....1

When the plate separation is increased after the capacitor has been disconnected, the charge Q of the capacitor remain constant.

The final energy stored in the capacitor at d = 2d₀ is

U = 2d₀Q^2/2ε₀ A ...2

The factor U/U₀ can be derived by substituting equation 1 and 2

U/U₀ = (2d₀Q^2/2ε₀ A)/( d₀Q^2/2ε₀ A )

Simplifying we have;

U/U₀ = 2

U = 2U₀

Therefore, the energy stored in the capacitor is doubled when the plate separation is doubled while the capacitor has been disconnected.

Ken Griffey, Jr's warehouse shot in the 1933 home run derby travelled 93 feet per second for 5 seconds. How far did he hit the ball?

Answers

Answer:

465 feet because 93*5 = 465, btw that was 1993 not 1933

Explanation:

A current cannot produce a magnetic field. *

True or false

Answers

Answer:

False

Explanation:

False, as a magnetic field is generated whenever current travels through a conductor.

An electromagnet consists of a coil of wire wrapped around a bar of iron. The coil and iron bar get magnetized when electric current flows through the wire. An electromagnet also has north and south magnetic poles. The magnetic field is strongest at either pole of the magnet.

When a 20.0-ohm resistor is connected across the terminals of a 12.0-V battery, the voltage across the terminals of the battery falls by 0.300 V. What is the internal resistance of this battery

Answers

The internal resistance of the battery is 0.5 ohms.

To calculate the internal resistance of the battery, we use the formula below

Formula:

  • (V/R)r = V'............. Equation 1

Where:

  • V = Voltage across the terminal of the battery
  • R = Resistance connected across the battery
  • r = internal resistance of the battery
  • V' = voltage drop of the battery.

Make r the subject of the equation

  • r = V'R/V............ Equation 2

From the question,

Given:

  • V = 12 V
  • R = 20 ohms
  • V' = 0.3 V

Substitute these values into equation 2

  • r = (0.3×20)/12
  • r = 6/12
  • r = 0.5 ohms.

Hence, The internal resistance of the battery is 0.5 ohms.

Learn more about internal resistance here: brainly.com/question/14883923

Answer:

The  internal resistance is  r =  0.5 \ \Omega

Explanation:

From the question we are told that the resistance of

   The  resistance of the resistor is  R  =  20.0\  \Omega

    The  voltage is V  = 12.0 \ V

     The magnitude of the voltage fall is  e   =  0.300\  V

Generally the current flowing through the terminal due to the voltage of the battery  is  mathematically represented as

        I  =  (V)/(R)

substituting values

        I  =  (12.0 )/(20 )

       I  =  0.6 \ A

The internal resistance of the battery is mathematically represented as

      r =  (e)/(I)

substituting values

     r =  (0.300)/( 0.6 )

    r =  0.5 \ \Omega

Determine whether the following statements are true and give an explanation or counterexample.(A) If the acceleration of an object remains constant, its velocity is constant.
(B) If the acceleration of object moving along a line is always 0, then its velocity is constant.
(C) It is impossible for the instantaneous velocity at all times a(D) A moving object can have negative acceleration and increasing speed.

Answers

Answer:

Explanation:(A)if a body is accelerating then it's velocity can't be constant since an object is said to be accelerating if it is changing velocity (B)if the acceleration of an object moving along a line is 0 then it's velocity will be constant since there is no change in direction or speed(C)No.it is not possible for a moving body to have an instantaneous velocity at all times since instantaneous velocity is the velocity of a body at a certain instant of time..(D)Yes a moving object can have a negative acceleration and increasing speed,it can also have a positive acceleration with decreasing speed.

A 0.320 kg ball approaches a bat horizontally with a speed of 14.0 m/s and after getting hit by the bat, the ball moves in the opposite direction with a speed of 22 m/s. If the ball is in contact with the bat for 0.0600 s, determine the magnitude of the average force exerted on the bat.

Answers

Answer:

42.67N

Explanation:

Step one:

Given

mass m= 0.32kg

intital velocity, u= 14m/s

final velocity v= 22m/s

time= 0.06s

Step two:

Required

Force F

the expression for the force is

F=mΔv/t

F=0.32*(22-14)/0.06

F=(0.32*8)/0.06

F=2.56/0.06

F=42.67N

The average force exerted on the bat 42.67N