Answer:
We have learned that refraction occurs as light passes across the boundary between two media. Refraction is merely one of several possible boundary behaviors by which a light wave could behave when it encounters a new medium or an obstacle in its path.
It would take about 2 thirds of a second or .66666666 repeating of a second. please give brainliest?
The only graph that accurately depict the given motion is graph D.
The given parameters;
- initial position of the man = 0
- direction of the man's first displacement = backward
- time of first motion, t₁ = 6 seconds
- velocity of this first displacement = v₁
- time without any motion (<em>zero movement</em>) = 6 seconds
- direction of the second displacement = forward
- velocity of second displacement = 2v₁
Let the acceleration of the first displacement = a
Acceleration of the second displacement = 2a
From the given graphs we can eliminate every graph without initial decrease or motion towards the negative direction.
The only options with initial motion towards the negative direction are;
The difference between graph B and D;
- in graph B there is a uniform motion for 6 seconds
- in graph D there is no motion for 6 seconds (<em>this is obvious as the line fall directly on top of the horizontal axis maintaining a value of zero for 6 seconds</em>).
Thus, the only graph that accurately depict the given motion is graph D.
Learn more here: brainly.com/question/21095906
The magnitude of the induced emf is given by:
ℰ = |Δφ/Δt|
ℰ = emf, Δφ = change in magnetic flux, Δt = elapsed time
The magnetic field is perpendicular to the loop, so the magnetic flux φ is given by:
φ = BA
B = magnetic field strength, A = loop area
The area of the loop A is given by:
A = πr²
r = loop radius
Make a substitution:
φ = B2πr²
Since the strength of the magnetic field is changing while the radius of the loop isn't changing, the change in magnetic flux Δφ is given by:
Δφ = ΔB2πr²
ΔB = change in magnetic field strength
Make another substitution:
ℰ = |ΔB2πr²/Δt|
Given values:
ΔB = 0.20T - 0.40T = -0.20T, r = 0.50m, Δt = 2.5s
Plug in and solve for ℰ:
ℰ = |(-0.20)(2π)(0.50)²/2.5|
ℰ = 0.13V
