) The magnitude of scalar product of two unit vectors perpendicular to each other is
1 answer:
This question involves the concept of the scalar product.
The magnitude of the scalar product will be "0".
<h3>SCALAR PRODUCT</h3>The scalar product, also known as the dot product of the two vectors is given by the following formula:
where,
- A.B = Scalar product = ?
- |A| = Mangnitude of vector A = 1 unit
- |B| = Magnitude of Vector B = 1 unit
- θ = Angle between vectors = 90°
Therefore,
A.B = 0
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Answer:
R = 1.2295 10⁵ m
Explanation:
After reading your problem they give us the diameter of the lens d = 4.50 cm = 0.0450 m, therefore if we use the Rayleigh criterion for the resolution in the diffraction phenomenon, we have that the minimum separation occurs in the first minimum of diffraction of one of the bodies m = 1 coincides with the central maximum of the other body
θ = 1.22 λ / D
where the constant 1.22 leaves the resolution in polar coordinates and D is the lens aperture
how angles are measured in radians
θ = y / R
where y is the separation of the two bodies (bulbs) y = 2 m and R the distance from the bulbs to the lens
R =
let's calculate
R =
R = 1.2295 10⁵ m
Hence ,From the Guide there are other parameters which with this equation will give the exact time the athlete's walk back
From the question we are told
If the average velocity during the athlete's walk back to the starting line in Guided Example 2.5 is – 1.50 m/s,
Generally the equation Time spent is mathematically given as
T=\frac{d}{v}
Therefore
Hence ,From the Guide there are other parameters which with this equation will give the exact time the athlete's walk back
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Answer:
The appropriate solution is "2.78 mm".
Explanation:
Given:
or,
or,
As we know,
Fringe width is:
⇒
hence,
Separation between second and third bright fringes will be:
⇒
or,