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ser-zykov [4K]
3 years ago
10

15. In Young’s double-slit experiment using monochromatic light, the interference pattern consists of a central _____.

Physics
1 answer:
levacccp [35]3 years ago
4 0

Answer: In young's double slit experiment that uses monochromatic light the interference pattern formed has Central bright band with alternate dark and bright band That is option B.

Explanation: In this young experiment two small slits Namely a and b  are formed on the screen and a monochromatic light is focused on them. Wavelets come out of of this lets scintillating and overlapping each other. This we get from huyginns principal, thus forming alternate dark and bright bands with bright band at center and all bands are about one meter apart.

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A sprinter must average 24.0 mi/h to win a 100-m dash in 9.30 s. What is his wavelength at this speed if his mass is 84.5 kg?
crimeas [40]

Answer:

Wavelength λ = 7.31 × 10^-37 m

Explanation:

From De Broglie's equation;

λ = h/mv

Where;

λ = wavelength in meters

h = plank's constant = 6.626×10^-34 m^2 kg/s

m = mass in kg

v = velocity in m/s

Given;

v = 24 mi/h

Converting to m/s

v = 24mi/h × 0.447 m/s ×1/(mi/h)

v = 10.73m/s

m = 84.5kg

Substituting the values into the equation;

λ = (6.626×10^-34 m^2 kg/s)/(84.5kg × 10.73m/s)

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7 0
4 years ago
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(b) Find a point between the two charges on the horizontal line where the electric potential is zero. (Enter your answer as meas
aleksley [76]

Complete Question: A charge q1 = 2.2 uC is at a distance d= 1.63m from a second charge q2= -5.67 uC. (b) Find a point between the two charges on the horizontal line where the electric potential is zero. (Enter your answer as measured from q1.)

Answer:

d= 0.46 m

Explanation:

The electric potential is defined as the work needed, per unit charge, to bring a positive test charge from infinity to the point of interest.

For a point charge, the electric potential, at a distance r from it, according to Coulomb´s Law and the definition of potential, can be expressed as follows:

V = \frac{k*q}{r}

We have two charges, q₁ and q₂, and we need to find a point between them, where the electric potential due to them, be zero.

If we call x to the distance from q₁, the distance from q₂, will be the distance between both charges, minus x.

So, we can find the value of x, adding the potentials due to q₁ and q₂, in such a way that both add to zero:

V = \frac{k*q1}{x} +\frac{k*q2}{(1.63m-x)} = 0

⇒k*q1* (1.63m - x) = -k*q2*x:

Replacing by the values of q1, q2, and k, and solving for x, we get:

⇒ x = (2.22 μC* 1.63 m) / 7.89 μC = 0.46 m from q1.

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3 years ago
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Explanation:

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