Three basic electrical effects occur when an electric current flows in a conductor: A magnetic field is set up around the conduc
1 answer:
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Answer:
a)
b)
c)
d)
e) &
f)
Explanation:
From the question we are told that:
Stretch Length
Mass
Total stretch length
a)
Generally the equation for Force F on the spring is mathematically given by
b)Generally the equation for Max Velocity of Mass on the spring is mathematically given by
Where
A=Amplitude
And
Therefore
c)
Generally the equation for Max Acceleration of Mass on the spring is mathematically given by
d)
Generally the equation for Total mechanical energy of Mass on the spring is mathematically given by
e)
Generally the equation for the period T is mathematically given by
Generally the equation for the Frequency is mathematically given by
f)
Generally the Equation of time-dependent vertical position of the mass is mathematically given by
Where
'= signify order of differentiation
Answer:
Yes, it will produce a diffraction pattern.
a. 3.9 mm b. 1.95 mm
Explanation:
The light shined from a single slit will produce a diffraction pattern because, the wavefront act as wavelets which generates its own wave according to Huygens principle. This therefore causes the diffraction pattern.
Given
wavelength of green light, λ = 520 nm = 520 × 10⁻⁹ m = 5.20 × 10⁻⁷ m
width of slit, d = 0.440 mm = 0.44 × 10⁻³ m = 4.4 × 10⁻⁴ m
Distance of slit from central maximum , D = 1.65 m
Distance of first minimum from central maximum, y = ?
a. The relationship between the slit width and wavelength is given by [tex} dsinθ = mλ [/tex]where d = slit width, θ = angular distance from central maximum, λ = wavelength of light and m = ±1, ±2, ±3...
The relationship between y and D is given by
Since θ is small, sinθ ≈ θ ≈ tanθ
so, dθ = mλ ⇒ θ = mλ/d = y/D
Therefore, y = mλD/d
Now, for the first minimum above the slit, m = +1 and for the first minimum below the slit, m = -1. So, y₁ = λD/d and y₋₁ = -λD/d. So, the width of the central maximum Δy is the difference between the first minima below and above the central maximum. So, Δy = y₁ - y₋₁ = λD/d -(-λD/d) = 2λD/d
Substituting the values from above, Δy= 2 × 5.20 × 10⁻⁷ × 1.65/4.4 × 10⁻⁴ = 3900 × 10⁻⁶ m = 3.9 × 10⁻³ m = 3.9 mm
b. The first order fringe is the fringe located between the first minimum and the second minimum. From dsinθ = mλ and tanθ = y/D when θ is small, sinθ ≈ θ ≈ tanθ. So, y = mλD/d. Let m= 1 and m=2 be the first and second minima respectively. So,y₁ = λD/d and y₂ = 2λD/d. The difference Δy₁ = y₂ - y₁ is the width of the first order fringe. Therefore, Δy₁ = 2λD/d - λD/d= λD/d. Substituting the values from above, we have
λD/d= 5.20 × 10⁻⁷ × 1.65/4.4 × 10⁻⁴= 1.95 × 10⁻³ m = 1.95 mm