Answer:
V₀ₓ = 10.94 m/s
V₀y = 18.87 m/s
Explanation:
To find the launch velocity, we use 1st equation of motion.
Vf = Vi + at
where,
Vf = Final Velocity of Ball = Launch Speed = V₀ = ?
Vi = Initial Velocity = 0 m/s (Since ball was initially at rest)
a = acceleration = 376 m/s²
t = time = 0.058 s
Therefore,
V₀ = 0 m/s + (376 m/s²)(0.058 s)
V₀ = 21.81 m/s
Now, for x-component:
V₀ₓ = V₀ Cos θ
where,
V₀ₓ = x-component of launch velocity = ?
θ = Angle with horizontal = 59.9⁰
V₀ₓ = (21.81 m/s)(Cos 59.9°)
<u>V₀ₓ = 10.94 m/s</u>
<u></u>
for y-component:
V₀ₓ = V₀ Sin θ
where,
V₀y = y-component of launch velocity = ?
θ = Angle with horizontal = 59.9⁰
V₀y = (21.81 m/s)(Sin 59.9°)
<u>V₀y = 18.87 m/s</u>
<u></u>
Angular width is 3 x 10^-3
Let D be the distance between source and screed d the distance between coherent source then for central diffraction maxima,
where λ is wavelength
Given:
λ = 450 nm = 450×10^−9m
d = 0.3x10^−3m, D = 1m
W = 2 x 450×10^−9/0.3x10^−3*1
To Find:
Angular width
Solution: The width of the central maxima is nothing but the difference between the positions of the first two minima. Hence we will use the expression for the position of minima and accordingly obtain the expression of the width of central maxima and secondary maxima
θ = W/D
θ = 2 x 450×10^−9/0.3x10^−3*1/1 = 3 x 10^-3
Hence, angular width is 3 x 10^-3
Learn more about Angular width here:
brainly.com/question/25292087
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It’s red shift and blue shift because This only occurs when the frequency of the wave is made longer or shorter due to the movement of the source relative to the observer
