Use equations of motion to find the velocity just before it hits the floor:
<span>Vf^2 = Vi^2 + 2gx </span>
<span>Final velocity = 4.42m/s </span>
<span>Impulse is change in momentum so: </span>
<span>m(Vf - Vi) = 0.05(0 - 4.42) </span>
<span>= - 0.221 kg.m/s
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** Missing info: Lines per mm = 500 **
Ans: The wavelength is = λ = 1414.21 nm
Explanation:
The formula for diffraction grading is:
dsinθ = mλ --- (1)
Where
d = 1/lines-per-meter = (1/500)*10^-3 = 2 * 10^-6
m = order = 1
λ = wavelength
θ = 45°
Plug in the values in (1):
(1) => 2*10^-6*sin(45°) = (1)λ
=> λ = 1414.21 nm
Complete Question:
A hollow cylinder with an inner radius of 4.0mm and an outer radius of 30mm conducts a 3.0-A current flowing parallel to the axis of the cylinder. If the current density is uniform throughout the wire, what is the magnitude of the magnetic field at a point 12mm from its center?
Answer:
The magnitude of the magnetic field = 7.24 μT
Explanation:
Inner radius, a = 4.0 mm = 0.004 m
Outer radius, b = 30 mm = 0.03 m
Radius, r = 12 mm = 0.012 m
let h² = b² - a²
h² = 0.03² - 0.004²
h² = 0.000884
Let d² = r² - a²
d² = 0.012² - 0.004²
d² = 0.000128
Current I = 3A
μ = 4π * 10⁻⁷
The magnitude of the magnetic field is given by:

B = 7.24 * 10⁻⁶T
B = 7.24 μT