Answer:
Electromagnetic force creates electromagnetic field.
Explanation:
Electromagnetic force acts between charged particles and this force is the combination of all electrical and magnetic forces. On the other hand, the electromagnetic field is detected by the force it exerts on other magnetic particles and moving electric charges. So, electromagnetic field is created by electromagnetic force. Without this force there cannot be any electromagnetic field.
B. a covalent bond because a covalent bond is when 2 atoms share either 1 or more than 1 pairs of electrons, and valence electrons are the type of electrons that are shared in a covalent bond.
Answer:
Ek = 1705.28 [J]
Explanation:
In order to solve this problem, we must remember that kinetic energy can be calculated by means of the following equation.
where:
m = mass [kg]
v = velocity [m/s]
Ek = kinetic energy [J] (Units of Joules)
<u>For the person running</u>
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<u>For the bullet</u>
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The difference in Kinetic energy is equal to:
Ek = 2025 - 319.72
Ek = 1705.28 [J]
The right answer for the question that is being asked and shown above is that: "A crest with an amplitude of 9 will form." Wave X has an amplitude of 5 and wave Y has an amplitude of 4. the crest of wave X meets the crest of wave Y is that <span>A crest with an amplitude of 9 will form.</span>
Answer:
These are Diffraction Grating Questions.
Q1. To determine the width of the slit in micrometers (μm), we will need to use the expression for distance along the screen from the center maximum to the nth minimum on one side:
Given as
y = nDλ/w Eqn 1
where
w = width of slit
D = distance to screen
λ = wavelength of light
n = order number
Making x the subject of the formula gives,
w = nDλ/y
Given
y = 0.0149 m
D = 0.555 m
λ = 588 x 10-9 m
and n = 3
w = 6.6x10⁻⁵m
Hence, the width of the slit w, in micrometers (μm) = 66μm
Q2. To determine the linear distance Δx, between the ninth order maximum and the fifth order maximum on the screen
i.e we have to find the difference between distance along the screen (y₉-y₅) = Δx
Recall Eqn 1, y = nDλ/w
given, D = 27cm = 0.27m
λ = 632 x 10-9 m
w = 0.1mm = 1.0x10⁻⁴m
For the 9th order, n = 9,
y₉ = 9 x 0.27 x 632 x 10-9/ 1.0x10⁻⁴m = 0.015m
Similarly, for n = 5,
y₅ = 5x 0.27 x 632 x 10-9/ 1.0x10⁻⁴m = 0.0085m
Recall, Δx = (y₉-y₅) = 0.015 - 0.0085 = 0.0065m
Hence, the linear distance Δx between the ninth order maximum and the fifth order maximum on the screen = 6.5mm