Answer:
A negatively charged particle -q is placed at the center of a uniformly charged ring, where the ring has a total positive charge Q as shown in the following figure. The particle, confined to move along the x axis, is moved a small distance x along the axis ( where x << a) and released. Show that the particle oscillates in simple harmonic motion with a frequency given by,
(a) The velocity ratio of the screw is 1570.8.
(b) The mechanical advantage of the screw is 785.39.
<h3>
Velocity ratio of the screw</h3>
The velocity ratio of the screw is calculated as follows;
V.R = 2πr/P
where;
- P is the pitch = 1/10 cm = 0.1 cm = 0.001 m
- r is radius = 25 cm = 0.25 m
V.R = (2π x 0.25)/(0.001)
V.R = 1570.8
<h3>Mechanical advantage of the screw</h3>
E = MA/VR x 100%
0.5 = MA/1570.8
MA = 785.39
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Answer:
a) i₈ = 0.5 i₄, b) i₁₀ = 0.3 i₃, i₁₀ = 0.8 i₈
Explanation:
For this exercise we use ohm's law
V = i R
i = V / R
we assume that the applied voltage is the same in all cases
let's find the current for each resistance
R = 4 Ω
i₄ = V / 4
R = 8 Ω
i₈ = V / 8
we look for the relationship between these two currents
i₈ /i₄ = 4/8 = ½
i₈ = 0.5 i₄
R = 3 Ω
i₃ = V3
R = 10 Ω
i₁₀ = V / 10
we look for relationships
i₁₀ / 1₃ = 3/10
i₁₀ = 0.3 i₃
i₁₀ / 1₈ = 8/10
i₁₀ = 0.8 i₈
The answer is D. <span> There would be a decrease in the population of marine organisms.
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