The speed of the earth's surface located at 2/3 of the length of the arc between the pole which measure from the equator is 232.5 m/s.
Solution:
So the givens are, earth's radius = 6.37X10^6m, and the angular distance from the pole is 90 degrees. So 60 degrees is the 2/3.
r = 6.37x10^6 * cos(60) = 3.185x10^6m
since v = wr
v = 7.3x10^-5 * 3.185x10^6
v - 232.5 m/s
Answer:
298rad/s and 116.96 ohms
Explanation:
Given an L-R-C series circuit where
L = 0.450 H,
C=2.50×10^−5F, and resistance R= 0
In this situation we have a simple LC circuit with angular frequency
Wo = 1√LC
= 1/√(0.450)(2.50×10^-5)
= 1/√0.00001125
= 1/0.003354
= 298rad/s
B) Now we need to find the value of R such that it gives a 10% decrease in angular frequency.
Wi/W° = (100-10)/100
Wi/W° = 90/100
Wi/W° = 0.90 ............... 1
Angular frequency of oscillation
The complete aspect of the solution is attached, please check.
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The vertical velocity of the projectile upon returning to its original is 17. 74 m/s
<h3>
How to determine the vertical velocity</h3>
Using the formula:
Vertical velocity component , Vy = V * sin(α)
Where
V = initial velocity = 36. 6 m/s
α = angle of projectile = 29°
Substitute into the formula
Vy = 36. 6 * sin ( 29°)
Vy = 36. 6 * 0. 4848
Vy = 17. 74 m/s
Thus, the vertical velocity of the projectile upon returning to its original is 17. 74 m/s
Learn more about vertical velocity here:
brainly.com/question/24949996
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