From January 26, 1977, to September 18, 1983, George Meegan of Great Britain walked from Ushuaia, at the southern tip of South A
merica, to Prudhoe Bay in Alaska, covering 30 600 km. In meters per second, what was the magnitude of his average velocity during that time period?
1 answer:
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Answer:
1) The maximum jump height is reached at A.
2) The maximum center of mass height off of the ground is B.
3) The time of flight is C.
4) The distance of jump is B.
Explanation:
First of all we need to decompose velocity in its rectangular components, so
1) We use, , as we clear it for
and using the fact that
at max height, we obtain
2) We can use the formula for
, so
3) We can use the formula , to find total time of fligth, so
, as it is a second-grade polynomial, we find that its positive root is
4) Finally, we use , as it has an additional displacement of
due the leg extension we obtain,
, aprox
Answer:
L= 0.059 mH
Explanation:
Given that
R = 855 Ω and C = 6.25 μF
V= 84 V
Frequency
ω = 51900 1/s
We know that
L=Inductance
C=Capacitance
ω =angular Frequency
ω² L C =1
(51900)² x L x 6.25 x 10⁻⁶ = 1
L= 5.99 x 10⁻⁵ H
L= 0.059 mH