"v0" means that there are no friction forces at that speed
<span>mgsinΘ = (mv0²/r)cosΘ → the variable m cancels </span>
<span>sinΘ/cosΘ = tanΘ = v0² / gr
</span><span>Θ = arctan(v0² / gr) </span>
<span>When v > v0, friction points downslope: </span>
<span>mgsinΘ + µ(mgcosΘ + (mv²/r)sinΘ) = (mv²/r)cosΘ → m cancels: </span>
<span>gsinΘ + µ(gcosΘ + (v²/r)sinΘ) = (v²/r)cosΘ </span>
<span>µ = ((v²/r)cosΘ - gsinΘ) / (gcosΘ + (v²/r)sinΘ) </span>
<span>where Θ is defined above. </span>
<span>When v > v0, friction points upslope: </span>
<span>mgsinΘ - µ(mgcosΘ + (mv²/r)sinΘ) = (mv²/r)cosΘ → m cancels: </span>
<span>gsinΘ - µ(gcosΘ + (v²/r)sinΘ) = (v²/r)cosΘ </span>
<span>µ = (gsinΘ - (v²/r)cosΘ) / (gcosΘ + (v²/r)sinΘ) </span>
<span>where Θ is defined above. </span>
Behavior has at least six dimensions, which are: frequency, duration, latency, topography, locus, and force. Since the coach is recording how long it takes, the track coach is recording the duration behavior because duration is a synonym for time. Duration is your answer.
Explanation:
Given:
v₀ = 250 mph
v = 0 mph
t = 25 s
Find: a
v = at + v₀
(0 mph) = a (25 s) + (250 mph)
a = -10 mph/s
Answer:
f1 = 58.3Hz, f2 = 175Hz, f3 = 291.6Hz
Explanation:
lets assume speed of sound is 350 m/s.
frequencies of a standing wave modes of an open-close tube of length L
fm = m(v/4L)
where m is 1,3,5,7......
and fm = mf1
where f1 = fundamental frequency
so therefore: f1 = 350 x 4 / 1.5
f1 = 58.3Hz
f2 = 3 x 58.3
f2 = 175Hz
f3 = 5 x 58.3
f3 = 291.6Hz
