A city engineer knows that she will need $25 million in 3 years to replace toll booths on a toll road in the city. Traffic on th
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Answer:
1) 1.71 rad/s
2) -6.22 rad/s²
Explanation:
Choose point C to be the origin.
Using geometry, we can show that the coordinates of point A are:
(a cos 30°, a sin 30° − b)
Therefore, the coordinates of point D at time t are:
(a cos 30° − b sin(ωt), a sin 30° − b + b cos(ωt))
The angle formed by CB with the x-axis is therefore:
tan θ = (a sin 30° − b + b cos(ωt)) / (a cos 30° − b sin(ωt))
1) Taking the derivative with respect to time, we can find the angular velocity:
sec² θ dθ/dt = [(a cos 30° − b sin(ωt)) (-bω sin(ωt)) − (a sin 30° − b + b cos(ωt)) (-bω cos(ωt))] / (a cos 30° − b sin(ωt))²
sec² θ dθ/dt = -bω [(a cos 30° − b sin(ωt)) sin(ωt) − (a sin 30° − b + b cos(ωt)) cos(ωt)] / (a cos 30° − b sin(ωt))²
sec² θ dθ/dt = -bω [(a cos 30° sin(ωt) − b sin²(ωt)) − (a sin 30° cos(ωt) − b + b cos²(ωt))] / (a cos 30° − b sin(ωt))²
sec² θ dθ/dt = -bω (a cos 30° sin(ωt) − b sin²(ωt) − a sin 30° cos(ωt) + b − b cos²(ωt)) / (a cos 30° − b sin(ωt))²
sec² θ dθ/dt = -bω (a cos 30° sin(ωt) − a sin 30° cos(ωt)) / (a cos 30° − b sin(ωt))²
sec² θ dθ/dt = -abω (cos 30° sin(ωt) − sin 30° cos(ωt)) / (a cos 30° − b sin(ωt))²
We know at the moment shown, a = 350 mm, b = 200 mm, θ = 30°, ω = 6 rad/s, and t = 0 s.
sec² 30° dθ/dt = -(350) (200) (6) (cos 30° sin(0) − sin 30° cos(0)) / (350 cos 30° − 200 sin(0))²
sec² 30° dθ/dt = -(350) (200) (6) (-sin 30°) / (350 cos 30°)²
dθ/dt = (200) (6) (1/2) / 350
dθ/dt = 600 / 350
dθ/dt = 1.71 rad/s
2) Taking the second derivative of θ with respect to time, we can find the angular acceleration.
sec² θ d²θ/dt² + 2 sec² θ tan θ dθ/dt = -abω [(a cos 30° − b sin(ωt))² (ω cos 30° cos(ωt) + ω sin 30° sin(ωt)) − (cos 30° sin(ωt) − sin 30° cos(ωt)) (2 (a cos 30° − b sin(ωt)) (-bω cos(ωt)))] / (a cos 30° − b sin(ωt))⁴
At t = 0:
sec² θ d²θ/dt² + 2 sec² θ tan θ dθ/dt = -abω [(a cos 30°)² (ω cos 30°) − (0 − sin 30°) (2 (a cos 30°) (-bω))] / (a cos 30°)⁴
sec² θ d²θ/dt² + 2 sec² θ tan θ dθ/dt = -abω (a²ω cos³ 30° − 2abω sin 30° cos 30°) / (a⁴ cos⁴ 30°)
sec² θ d²θ/dt² + 2 sec² θ tan θ dθ/dt = -bω (aω cos² 30° − 2bω sin 30°) / (a² cos³ 30°)
d²θ/dt² + 2 tan θ dθ/dt = -bω² (a cos² 30° − b) / (a² cos 30°)
Plugging in values:
d²θ/dt² + 2 tan 30° dθ/dt = -(200) (6)² (350 cos² 30° − 200) / (350² cos 30°)
d²θ/dt² + 2 tan 30° dθ/dt = -7200 (262.5 − 200) / (350² cos 30°)
d²θ/dt² + 2 tan 30° (1.71) = -4.24
d²θ/dt² = -6.22 rad/s²
Explanation:
A.
H = Aeσ^4
Using the stefan Boltzmann law
When we differentiate
dH/dT = 4AeσT³
dH/dT = 4(0.15)(0.9)(5.67)(10^-8)(650)³
= 8.4085
Exact error = 8.4085x20
= 168.17
H(650) = 0.15(0.9)(5.67)(10^-8)(650)⁴
= 1366.376watts
B.
Verifying values
H(T+ΔT) = 0.15(0.9)(5.67)(10)^-8(670)⁴
= 1542.468
H(T+ΔT) = 0.15(0.9)(5.67)(10^-8)(630)⁴
= 1205.8104
Error = 1542.468-1205.8104/2
= 168.329
ΔT = 40
H(T+ΔT) = 0.15(0.9)(5.67)(10)^-8(690)⁴
= 1735.05
H(T-ΔT) = 0.15(0.9)(5.67)(10^-8)(610)⁴
= 1735.05-1059.83/2
= 675.22/2
= 337.61
