Answer:
Therefore % increase in velocity is 18.23 %
Explanation:
we use the equality of mass flow rate and the areas
The percentage increase in velocity is
Δ v% = 
100%
= 
.100%
= 
. 100%
= Therefore % increase in velocity is 18.23 %
Answer:
a. 21.68 rad/s b. 30.78 m/s c. 897 rev/min² d. 1085 revolutions
Explanation:
a. Its angular speed in radians per second ω = angular speed in rev/min × 2π/60 = 207 rev/min × 2π/60 = 21.68 rad/s
b. The linear speed of a point on the flywheel is gotten from v = rω where r = radius of flywheel = 1.42 m
So, v = rω = 1.42 m × 21.68 rad/s = 30.78 m/s
c. Using α = (ω₁ - ω)/t where α = angular acceleration of flywheel, ω = initial angular speed of wheel in rev/min = 21.68 rad/s = 207 rev/min, ω₁ = final angular speed of wheel in rev/min = 1410 rev/min = 147.65 rad/s, t = time in minutes = 80.5/60 min = 1.342 min
α = (ω₁ - ω)/t
= (1410 - 207)/(80.5/60)
= 60(1410 - 207)/80.5
= 60(1203)80.5
= 896.65 rev/min² ≅ 897 rev/min²
d. Using θ = ωt + 1/2αt²
where θ = number of revolutions of flywheel. Substituting the values of the variables from above, ω = 207 rev/min, α = 896.65 rev/min² and t = 80.5/60 min = 1.342 min
θ = ωt + 1/2αt²
= 207 × 1.342 + 1/2 × 896.65 × 1.342²
= 277.725 + 807.417
= 1085.14 revolutions ≅ 1085 revolutions
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Explanation:
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Answer:
Vprom = 0.00347[km/min]
Explanation:
We can calculate each of the average speeds and then perform the overall average between the two speeds.
V1 = 6/54
V1 = 0.111[km/min]
V2 = 1/16
V2 = 0.0625[km/min]
![V_{prom} = \frac{V_{1} + V_{2}}{2} \\V_{prom} = \frac{0.1111 + 0.0625}{2}\\V_{prom} = 0.00347 [km/min]](https://tex.z-dn.net/?f=V_%7Bprom%7D%20%3D%20%5Cfrac%7BV_%7B1%7D%20%2B%20V_%7B2%7D%7D%7B2%7D%20%20%5C%5CV_%7Bprom%7D%20%3D%20%5Cfrac%7B0.1111%20%2B%200.0625%7D%7B2%7D%5C%5CV_%7Bprom%7D%20%3D%200.00347%20%5Bkm%2Fmin%5D)
