Annabella time for a race was 24.2 seconds. Another runnee's time was 6.43 seconds faster write and evaluate an equation with to
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The attached figure represents the relation between ω (rpm) and t (seconds)
To find the blade's angular position in radians ⇒ ω will be converted from (rpm) to (rad/s)
ω = 250 (rpm) = 250 * (2π/60) = (25/3)π rad/s
ω = 100 (rpm) = 100 * (2π/60) = (10/3)π rad/s
and from the figure it is clear that the operation is at constant speed but with variable levels
⇒ ω = dθ/dt ⇒ dθ = ω dt
∴ θ = ∫₀²⁰ ω dt
while ω is not fixed from (t = 0) to (t =20)
the integral will divided to 3 integrals as follow;
ω = 0 from t = 0 to t = 5
ω = 250 (rpm) = (25/3)π from t = 5 to t = 15
ω = 100 (rpm) = (10/3)π from t = 15 to t = 20
∴ θ = ∫₀⁵ (0) dt + ∫₅¹⁵ (25/3)π dt + ∫₁₅²⁰ (10/3)π dt
the first integral = 0
the second integral = (25/3)π t = (25/3)π (15-5) = (250/3)π
the third integral = (10/3)π t = (25/3)π (20-15) = (50/3)π
∴ θ = 0 + (250/3)π + (50/3)π = 100 π
while the complete revolution = 2π
so instantaneously at t = 20
∴ θ = 100 π - 50 * 2 π = 0 rad
Which mean:
the blade will be at zero position making no of revolution = (100π)/(2π) = 50
To find the blade's angular position in radians ⇒ ω will be converted from (rpm) to (rad/s)
ω = 250 (rpm) = 250 * (2π/60) = (25/3)π rad/s
ω = 100 (rpm) = 100 * (2π/60) = (10/3)π rad/s
and from the figure it is clear that the operation is at constant speed but with variable levels
⇒ ω = dθ/dt ⇒ dθ = ω dt
∴ θ = ∫₀²⁰ ω dt
while ω is not fixed from (t = 0) to (t =20)
the integral will divided to 3 integrals as follow;
ω = 0 from t = 0 to t = 5
ω = 250 (rpm) = (25/3)π from t = 5 to t = 15
ω = 100 (rpm) = (10/3)π from t = 15 to t = 20
∴ θ = ∫₀⁵ (0) dt + ∫₅¹⁵ (25/3)π dt + ∫₁₅²⁰ (10/3)π dt
the first integral = 0
the second integral = (25/3)π t = (25/3)π (15-5) = (250/3)π
the third integral = (10/3)π t = (25/3)π (20-15) = (50/3)π
∴ θ = 0 + (250/3)π + (50/3)π = 100 π
while the complete revolution = 2π
so instantaneously at t = 20
∴ θ = 100 π - 50 * 2 π = 0 rad
Which mean:
the blade will be at zero position making no of revolution = (100π)/(2π) = 50
If Sa=2πrh+2πv=π
then the surface area is π
and volume is
(rh-2h)/2r.
Given Sa=2πrh+2π=π
.
We have to find surface area and volume from the given expression.
Volume is basically amount of substance a container can hold in its capacity.
First we will find the value of v from the expression. Because they are in equal to each other, we can easily find the value of v.
2πrh+2πv=π
h
Keeping the term containing v at left side and take all other to right side.
2πv=π
-2πrh
v=(πh-2πrh)/2π
v=π/2π
-2πrh/2π
v=h/2-h/r
v=h(r-2)/2r
Put the value of v in Sa=2πrh+2π
Sa=2πrh+2π*h(r-2)/2r
=2πrh+2πrh(r-2)/2
=2πrh+πrh(r-2)
=2πrh+πh-2πrh
=πh
Hence surface area is πh and volume is h(r-2)/2.
Learn more about surface area at brainly.com/question/16519513
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