A tank with some water in it begins to drain. The function v ( t ) = 46 − 1.5 t determines the volume of the water in the tank (
in gallons) given a number of minutes t since the water began draining. What is the vertical intercept of f ? What does your answer to part (a) represent? Select all that apply. How many minutes it takes for all of the water to drain from the tank. The number of gallons of water in the tank when it starts draining. The weight of the tank when it is empty. List all horizontal intercepts (or "roots") of f . What does your answer to part (b) represent? Select all that apply. The weight of the tank when it is empty. How many minutes it takes for all of the water to drain from the tank. The number of gallons of water in the tank when it starts draining.
1 answer:
Answer:
(a) The number of gallons of water in the tank when it starts draining.
(b) How many minutes it takes for all of the water to drain from the tank.
Explanation:
So, basically v(t) defines the volume of tank as a function of time. When we evaluate t=0, we find the intercept with the y-axis which represent the initial water volume of the tank in gallons:
now we need to know when v(t)=0, so:
Solving for t:
This represents the amount of time in minutes required to drain all of the water from the tank, in other words v=0gal
I attached a picture where you can see the graph that represents the function:
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Answer:
Explanation:
Given the data in the question;
L = 46 in
Ga = 5 × 10³ ksi
Gs = 11 × 10³ ksi
Outside diameter da = 5 in
ds = 4 in
Tb = 3 kip.in
Now,
Ja = polar moment of Inertia of Aluminum;
Ja ⇒ π/32( 5⁴ - 4⁴ ) = π/32( 625 - 256 ) = π/32( 369 )
Js = polar moment of inertia of steel
Js ⇒ π/32 ds⁴ = π/32( 4⁴ ) = π/32( 256 )
Ta is torque transmitted by Aluminum
Ts is torque transmitted by steel
{composite member }
T = Ta + Ts ------ let this be equation m1
Now, we use the relation;
T/J = G∅/L
JG∅ = TL
∅ = TL/GJ
so, for aluminum rod ∅ = TaLa/GaJa
for steel rod ∅ = TsLs/GsJs
but we know that, ∅a = ∅s = ∅
so
[TaLa/GaJa] = [TsLs/GsJs]
also, we know that, La = Ls = L
∴ [Ta/GaJa] = [Ts/GsJs]
we solve for Ta
TaGsJs = TsGaJa
Ta = TsGaJa / GsJs
we substitute
Ta = [Ts(5 × 10³)( π/32( 369) )] / [ (11 × 10³)( π/32( 256 ) ) ]
Ta = 0.66Ts
now, we substitute 0.66Ts for Ta and 3 for T in equation 1
T = Ta + Ts
3 = 0.66Ts + Ts
3 = 1.66Ts
Ts = 3 / 1.66
Ts = 1.8072 ≈ 1.81 kip-in
so
∅ = TsLs / GsJs
we substitute
∅ = (1.81 × 46 ) / ( 11 × 10³ × π/32( 256 ) )
∅ = 83.26 / 276460.1535
∅ = 0.000301
∅ = 3.01 × 10⁻⁴ rad
so
∅ = ∅
= 3.01 × 10⁻⁴ rad
Therefore, the magnitude of the angle of twist at end B is 3.01 × 10⁻⁴ rad