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What is the minimum efficiency of a functioning current-model catalytic converter? a. 60% b. 75% c. 80% d. 90%
1 answer:
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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
Answer:
Both series circuits provide a total voltage of 9 volts to the two bulbs connected in series and the voltage will be equally divided among two bulbs and they will have same brightness. Therefore, both circuits will have same characteristics.
Explanation:
We are asked to compare two series circuits having equal number of light bulbs.
1st circuit is powered by 6 batteries each having a voltage of 1.5V
2nd circuit is powered by a single battery having a voltage of 9V.
The six batteries in the 1st circuit can be connected together in series or in parallel.
When the batteries are connected in series (positive terminal of one battery connected to negative terminal of another battery) their voltage gets added which means
Voltage of pack = number of batteries*voltage of each battery
Voltage of pack = 6*1.5
Voltage of pack = 9 volts
But the current remains same in the series connection since there is only path for the current to flow.
On the other hand, when the batteries are connected in parallel, the voltage remains same but the current increases.
Circuit 1:
In this circuit, we have 6 batteries each of 1.5 volts connected in series to provide a voltage of 9 volts.
We have connected 2 bulbs in this series circuit.
The voltage will be equally divided between two bulbs if both bulbs are identical in construction.
So there will be 4.5 volts across each bulb and both bulbs will have same brightness.
Circuit 2:
In this circuit, we have 1 battery which provide a voltage of 9 volts.
We have connected 2 bulbs in this series circuit just like in circuit 1.
The voltage will be equally divided between two bulbs if both bulbs are identical in construction.
So there will be 4.5 volts across each bulb and both bulbs will have same brightness.
Conclusion:
Both series circuits provide a total voltage of 9 volts to the two bulbs connected in series and the voltage will be equally divided among two bulbs and they will have same brightness. Therefore, both circuits will have same characteristics.
