Initialize the tuple team_names with the strings 'Rockets', 'Raptors', 'Warriors', and 'Celtics' (The top-4 2018 NBA teams at th
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Answer:
Work input =283.47 KJ
Explanation:
Given that
T=12°C=285 K
m= 2.5 kg
Given that this is the constant temperature process.
e know that work for isothermal process
So now putting the values
W=-283.47 KJ
Negative sign indicates that work is done on the system.
So work input =283.47 KJ
Answer:
-4.5 m/s
Explanation:
Assuming steady and incompressible flow and uniform properties at each section
Here V is velocity of flow and A is area, Q is flow rate out of the leak, subscript 1-4 represent different sections
At the surface, is negative hence the equation above will be
Making the subject of the formula then
Substituting the given values then
Answer:
hello your question lacks the required figures here is the complete question
A 1.25-in diameter shaft contains a 0.25-in deep U-shaped groove that has a 1/8-in radius at the bottom of the groove. The maximum shear stress in the shaft must be limited to 12000 psi . If the shaft rotates at a constant angular speed of 6Hz , determine the maximum power that may be delivered by the shaft.
Answer: max power delivered by shaft = 4.045 hp
Explanation:
Determine The maximum power that can be delivered by the shaft
using the given data
diameter of shaft ( D ) = 1.25 inches
depth of U-shaped groove = 0.25 inches
radius of U-shaped groove = 1/8 inches = 0.125 inches
maximum shear stress in shaft = 12000 psi
shaft angular speed at frequency of 6 Hz
firstly calculate
The minor diameter (d) = 1.25 - 2(0.25 ) = 0.5 inches
Ratio = radius of groove / minor diameter = 0.125 / 0.75 = 0.167
Ratio, = diameter of shaft / minor diameter = 1.25 / 0.75 = 1.667
k = 1.39 from stress concentration factors graph
calculate the maximum shear stress produced by the torque in the minor diameter of the shaft
<em>Tmax</em> = -----------equation 1
where Tc = 16T
J =
equation 1 becomes( Tmax ) =
also <em>Tmax</em> = K * <em>Tmin -------- equation 2</em>
<em> </em> 1.39 *
T ≤ 715.122 Ib-in
Tmax = 59.593 Ib-ft ( max shear stress )
Finally calculate the max power transmitted by the shaft
P max = 2fTmax = 2
* 6 * 59.593
therefore Pmax = 2246.6 Ib-ft/s
= 4.045 hp
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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.
