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Arturiano [62]
3 years ago
5

a. To measure the water current in an ocean, a marker is dropped onto it. Determine if the trajectory traced by the drifting mar

ker forms a streamline, streakline, or pathline.b. To learn more about the flow, a large number of similar markers are released at the same point in space in quick succession. Determine if the line connecting these markers at a later instant forms a streamline, streakline, or pathline.

Engineering
1 answer:
ValentinkaMS [17]3 years ago
5 0

Answer:

A) The trajectory traced by the drifting marker form a pathline. The pathline is the trajectory than a single particle does in the fluid (like in this case) along with all its travel.

b) The line connecting these markers released at the same point in space in quick succession is a Streakline. The streaklines are the lines formed by the loci of points in the fluid that have continuously pass throw particulars points in space in the past. The main difference between the pathline and the streakline is that the first one is not affected by flow changes over time, while the second one is. The Pathline is a movie that shows the travel of a single particle in time. The streakline is movie that shows how the particles change their path along time.

In the gif attached to this answer, you can see in red the pathline of particle, in blue the streakline of several particles released at the same point in space in quick succession and in grey the streamline.

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Consider the circuit below where R1 = R4 = 5 Ohms, R2 = R3 = 10 Ohms, Vs1 = 9V, and Vs2 = 6V. Use superposition to solve for the
VladimirAG [237]

Answer:

The value of v2 in each case is:

A) V2=3v for only Vs1

B) V2=2v for only Vs2

C) V2=5v for both Vs1 and Vs2

Explanation:

In the attached graphic we draw the currents in the circuit. If we consider only one of the batteries, we can consider the other shorted.

Also, what the problem asks is the value V2 in each case, where:

V_2=I_2R_2=V_{ab}

If we use superposition, we passivate a battery and consider the circuit affected only by the other battery.

In the first case we can use an equivalent resistance between R2 and R3:

V_{ab}'=I_1'R_{2||3}=I_1'\cdot(\frac{1}{R_2}+\frac{1}{R_3})^{-1}

And

V_{S1}-I_1'R_1-I_1'R_4-I_1'R_{2||3}=0 \rightarrow I_1'=0.6A

V_{ab}'=I_1'R_{2||3}=3V=V_{2}'

In the second case we can use an equivalent resistance between R2 and (R1+R4):

V_{ab}''=I_3'R_{2||1-4}=I_3'\cdot(\frac{1}{R_2}+\frac{1}{R_1+R_4})^{-1}

And

V_{S2}-I_3'R_3-I_3'R_{2||1-4}=0 \rightarrow I_3'=0.4A

V_{ab}''=I_3'R_{2||1-4}=2V

If we consider both batteries:

V_2=I_2R_2=V_{ab}=V_{ab}'+V_{ab}''=5V

7 0
4 years ago
3.
Andreyy89

Answer:

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Explanation:

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