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likoan [24]
3 years ago
8

A boat is able to move through still water at 20 m/s. It makes a round trip to a town 10 km upstream. If the river flows at 5m/s

, what is the time required for this round trip? (give your answer in seconds)​
Physics
1 answer:
Nataly_w [17]3 years ago
7 0

Answer:

The time required for this round trip is 1066.66 s

Explanation:

Given;

velocity of boat through still water, v₁ = 20 m/s

distance moved by the boat, d = 10 km = 10,000 m

velocity of the river, v₂ = 5 m/s

time = d / v

The boat made two journeys which formed the round trip.

During the first part of the journey, the boat moves upstream in opposite direction to the flow of the river and the resultant velocity is calculated as;

Resultant velocity = 20 - 5 = 15 m/s

Time for this journey, = 10000/ 15 = 666.66 s

During the second part of the journey, the boat moves downstream in the same direction to the flow of the river and the resultant velocity is calculated as;

Resultant velocity = 20 + 5 = 25 m/s

Time for this journey, = 10000/ 25 = 400 s

Total time for the trip = 666.66 s + 400 s = 1066.66 s

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Let's calculate first the electric field generated by the positive charge at that point:
E_1=k_e  \frac{Q_1}{r^2}=(8.99 \cdot 10^9 Nm^2C^{-2}) \frac{(2.0 \cdot 10^{-6} C)}{(0.05 m)^2} =+7.19 \cdot 10^6 N/C
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while the electric field generated by the negative charge is:
E_2=k_e \frac{Q_1}{r^2}=(8.99 \cdot 10^9 Nm^2C^{-2}) \frac{(-2.0 \cdot 10^{-6} C)}{(0.05 m)^2} =-7.19 \cdot 10^6 N/C
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If we assume that the positive charge is on the left and the negative charge is on the right, we see that E1 is directed to the right, and E2 is directed to the right as well. This means that the net electric field at the midpoint between the two charges is just the sum of the two fields:
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