A ferry is travelling at 10 m/s relative to the river towards the south. The river is flowing at 2 m/s relative to

Physics

1 answer:

mixer [17]4 years ago

8 0

Answer:

11 m/s south

Explanation:

The velocity of the passenger relative to the river bank is equal to the velocity of the passenger relative to the ferry, plus the velocity of the ferry relative to the river, plus the velocity of the river relative to the river bank.

v_passenger,bank = v_passenger,ferry + v_ferry,river + v_river,bank

If we take north to be positive and south to be negative:

v = 1.0 m/s + (-10 m/s) + (-2 m/s)

v = -11 m/s

v = 11 m/s south

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What is the work done when a 400.n force is use to lift a 400.n object 3.5 meters strait up

alexandr1967 [171]

Answer:

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Explanation:

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3 years ago

A 15.0 Ohms resistor is connected in series to a 120V generator and two 10.0 Ohms resistors that are connected in parallel to ea

Nostrana [21]

Hi there! :)

Reference the diagram below for clarification.

We must begin by knowing the following rules for resistors in series and parallel.

In series:
$R_T = R_1 + R_2 + ... + R_n$

In parallel:
$\frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2} + ... + \frac{1}{R_n}$

We can begin solving for the equivalent resistance of the two resistors in parallel using the parallel rules.

$\frac{1}{R_{T, parallel}} = \frac{1}{10} + \frac{1}{10}\\\\\frac{1}{R_{T, parallel}} = \frac{2}{10} = \frac{1}{5}\\\\R_{T, parallel} = 5\Omega$

Now that we have reduced the parallel resistors to a 'single' resistor, we can add their equivalent resistance with the other resistor in parallel (15 Ohm) using series rules:
$R_T = 15 + 5\\\\\boxed{R_T = 20 \Omega}$

We can use Ohm's law to solve for the current in the circuit.

$i = \frac{V}{R_T}\\\\i = \frac{120}{20} = \boxed{6 A}$

For resistors in series, both resistors receive the SAME current.

Therefore, the 15Ω resistor receives 6A, and the parallel COMBO (not each individual resistor, but the 5Ω equivalent when combined) receives 6A.

In this instance, since both of the resistors in parallel are equal, the current is SPLIT EQUALLY between the two. (Current in parallel ADDS UP). Therefore, an even split between 2 resistors of 6 A is <u>3A for each 10Ω resistor</u>.

Since the 15.0 Ω resistor receives 6A, we can use Ohm's Law to solve for voltage.

$V = iR\\\\V = (6)(15) = \boxed{90 V}$