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9 days ago

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The thrust of a certain boat’s engine generates a power of 10kW as the boat moves at constant speed 10ms through the water of a

lake. The magnitude of the drag force that is exerted on the boat’s hull as it is moving through the water is directly proportional to the boat's speed and is given by the equation F=kv. The increase in power needed for the boat to move through the lake at a constant speed of 12ms is : a. 0W b. 1440 W c. 4400 W 10,000 W e. 4,400 W

1 answer:

ValentinkaMS [1.1K]9 days ago

6 0

Answer:

The power increase is 4400 W.

Explanation:

Given data,

Power = 10 kW

Initial speed = 10 m/s

New speed = 12 m/s

The relevant equation is:

$F=kv$

It's known that,

The expression for power is,

$P=Fv$

Substituting the value of F into the formula gives

$P=(kv)v$

$P=kv^2$

$P\propto v^2$

Now we must determine the new power

By applying the power formula

$\dfrac{P}{P'}=\dfrac{v^2}{v'^2}$

Insert the value into the equation

$\dfrac{10}{P'}=(\dfrac{10}{12})^2$

$P'=(\dfrac{12}{10})^2\times10$

$P'=14.4\ kW$

Next, we calculate the change in power

Using the formula for power change

$\Delta P=P'-P$

Insert values into the equation

$\Delta P=14.4-10$

$\Delta P=4.4\ kW$

$\Delta P=4.4\times1000$

$\Delta P=4400\ W$

Thus, the change in power amounts to 4400 W.

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