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2 years 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:

Lunna [17]2 years ago

6 0

Answer:

The change in power is 4400 W.

Explanation:

Given that,

Power = 10 kW

Speed = 10 m/s

Increases speed = 12 m/s

Given equation is,

$F=kv$

We know that,

The power is,

$P=Fv$

Put the value of F into the formula

$P=(kv)v$

$P=kv^2$

$P\propto v^2$

We need to calculate the new power

Using formula for power

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

Put the value into the formula

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

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

$P'=14.4\ kW$

We need to calculate the change in power

Using formula of change in power

$\Delta P=P'-P$

Put the value into the formula

$\Delta P=14.4-10$

$\Delta P=4.4\ kW$

$\Delta P=4.4\times1000$

$\Delta P=4400\ W$

Hence, The change in power is 4400 W.

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1 year ago

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The closest answer would be C.
The choices given do not give the exact value.

To answer this, you just need to remember the main formula:

$d = Vit + \frac{1}{2}gt^{2}$

Where:
d = distance/displacement
g = acceleration due to gravity
t = time in flight
Vi = initial velocity.

With this formula, you derive all the formulas you need to look for certain components. You need to keep in mind of the following:

---if you are looking for a vertical component(y), you need to use values of vertical motion.

$dy = Viyt + \frac{1}{2}gt^{2}$

*Viy is always 0m/s at the beginning of a free-fall.

$dy = (0m/s)t + \frac{1}{2}gt^{2}$

$dy = \frac{1}{2}gt^{2}$

---If you are looking for a horizontal component(x), you need to use values of horizontal motion.

$dx = Vixt + \frac{1}{2}gt^{2}$

*g is always 0m/s² when taking horizontal motion into account.

$dx = Vixt + \frac{1}{2}(0m/s^{2})t^{2}$

$dx = Vixt$
--- time is the only value that is both vertical and horizontal.

Okay, let's get back to solving your problem. Let's see what your given is first:
dy = 137m (as long as it refers to height, it is vertical distance)
dx = 4km (the word, far or away usually indicates horizontal distance)
g = 1.63m/s²

The question is how fast was it going horizontally and we can derive it from our equation:

$dx = Vixt$

We use this because x means horizontal. But notice that we do not have time yet. So how are we going to solve this 2 variables missing? The key is that time is a horizontal and vertical component. Whatever time it took moving horizontally, it is the same vertically as well. So we use the vertical formula to derive time:

$dy = \frac{1}{2}gt^{2}$
$\frac{2dy}{g}=t^{2}$
$\sqrt{\frac{2dy}{g}} = \sqrt{t^{2}}$
$\sqrt{\frac{2dy}{g}} = t$

Now plug in what you know and solve for what you don't know:

$\sqrt{\frac{2(137m)}{1.63m/s^{2}}} = t$
$\sqrt{\frac{274m}{1.63m/s^{2}}} = t$
$\sqrt{168.098s^{2}}=t$
$12.965s = t$

The total time in flight is 12.965s.
Let's round it off to 13s.

Now that we know that, we can use this in the horizontal formula:
$dx = Vixt$
$4km = Vix(13s)$

Hold up! Look at the unit of the horizontal distance. It is in km but all our units are expressed in m so we need to convert that first.

1km = 1,000m
4km = 4,000m

Our new horizontal distance is 4,000m.

Okay, let's wrap this up by solving for what is asked for, using all the derived values.
$dx = Vixt$
$4,000m = Vix(13s)$
$\frac{4,000m}{13s} = Vix$
$308m/s= Vix$

The horizontal velocity is 308m/s.

Such a long explanation I know, but hopefully, you learned from it.

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1 year ago