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2 years ago

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Two speedboats are traveling at the same speed relative to the water in opposite directions in a moving river. An observer on th

e riverbank sees the boats moving at 4.9 m/s and 7.0 m/s.(a) What is the speed of the boats relative to the river? (b) How fast is the river moving relative to the shore?

1 answer:

DENIUS [597]2 years ago

3 0

Answer:

a) vboat = 5.95 m/s b) vriver= 1.05 m/s

Explanation:

a) As observed from the shore, the speed of the boats can be expressed as the vector sum, of the boat speed relative to the water and the river speed relative to the shore, as follows:

vb₁s = vb₁w + vrs

In one case, the boat is moving in the same direction as the water:

vb₁s = vb₁w + vrs = 7.0 m/s (1)

For the other boat, it is clear that is moving in an opposite direction:

vb₂s = vb₂w - vrs = 4.9 m/s (2)

As we know that vb₁w = vb₂w, adding both sides, we can remove the river speed from the equation, as follows:

vb₁w = vb₂w = $\frac{7.0 m/s + 4.9 m/s}{2} =5.95 m/s$

b) Replacing this value in (1) and solving for vriver, we have:

vriver = 7.0 m/s - 5.95 m/s = 1.05 m/s

(we could have arrived to the same result subtracting both sides in (1), and (2))

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A glass jar which is kept in freezing mixture but is in broken because

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Answer:

the glass contains air bubbles that expands and contracts as the glass is heated or froze. when they expand they may cause the glass to break or even explode

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2 years ago

A torus is formed when a circle of radius 3 centered at (5 comma 0 )is revolved about the y-axis. a. Use the shell method to wr

arlik [135]

Answer:

a) $V=4\pi\int\limits^8_2 {x\sqrt{9-(x-5)^{2}}} \, dx$

b) $V=20\pi\int\limits^3_{-3} {\sqrt{9-y^{2}}} \, dy$

c) $V=90\pi ^{2}$

Explanation

In order to solve these problems, we must start by sketching a drawing of what the graph of the problem looks like, this will help us analyze the drawing better and take have a better understanding of the problem (see attached pictures).

a)

On part A we must build an integral for the volume of the torus by using the shell method. The shell method formula looks like this:

$V=\int\limits^a_b {2\pi r y } \, dr$

Where r is the radius of the shell, y is the height of the shell and dr is the width of the wall of the shell.

So in this case, r=x so dr=dx.

y is given by the equation of the circle of radius 3 centered at (5,0) which is:

$(x-5)^{2}+y^{2}=9$

when solving for y we get that:

$y=\sqrt{9-(x-5)^{2}}$

we can now plug all these values into the shell method formula, so we get:

$V=\int\limits^8_2 {2\pi x \sqrt{9-(x-5)^{2}} } \, dx$

now there is a twist to this problem since that will be the formula for half a torus.Luckily for us the circle is symmetric about the x-axis, so we can just multiply this integral by 2 to get the whole volume of the torus, so the whole integral is:

$V=\int\limits^8_2 {4\pi x \sqrt{9-(x-5)^{2}} } \, dx$

we can take the constants out of the integral sign so we get the final answer to be:

$V=4\pi\int\limits^8_2 {x\sqrt{9-(x-5)^{2}}} \, dx$

b)

Now we need to build an integral equation of the torus by using the washer method. In this case the formula for the washer method looks like this:

$V=\int\limits^b_a{\pi(R^{2}-r^{2})} \, dy$

where R is the outer radius of the washer and r is the inner radius of the washer and dy is the width of the washer.

In this case both R and r are given by the x-equation of the circle. We start with the equation of the circle:

$(x-5)^{2}+y^{2}=9$

when solving for x we get that:

$x=\sqrt{9-y^{2}}+5$

the same thing happens here, the square root can either give you a positive or a negative value, so that will determine the difference between R and r, so we get that:

$R=\sqrt{9-y^{2}}+5$

and

$r=-\sqrt{9-y^{2}}+5$

we can now plug these into the volume formula:

$V=\pi \int\limits^3_{-3}{(5+\sqrt{9-y^{2}})^{2}-(5-\sqrt{9-y^{2}})^{2}} \, dy$

This can be simplified by expanding the perfect squares and when eliminating like terms we end up with:

$V=20\pi\int\limits^3_{-3} {\sqrt{9-y^{2}}} \, dy$

c) We are going to solve the integral we got by using the washer method for it to be easier for us to solve, so let's take the integral:

$V=20\pi\int\limits^3_{-3} {\sqrt{9-y^{2}}} \, dy$

This integral can be solved by using trigonometric substitution so first we set:

$y=3 sin \theta$

which means that:

$dy=3 cos \theta d\theta$

from this, we also know that:

$\theta=sin^{-1}(\frac{y}{3})$

so we can set the new limits of integration to be:

$\theta_{1}=sin^{-1}(\frac{-3}{3})$

$\theta_{1}=-\frac{\pi}{2}$

and

$\theta_{2}=sin^{-1}(\frac{3}{3})$

$\theta_{2}=\frac{\pi}{2}$

so we can rewrite our integral:

$V=20\pi\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}} {\sqrt{9-(3 sin \theta)^{2}}} \, 3 cos \theta d\theta$

which simplifies to:

$V=60\pi\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}} {(\sqrt{9-(3 sin \theta)^{2}}} \, cos \theta d\theta$

we can further simplify this integral like this:

$V=60\pi\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}} {(\sqrt{9-9 sin^{2} \theta}}} \, cos \theta d\theta$

$V=60\pi\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}} {3(\sqrt{1- sin^{2} \theta})}} \, cos \theta d\theta$

$V=180\pi\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}} {(\sqrt{1- sin^{2} \theta})}} \, cos \theta d\theta$

$V=180\pi\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}} {(\sqrt{cos^{2} \theta})}} \, cos \theta d\theta$

$V=180\pi\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}} {(cos \theta})} \, cos \theta d\theta$

$V=180\pi\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}} {cos^{2} \theta}} \, d\theta$

We can use trigonometric identities to simplify this so we get:

$V=180\pi\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}} {\frac{1+cos 2\theta}{2}}} \, d\theta$

$V=90\pi\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}} {1+cos 2\theta}}} \, d\theta$

we can solve this by using u-substitution so we get:

$u=2\theta$

$du=2d\theta$

and:

$u_{1}=2(-\frac{\pi}{2})=-\pi$

$u_{2}=2(\frac{\pi}{2})=\pi$

so when substituting we get that:

$V=45\pi\int\limits^{\pi}_{-\pi} {1+cos u}} \, du$

when integrating we get that:

$V=45\pi(u+sin u)\limit^{\pi}_{-\pi}$

when evaluating we get that:

$V=45\pi[(\pi+0)-(-\pi+0)]$

which yields:

$V=90\pi ^{2}$

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2 years ago

The final position minus the initial position is the

ycow [4]

Answer:

Displacement.

Explanation:

The change in between the starting position (initial position) and your final position is the amount of distance you <em>actually </em>traveled.

For example, you can use a triangle. The distance you travel for side length a & b is equal to the distance you travel on side c (the hypotenuse). You are pretty much solving for the hypotenuse in this case, or the displacement.

~

6 0

3 years ago

Read 2 more answers

Calculate the acceleration of a bottle rocket that is traveling at a speed of 74 m/s, and then descends at a speed of 89 m/s. It

Harlamova29_29 [7]

Answer:Falling objects form an interesting class of motion problems. For example, we can estimate the depth of a vertical mine shaft by dropping a rock into it and listening for the rock to hit the bottom. By applying the kinematics developed so far to falling objects, we can examine some interesting situations and learn much about gravity in the process.

Explanation:

5 0

2 years ago

A wooden object (conically shaped) has a diameter of 8cm and height of 14cm. It floats in oil with 6cm of its height above oil l

baherus [9]

Answer:

(a) The density of the object is 316/343 × the density of the oil

(b) The fraction of oil displaced after immersing the object is 0.461 of the oil volume

Explanation:

(a) The volume, V of a cone of height, h and base diameter, D = 2×r is given by the following equation;

$V = \dfrac{\pi r^{2} h}{3}$

The volume of the object is therefore;

$\dfrac{\pi \times 4^{2} \times 14}{3} = 74\tfrac{2}{3}\pi \, cm^3$

Where 6 cm is above the oil level we have;

$\dfrac{\pi \times \left (6 \times \dfrac{4}{14} \right )^{2} \times 6}{3} = 5\tfrac{43}{49}\pi \, cm^3$ above the oil level

Therefore, volume of the oil displaced = $68\tfrac{116}{147}\pi$ cm³ = 216.11 cm³

The density of the object is thus;

$\dfrac{68\tfrac{116}{147}\pi}{ 74\tfrac{2}{3}\pi} \times Density \ of \ the \ oil = \dfrac{316}{343} \right ) \times Density \ of \ the \ oil$

The density of the object = 316/343 × the density of the oil.

(b) The volume of the oil = 2 × Volume of the object = $2 \times 74\tfrac{2}{3}\pi \, cm^3 = 149\tfrac{1}{3}\pi \, cm^3$

The fraction of the volume displaced, x, after immersing the object is given as follows;

$x = \dfrac{68\tfrac{116}{147}\pi}{ 149\tfrac{1}{3}\pi} = \dfrac{158}{343} = 0.461$

The fraction of oil displaced after immersing the object = 0.461 of the volume of the oil

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3 years ago