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3 years ago

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An elephant's legs have a reasonably uniform cross section from top to bottom, and they are quite long, pivoting high on the ani

mal's body. When an elephant moves at a walk, it uses very little energy to bring its legs forward, simply allowing them to swing like pendulums. For fluid walking motion, this time should be half the time for a complete stride; as soon as the right leg finishes swinging forward, the elephant plants the right foot and begins swinging the left leg forward.

1 answer:

xz_007 [3.2K]3 years ago

8 0

Answer:

is this the full question?

Explanation:

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A force of 100N is applied to move an object a horizontal distance of 20m to the right. The work done by this force on the objec

horsena [70]

WORKDONE = FORCE * DISPLACEMENT
W=F*S
HERE, THE FORCE = 100N AND DISTANCE = 20M
WORKDONE = 100*20
WORKDONE=2000
ITS S.I UNIT IS JOULE OR J
SO, 2000J

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3 years ago

Ray creates an energy transfer diagram for a hair dryer. However, the diagram contains an error that could be corrected in sever

Tresset [83]

If the energy transfer diagram created by Ray has error in it, it can be corrected by applying the principle of energy conservation.

<h3>What is energy diagram?</h3>

Energy diagram is sketch of the energy flow process or energy transfer or energy conversion process.

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Thus, if the energy transfer diagram created by Ray has error in it, it can be corrected by applying the principle of energy conservation.

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2 years ago

How will you define motion?

Naddik [55]

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3 years ago

When an RL circuit is connected to a battery, which is true about its initial and final states?

zubka84 [21]

Answer:A

Explanation:

In R-L circuit current is given by

$i=i_0\left [ 1-e^{\frac{-t}{L/R}}\right ]$

where i=current at any time t

$i_0=maximum\ current$

R=resistance

L=Inductance

at t=0 $e^{\frac{-t}{L/R}}$ approaches to 1

therefore $i=i_0\left [ 1-1\right ]$

i=0

when t approaches to $\infty$ , $e^{\frac{-t}{L/R}}$ approaches to zero

thus $i=i_0$

thus we can say that initially circuit act as broken wire with zero current

and it increases exponentially with time and act as ordinary connecting wire

7 0

3 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:

$V=90\pi ^{2}$

8 0

3 years ago