Ask question

Ask question

All categories

2 years ago

8

Question 5 of 20 Engineers are using computer models to study train collisions to design safer train cars. They start by modelin

g an elastic collision between two train cars traveling toward each other. Car 1 is traveling south at 18 m/s and has a full load, giving it a total mass of 14,650 kg. Car 2 is traveling north at 11 m/s and has a mass of 3,825 kg. After the collision car 1 has a final velocity of 6 m/s south. What is the final velocity of car 2? A. 81 m/s north O B. 81 m/s south C. 35 m/s south D. 35 m/s north

2 answers:

Sonbull [250]2 years ago

8 0

Answer:

It's C

Explanation:

Your welcome!

aniked [119]2 years ago

6 0

Answer:

35m/s south

Explanation:

You might be interested in

Tyler drives 50km north. Tyler then drives back 30km south. What distance did he cover? What was his displacement?

Luda [366]

Answer:

he covered 80km his displacement was 20km

Explanation:

displacement is the distance from the starting point so in this case its 20 (50-30) and total distance covered is how many kilometers he drove in total

7 0

2 years ago

When the displacement in SHM is equal to 1/3 of the amplitude xm, what fraction of the total energy is (a) kinetic energy and (b

Nesterboy [21]

Answer:

Explanation:

Given

Displacement is $\frac{1}{3}$ of Amplitude

i.e. $x=\frac{A}{3}$ , where A is maximum amplitude

Potential Energy is given by

$U=\frac{1}{2}kx^2$

$U=\frac{1}{2}k(\frac{A}{3})^2$

$U=\frac{1}{18}kA^2$

Total Energy of SHM is given by

$T.E.=\frac{1}{2}kA^2$

Total Energy=kinetic Energy+Potential Energy

$K.E.=\frac{1}{2}kA^2 -\frac{1}{18}kA^2$

$K.E.=\frac{8}{18}kA^2$

Potential Energy is $\frac{1}{8}$ th of Total Energy

Kinetic Energy is $\frac{8}{9}$ of Total Energy

(c)Kinetic Energy is $0.5\times \frac{1}{2}kA^2$

$P.E.=\frac{1}{4}kA^2$

$\frac{1}{2}kx^2=\frac{1}{4}kA^2$

$x=\frac{A}{\sqrt{2}}$

7 0

3 years ago

Read 2 more answers

madam [21]

The Correct answer to number 1 is A or D

The correct answer to number 2 is C because transmission meaning is to travel.

PS. I think number 1 is D

3 0

3 years ago

I WILL GIVE BRAINLIEST 50!!!!!

Inessa05 [86]

Transfer of heat through objects touching source

4 0

3 years ago

Read 2 more answers

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