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2 years ago

The faster am igneous rock cools the_ the crystal size

Physics

1 answer:

HACTEHA [7]2 years ago

7 0

Bigger is the correct answer. the faster an igneous rock cools the bigger the bigger the crystal size will be

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1.(16 pts.) Find the volume of the solid obtained by revolving the region enclosed by y = xex , y = 0 and x = 1 about the x-axis

MrRa [10]

Answer:

<em>The Volume is 5.018 cubic units</em>

Explanation:

<u>Volume Of A Solid Of Revolution</u>

Let f(x) be a continuous function defined in an interval [a,b], if we take the area enclosed by f(x) between x=a, x=b and revolve it around the x-axis, we get a solid whose volume can be computed as

$\displaystyle V=\pi \int_a^bf^2(x)dx$

It's called the disk method. There are other available methods to compute the volume.

We have

$f(x)=xe^x$

And the boundaries defined as x=1, y=0 and revolved around the x-axis. The left endpoint of the integral is easily identified as x=0, because it defines the beginning of the region to revolve. So we need to compute

$\displaystyle V=\pi \int_0^1(xe^x)^2dx=\pi \int_0^1x^2e^{2x}dx$

We need to first determine the antiderivative

$\displaystyle I=\int x^2e^{2x}dx$

Let's integrate by parts using the formula

$\displaystyle \int u.dv=u.v-\int v.du$

We pick $u=x^2,\ dv=e^{2x}dx$

Then $du=2xdx,\ v=\frac{e^{2x}}{2}$

Applying by parts:

$\displaystyle I=x^2\frac{e^{2x}}{2}-\int 2x\frac{e^{2x}}{2}dx$

$\displaystyle I=\frac{x^2e^{2x}}{2}-\int xe^{2x}dx$

Now we solve

$\displaystyle I_1=\int xe^{2x}dx$

Making $u=x,\ dv=e^{2x}dx$

$\displaystyle du=dx,\ v=\frac{e^{2x}}{2}$

Applying by parts again:

$\displaystyle I_1=x\frac{e^{2x}}{2}-\int \frac{e^{2x}}{2}dx$

$\displaystyle I_1=\frac{xe^{2x}}{2}-\frac{1}{2}\int e^{2x}dx$

The last integral is directly computed

$\displaystyle \int e^{2x}dx=\frac{e^{2x}}{2}$

Replacing every integral computed above

$\displaystyle I=\frac{x^2e^{2x}}{2}-\left(\frac{xe^{2x}}{2}-\frac{1}{2}\frac{e^{2x}}{2}\right)$

Simplifying

$\displaystyle I=\dfrac{\left(2x^2-2x+1\right)\mathrm{e}^{2x}}{4}$

Now we compute the definite integral as the volume

$V=\pi \left[\dfrac{\left(2(1)^2-2(1)+1\right)\mathrm{e}^{2(1)}-\left(2(0)^2-2(0)+1\right)\mathrm{e}^{2(0)}}{4}\right]$

Finally

$V=\pi \dfrac{\mathrm{e}^2-1}{4}=5.018$

The Volume is 5.018 cubic units

8 0

3 years ago

How much energy becomes unavailable for work in an isothermal process at 440 k, if the entropy increase is 25 j/k?

just olya [345]

Answer: 11,000 J

Explanation:

In an isothermal process,

$\text{entropy increase} = \frac{\text{amount of energy in heat transfer}}{\text{temperature} }$ (1)

Note that, the energy used in heat transfer is not available for work. So, the amount of energy unavailable for work is equal to the energy used in heat transfer.

To obtain the amount of energy in heat transfer, we multiply both sides of equation (1) by the denominator of the right side of (1) so that

amount of energy in heat transfer = (entropy increase)(temperature)
= (25 J/K)(440 K)
= 11,000 J

Since the amount of energy unavailable for work is equal to the amount of energy in the heat transfer, therefore the amount of energy unavailable for work is 11,000 J.

8 0

3 years ago

Help me with this plzzz

Marysya12 [62]

Answer:

yeah I'm Pretty sure it's b

4 0

3 years ago

Read 2 more answers

What is the equation of the line that is perpendicular to y= -4x +5 and passes through the point (4,-3)?

Ostrovityanka [42]

Y=4x-19

Perpendicular because it has an opposite reciprocal slope