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2 years ago

13

A bicyclist with a mass of 50 kg is traveling at a rate of 30 m/s. It accelerates to a rate of 50 m/s in 5 seconds. What is the

net force acting on the bicycle? *

1 answer:

balandron [24]2 years ago

8 0

Answer:

F=m*(v^2/r)

F=82*(8^2/30)

F=174.9N

Explanation:

brainlest pls

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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

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3 years ago

What is neccessary for a magnetic field to create electric current in a copper coil?

Viktor [21]

A, Lenz' Law. There need to be a difference of flux, so if you use AC you will get a current too.

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3 years ago

A skier moving at 4.75 m/s encounters a long, rough, horizontal patch of snow having a coefficient of kinetic friction of 0.220

disa [49]

First we need to find the acceleration of the skier on the rough patch of snow.
We are only concerned with the horizontal direction, since the skier is moving in this direction, so we can neglect forces that do not act in this direction. So we have only one horizontal force acting on the skier: the frictional force, $\mu m g$ . For Newton's second law, the resultant of the forces acting on the skier must be equal to ma (mass per acceleration), so we can write:
$ma=-\mu m g$
Where the negative sign is due to the fact the friction is directed against the motion of the skier.
Simplifying and solving, we find the value of the acceleration:
$a=-(0.220)(9.81 m/s^2)=-2.16 m/s^2$

Now we can use the following relationship to find the distance covered by the skier before stopping, S:
$2aS=v_f^2-v_i^2$
where $v_f=0$ is the final speed of the skier and $v_i=4.75 m/s$ is the initial speed. Substituting numbers, we find:
$S=- \frac{v_i^2}{2a}=- \frac{(4.75 m/s)^2}{2(-2.16 m/s^2)}=5.23 m$

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3 years ago

Which is the BEST description of how eyeglasses work?

Sonbull [250]

Answer:

C I think.

Explanation:

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3 years ago

A block is attached to the end of a horizontal ideal spring and rests on a frictionless surface. The block is pulled so that the

bezimeni [28]

Answer:

The answer is "a, c and b"

Explanation:

Its total block power is equal to the amount of potential energy and kinetic energy.
Because the original block expansion in all situations will be the same, its potential power in all cases is the same.
Because the block in the first case has no initial speed, the block has zero film energy.
For both the second example, it also has the $v_o$ velocity, but the kinetic energy is higher among the three because its potential and kinetic energy are higher.
While over the last case the kinetic speed is greater and lower than in the first case, the total energy is also higher than the first lower than that of the second.
The greater the amplitude was its greater the total energy, therefore lower the second, during the first case the higher the amplitude.

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2 years ago