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Mama L [17]
4 years ago
14

A cylindrical shell of radius 7.00 cm and length 2.21 m has its charge uniformly distributed on its curved surface. The magnitud

e of the electric field at a point 15.2 cm radially outward from its axis (measured from the midpoint of the shell) is 36.0 kN/C. (a) Find the net charge on the shell.

Physics
1 answer:
polet [3.4K]4 years ago
8 0

Answer:

The net charge on the shell is 30x10^-9C

Explanation:

Pls see attached file

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According to the answer, the speed of light is being achieved.

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The velocity of the light is:

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Where E = energy = 3.03x10⁻¹⁹J

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Replacing:

C=\frac{3.03x10^{-19} }{1.01x10^{-27} } =3x10^{8} m/s

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A lithium oxide compound is represented by which formula?
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A graph titled Position versus time for with horizontal axis time (seconds) and vertical axis position (meters). A straight blue
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two forces are applied on a soccer ball. Which of the following statements explains why the ball acellerates
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3 years ago
A 2.00 kg block situated on a rough incline is connected to aspring
garri49 [273]

Figure P5.76 is missing from this problem, but I found a similar problem whose figure I uploaded with this answer. The angle of the incline might be different though, but the procedure to follow is the same.

Answer:

\mu_{k}=0.0708

Explanation:

In order to solve this problem we must first do a drawing of the situation so we can analyze it better. (See attached picture)

Once we have the drawing of the problem, we can go ahead and draw a free body diagram, which will help us determine what forces are acting upon the object. There are different approaches we can take to solve this problem, but I will use an energy balance to do so.

So first we do a sum of forces on the y-axis. This is for us to find what the Normal force is, which will be used to find the force of friction, so we get:

\sum F_{y}=0

by looking at the free body diagram we get the sum of forces to be:

N-W_{y}=0

when solving for the normal force we get that:

N=W_{y}

we also know from the free body diagram that:

W_{y}=Wcos \theta

and

W_{y}=mgcos \theta

So:

N=mgcos \theta

we can also determine the height of the block at the time it is released by analyzing the triangle find in the uploaded figure,, so we get that:

sin(37^{o})=\frac{h}{21.4cm}

so:

h=21.4cm sin(37°)

h=12.88cm

Once we got this, we can go ahead and do an energy balance on the system, so we get that:

U_{0}+K_{0}+E_{s0}=U_{f}+K_{f}+W_{f}+E_{sf}

where:

U=potential energy

K=Kinetic energy

W_{f}=Work of friction

E_{s}=Potential energy of the spring

Since the object is released from rest we know the initial kinetic energy is zero, just like the initial potential energy of the spring since when the block is released, the spring is unstretched. We also know the final potential energy of the block is zero because it reached its lowest point, while the kinetic energy of the block is also zero because it came to rest at that point. This simplifies our energy balance so we get:

U_{0}=W_{f}+E_{sf}

we can now determine each part of the equation so:

U_{0}=mgh

W_{f}=fx

we know that friction is given gy the equation:

f=\mu_{k}N

and that

N=mgcos \theta

so:

f=\mu_{k}mgcos \theta

and

W_{f}=\mu_{k}mgxcos \theta

and finally:

E_{s}=\frac{1}{2}kx^{2}

Once we got all these equations we can substitute them into our balance of energy, so we get:

mgh=\mu_{k}mgx cos \theta + \frac{1}{2}kx^{2}

we can now solve this for \mu_{k} so we get:

mgh - \frac{1}{2}kx^{2}=\mu_{k}mgx cos \theta

and:

\mu_{k}=\frac{2mgh-kx^{2}}{2mgxcos\theta}

now we can substitute the given data, make sure to use the correct units:

\mu_{k}=\frac{2(2kg)(9.81m/s^{2})(12.88x10^{-2}m)-(100N/m)(21.4x10^{-2})^{2}}{2(2kg)(9.81m/s^{2})(21.4x10^{-2})cos(37^{o})}

which solves to:

\mu_{k}=0.0708

5 0
4 years ago
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