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meriva
1 year ago
13

Compute the velocity of an electron that has been accelerated through a difference of potential of 100 volts. express your answe

r in meters per second
Physics
1 answer:
Elodia [21]1 year ago
3 0

The velocity of an electron that has been accelerated through a difference of potential of 100 volts will be 5.93 * 10^{6} m/s

Electrons move because they get pushed by some external force. There are several energy sources that can force electrons to move. Voltage is the amount of push or pressure that is being applied to the electrons.

By conservation of energy, the kinetic energy has to equal the change in potential energy, so KE=q*V. The energy of the electron in electron-volts is numerically the same as the voltage between the plates.

given

charge of electron = 1.6 × 10^{-19} C

mass of electron  = 9.1 × 10^{-31} kg

Force in an electric field = q*E

potential energy is stored in the form of work done

potential energy = work done = Force * displacement

                                                   = q * (E * d)  

                                                   = q * (V) = 1.6 × 10^{-19} * 100

stored potential energy = kinetic energy in electric field

kinetic energy = 1/2 * m * v^{2}

                        = 1/2 *  9.1 × 10^{-31} *  v^{2}

equation both the equations

1/2 *  9.1 × 10^{-31} *  v^{2} = 1.6 × 10^{-17}

v^{2} = 0.352 * 10^{14} m/s

v^{2} = 35.2 * 10^{12}

    = 5.93 * 10^{6} m/s

To learn more about  kinetic energy in electric field  here

brainly.com/question/8666051

#SPJ4

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Answer:

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2 years ago
How do I do these? My teacher didn’t show us how.
melisa1 [442]

Explanation:

Displacement is simply the change in position.  So in the first part of problem 1, looking at the graph between 0 s and 2 s, the position changes from 0 m to -4 m.  So the displacement is:

Δx =  -4 m − 0 m

Δx = -4 m

Between 2 s and 4 s, the position stays at -4 m.  The displacement is:

Δx = -4 m − (-4 m)

Δx = 0 m

Finally, between 4 s and 6 s, the position goes from -4 m to 6 m.  The displacement is:

Δx = 6 m − (-4 m)

Δx = 10 m

The net displacement is the change in position from 0 s to 6 s:

Δx = 6 m − 0 m

Δx = 6 m

In the second part of problem 1, we have a velocity vs time graph.

Car 1 starts with 0 velocity and ends with a velocity of 6 m/s, so it is accelerating and constantly moving to the right.

Car 2 starts with a velocity of -6 m/s and ends with a velocity of 6 m/s.  It is also accelerating, but first it is moving to the left, comes to a stop at t = 3 s, then moves to the right.

Car 3 starts with a velocity of 2 m/s and ends with a velocity of 2 m/s.  So it is moving constantly to the right, but never speeds up or slows down.

We want to know when two of the cars meet.  Unfortunately, this isn't as easy as looking for where the lines cross on the graph.  We need to calculate their displacements.  We can do this by finding the area under the graph (assuming all the cars start from the same point).

Let's start with Car 2.  Half of the area is below the x-axis, and half is above.  Without doing calculations, we can say the total displacement for this car is 0.  This means it ends back up where it started, and that it never meets either of the other cars, both of which have positive displacements.

So we know Car 1 and Car 3 meet, we just have to find where and when.  For Car 1, the area under the curve is a triangle.  So its displacement is:

Δx = ½ t v(t)

where t is the time and v(t) is the velocity of Car 1 at that time.  Since the line has a slope of 1 and y intercept of 0, we know v(t) = t.  So:

Δx = ½ t²

Now look at Car 3.  The area under the curve is a rectangle.  So its displacement is:

Δx = 2t

When the two cars have the same displacement:

½ t² = 2t

t² = 4t

t² − 4t = 0

t (t − 4) = 0

t = 0, 4

t = 0 refers to the time when both cars are at the starting point, so t = 4 is the answer we're looking for.  Where are the cars at this time?  Simply plug in t = 4 into either of the equations we found:

Δx = 8

So Cars 1 and 3 meet at 4 s and 8 m.

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Is it possible for an object to have a charge of 4.8x10-9 C? Why or why not?
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A capacitor is formed from two concentric spherical conducting shells separated by vacuum. The inner sphere has radius 11.0 cm ,
viktelen [127]
Part A)
First of all, let's convert the radii of the inner and the outer sphere:
r_A = 11.0 cm = 0.110 m
r_B = 16.5 cm=0.165 m
The capacitance of a spherical capacitor which consist of two shells with radius rA and rB is
C=4 \pi \epsilon _0  \frac{r_A r_B}{r_B- r_A}=4\pi(8.85 \cdot 10^{-12}C^2m^{-2}N^{-1}) \frac{(0.110m)(0.165m)}{0.165m-0.110m}=
=3.67\cdot 10^{-11}F

Then, from the usual relationship between capacitance and voltage, we can find the charge Q on each sphere of the capacitor:
Q=CV=(3.67\cdot 10^{-11}F)(100 V)=3.67\cdot 10^{-9}C

Now, we can find the electric field at any point r located between the two spheres, by using Gauss theorem:
E\cdot (4 \pi r^2) =  \frac{Q}{\epsilon _0}
from which
E(r) =  \frac{Q}{4 \pi \epsilon_0 r^2}
In part A of the problem, we want to find the electric field at r=11.1 cm=0.111 m. Substituting this number into the previous formula, we get
E(0.111m)=2680 N/C

And so, the energy density at r=0.111 m is
U= \frac{1}{2} \epsilon _0 E^2 =  \frac{1}{2} (8.85\cdot 10^{-12}C^2m^{-2}N^{-1})(2680 N/C)^2=3.17 \cdot 10^{-5}J/m^3

Part B) The solution of this part is the same as part A), since we already know the charge of the capacitor: Q=3.67 \cdot 10^{-9}C. We just need to calculate the electric field E at a different value of r: r=16.4 cm=0.164 m, so
E(0.164 m)= \frac{Q}{4 \pi \epsilon_0 r^2}=1228 N/C

And therefore, the energy density at this distance from the center is
U= \frac{1}{2}\epsilon_0 E^2 =  \frac{1}{2} (8.85\cdot 10^{-12}C^2m^{-2}N^{-1})(1228 N/C)^2=6.68 \cdot 10^{-6}J/m^3
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3 years ago
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