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1 year ago

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A 25 pF parallel-plate capacitor with an air gap between the plates is connected to a 100 V battery. A Teflon slab is then inser

ted between the plates, and completely fills the gap. What is the change in the charge on the positive plate when the Teflon is inserted?

1 answer:

fgiga [73]1 year ago

6 0

Answer:

5.3 nC

Explanation:

The initial charge stored on the capacitor is given by:

$Q=CV$ (1)

where

$C=25 pF = 25\cdot 10^{-12}F$ is the capacitance of the capacitor

$V=100 V$ is the potential difference across the capacitor

Substituting numbers into the equation, we have

$Q=(25\cdot 10^{-12} F)(100 V)=25\cdot 10^{-10}C=2.5 nC$

When the Teflon slab is inserted between the plates, the capacitance of the capacitor is increased as follows:

$C'=kC$

where k=2.1 is the dielectric constant of the Teflon. Since the voltage V remains constant, this means that the new charge stored by the capacitor (1) will be

$Q'=C'V=kCV=kQ$

and so

$Q'=(2.1)(2.5 nC)=5.3 nC$

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Answer:

C) rift valley

Explanation:

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1 year ago

Read 2 more answers

1)After catching the ball, Sarah throws it back to Julie. However, Sarah throws it too hard so it is over Julie's head when it r

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Answer:

1)

$v_{oy}=11.29\ m/s$

2)

$y=7.39\ m$

Explanation:

<u>Projectile Motion</u>

When an object is launched near the Earth's surface forming an angle $\theta$ with the horizontal plane, it describes a well-known path called a parabola. The only force acting (neglecting the effects of the wind) is the gravity, which acts on the vertical axis.

The heigh of an object can be computed as

$\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}$

Where $y_o$ is the initial height above the ground level, $v_{oy}$ is the vertical component of the initial velocity and t is the time

The y-component of the speed is

$v_y=v_{oy}-gt$

1) We'll find the vertical component of the initial speed since we have not enough data to compute the magnitude of $v_o$

The object will reach the maximum height when $v_y=0$ . It allows us to compute the time to reach that point

$v_{oy}-gt_m=0$

Solving for $t_m$

$\displaystyle t_m=\frac{v_{oy}}{g}$

Thus, the maximum heigh is

$\displaystyle y_m=y_o+\frac{v_{oy}^2}{2g}$

We know this value is 8 meters

$\displaystyle y_o+\frac{v_{oy}^2}{2g}=8$

Solving for $v_{oy}$

$\displaystyle v_{oy}=\sqrt{2g(8-y_o)}$

Replacing the known values

$\displaystyle v_{oy}=\sqrt{2(9.8)(8-1.5)}$

$\displaystyle v_{oy}=11.29\ m/s$

2) We know at t=1.505 sec the ball is above Julie's head, we can compute

$\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}$

$\displaystyle y=1.5+(11.29)(1.505)-\frac{9.8(1.505)^2}{2}$

$\displaystyle y=1.5\ m+16,991\ m-11.098\ m$

$y=7.39\ m$

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2 years ago

A gas has an initial volume of 168 cm3 at a temperature of 255 K and a pressure of 1.6 atm. The pressure of the gas decreases to

erastovalidia [21]

Would presume you are asked to find the volume, since there is no second volume.

By General Gas Law:

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1.6 * 168 /255 = 1.3*V₂/285

V₂ = 1.6 * 168 * 285 / (1.3*255)

V₂ = 231.095

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2 years ago

Read 2 more answers

A proton is released from rest at the positive plate of a parallel-plate capacitor. It crosses the capacitor and reaches the neg

WINSTONCH [101]

Answer:

$=2,012,319.36 \ m/s$

Explanation:

-The only relevant force is the electrostatic force

-The formula for the electrostatic force is:

$F = Eq$

E is the electric field and q is the magnitude of the charge.

#Since the electric field is the same in both cases, and the charge of the protons and electrons have the same magnitude, you can state that the magnitude of the electric forces acting in both proton and electron are the same.

$F_e = F_p\\\\F_e= Force \ on \ electron\\F_p = Force \ on \ proton$

-Applying Newton's 2nd Law:

$F=ma$

$F_e=M_ea_e$

$F_p=M_pa_p$

#equate the two forces:

$F_e = F_p\\\\M_ea_e=M_pa_p\\\\a_e=\frac{M_pa_p}{M_e}$

#The equations for velocity in uniform acceleration:

$V_f^2=V_o^2+2ad\\\\V_o^2=0\\\\\therefore V_f^2=2ad$

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$V_f^2=2a_pd\\\\a_p=\frac{V_f^2}{2d}\\\\a_p=\frac{47000m/s)^2}{2d}$

#For the electron:

$V_f^2=2{a_e}^2\times 2d\\\\A_e=M_p\times A_p/M_e\\\\V_f^2=M_p\times (47000m/s)^2/2d\times2d/M_e\\\\V_f^2=M_p\times (47000m/s)^2/M_e\\\\V_f=47000m/s\times\sqrt{\frac{M_p}{M_e}}$

The mass values of the proton and electron are:

$M_p=1.67\times 10^{-27} kg\\\\M_e=9.11\times10^{-31}kg$

The speed of the ion is therefore calculated as:

$V_f=47000m/s\times\sqrt{\frac{M_p}{M_e}}\\\\=47000m/s\times\sqrt{\frac{1.67\times10^{-27}}{9.11\times10^{-31}}\\\\=2,012,319.36 \ m/s$

Hence, the ion's speed at the negative plate is $=2,012,319.36 \ m/s$

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2 years ago

The velocity versus time graph of particle A is tangent to the velocity versus time graph for particle B at point O. What is the

lara [203]

As velocities are tangent, the value of both Particle A and Particle B would be same for that point O (Intersecting point)

a = v / t
Here, v = 7, t = 6
So, a = 7/6
a = 1.17
As the graph is decreasing, value of acceleration would be negative.
So, a = -1.17 m/s²

In short, Your Answer would be Option C

Hope this helps!

7 0

1 year ago