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avanturin [10]
3 years ago
6

Two oppositely charged but otherwise identical conducting plates of area 2.50 square centimeters are separated by a dielectric 1

.80 millimeters thick, with a dielectric constant of K=3.60. The resultant electric field in the dielectric is 1.20×106 volts per meter. Part A Compute the magnitude of the charge per unit area σ on the conducting plate. Express your answer in coulombs per square meter to three significant figures.
Physics
1 answer:
algol133 years ago
7 0

Answer: 38.2 μC

Explanation: In order to solve this problem we have to use the relationship for a two plate capacitor with a dielectric  so:

C= Q/V= we also know that for two paralel plates C=εo*k*A/d and V=E/d

where k is the dielectric constant, A plate area, V is potential difference; E electric field and d the separation between the plates.

reorganizing we have:

Q/A=σ= E*k/εo= 1.2 * 10^6*3.6/8.85 * 10^-12=38.2μC

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As the proton approaches the uranium nucleus, the repulsive force slows down the proton until it comes momentarily to rest, afte
Klio2033 [76]

Answer:

 r^2 = \frac{ 2 k  \ Ze^2}{ 2m}

Explanation:

For this exercise we must use the principle of conservation of energy

starting point. The proton very far from the nucleus

          Em₀ = K = ½ m v²

final point. The point where the proton is stopped (v = 0)

          Em_f = U = q V

where the potential is

          V = k Ze / r²

Let us consider that all the charge of the nucleus is in the center, therefore r is the distance from this point to the proton that is approaching

Energy is conserved

          Em₀ = Em_f

           ½ m v² = e (k \frac{Ze}{r^2})

          r^2 = \frac{ 2 k  \ Ze^2}{ 2m}

with this expression we can find the closest approach distance (r)

3 0
3 years ago
Hey guys, i need some help. I'm having a physics test tmmrow and I understand nothing :(. Can anyone plz explain or give me a br
professor190 [17]

We think of sound as something we hear—something that makes noise. But in pure physics terms, sound is just a vibration going through matter.

The way a vibration “goes through” matter is in the form of a sound wave. When you think of sound waves, you probably think of something like this:1

But that’s not how sound waves work. A wave like that is called a transverse wave, where each individual particle moves up and down to create a snake situation.

A sound wave is more like an earthworm situation:2

Like an earthworm, sound moves by compressing and decompressing. This is called a longitudinal wave. A slinky can do both kinds of waves:13

Sound starts with a vibration of some kind creating a longitudinal wave through matter. Check this out:4

That’s what sound looks like—except picture an expanding ripple of spheres doing that. In this animation, the sound wave is being generated by that vibrating grey bar on the left. The bar might be your vocal chords, a guitar string, or a waterfall continually pounding down into the river below. By looking at the red dots, you can see that even though the wave moves in one direction, each individual particle only moves back and forth, mimicking the vibration of the gray bar.

So instead of a curvy snake wave, sound is a pressure wave, which causes each piece of the air to be at either higher-than-normal pressure or lower-than-normal pressure. So when you see a snake-like illustration of a sound wave, it’s referring to the measure of pressure, not the literal path of movement of the particles:5

6 0
3 years ago
A bungee cord is 30m long and when stretched a distance x itexerts a restoring force of magnitude kx. Your father-in-law (mass95
madam [21]

Answer:

The distance the bungee cord that would be stretched 0.602 m, should be selected when pulled by a force of 380 N.

Explanation:

As from the given data

the length of the rope is given as l=30 m

the stretched length is given as l'=41m

the stretched length required is give as  y=l'-l=41-30=11m

the mass is m=95 kg

the  force is  F=380 N

the gravitational acceleration is g=9.8 m/s2

The equation of  k is given by equating the energy at the equilibrium point which is given as

U_{potential}=U_{elastic}\\mgh=\dfrac{1}{2} k y^2\\k=\dfrac{2mgh}{y^2}

Here

m=95 kg, g=9.8 m/s2, h=41 m, y=11 m so

k=\dfrac{2mgh}{y^2}\\k=\dfrac{2\times 95\times 9.8\times 41}{11^2}\\k=630.92 N/m

Now the force is

F=kx\\ or

x=\dfrac{F}{k}\\

So here F=380 N, k=630.92 N/m

x=\dfrac{F}{k}\\x=\dfrac{380}{630.92}\\x=0.602 m

So the distance is 0.602 m

6 0
3 years ago
A merry go round (mass of 200kg and radius 5m) is rotating so that the outside edge is moving 10m/s. A giant bug of 6kd at a dis
romanna [79]

Answer:

2 rad/s

Explanation:

For a rotating object, the linear velocity is given by

v=\omega r

where \omega is the angular velocity and r is the radius.

\omega=\dfrac{v}{r}

The edge has a linear velocity of 10 m/s and the radius at the edge is 5 m.

\omega=\dfrac{10 \text{ m/s}}{5 \text{ m}} = 2 \text{ rad/s}

8 0
3 years ago
A sample of gas is enclosed in a container of fixed volume. Identify which of the following statements are true. Check all that
motikmotik

Answer:

B. If the container is cooled, the gas particles will lose kinetic energy and temperature will decrease.

C. If the gas particles move more quickly, they will collide more frequently with the walls of the container and pressure will increase.

E. If the gas particles move more quickly, they will collide with the walls of the container more often and with more force, and pressure will increase.

#FreeMelvin

7 0
3 years ago
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