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ohaa [14]
3 years ago
14

A battery charges a parallel-plate capacitor fully and then is removed. The plates are immediately pulled apart. (With the batte

ry disconnected, the amount of charge on the plates remains constant.) What happens to the potential difference between the plates as they are being separated?
Physics
1 answer:
Assoli18 [71]3 years ago
4 0

Answer:

<em>There will be an increase in potential difference.</em>

Explanation:

As we know that the potential difference depends upon the capacitance.

ΔV = Q/C

When battery is disconnected the charge remains constant on the plates but the capacitance decreases. As the capacitance has an inverse relation with the potential difference, there will be an increase in it.

In addition to that the potential difference can also be defined as the product of field and distance between the plates. As the charge is constant so the field is constant. Upon increasing the separation between the plates the potential difference will also increased.

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Which literary device portrays a character's reflections?
DochEvi [55]
Of the literary devices you provided to choose from, snap hots and thought shots are not even devices so they fall off the list.

sensory language can explain the senses the character is using - helping figure out his reflections,

but figuratve language can helop the best. the character says something figuratively and we can analyze what they meant to find out what they were thinking
6 0
3 years ago
A tetrahedron has an equilateral triangle base with 25.0-cm-long edges and three equilateral triangle sides. The base is paralle
djverab [1.8K]

Answer:

a. 7.046 Nm²/C

b. 2.348 Nm²/C

Explanation:

Data given:

Base of equilateral triangle = 25.0 cm = 0.25 m

Strength of electric field = 260 N/C

In order to find the electric flux we first have to find out the area of triangle.

Area of triangle = \frac{\sqrt{3} }{4} a^{2}

                         = \frac{\sqrt{3} }{4} (0.25)^{2}

                         = 0.0271 m³

Lets find electric flux,

      Electric Flux = E. A

                          = 260×0.0271

                          = 7.046 Nm²/C

Now we can find the electric flux through each of the three sides.

Electric flux through three sides = \frac{7.046}{3}

                                                = 2.348 N m²/C

       

3 0
3 years ago
What is a rubens tube
Free_Kalibri [48]

Answer:

its an antique physics apparatus for demonstrating acoustic standing waves in a tube.

6 0
3 years ago
Suppose that you hear a clap of thunder 16 s after seeing the associated lightning strike. How far are you from the lightning st
riadik2000 [5.3K]

Answer:

d=5.376km

Explanation:

Since <em>light is so fast</em> we can assume no time passes between the lightning strikes and we observe it. We want to know then how far away did the strike occur for the sound to take 16s to reach our ears. Since the definition of velocity tells us that v=d/t, we can write d=vt=(336m/s)(16s)=5376m=5.376km

4 0
3 years ago
A narrow beam of light from a laser travels through air (n = 1.00) and strikes the surface of the water (n = 1.33) in a lake at
Natalka [10]

Answer:

A) d = 11.8m

B) d = 4.293 m

Explanation:

A) We are told that the angle of incidence;θ_i = 70°.

Now, if refraction doesn't occur, the angle of the light continues to be 70° in the water relative to the normal. Thus;

tan 70° = d/4.3m

Where d is the distance from point B at which the laser beam would strike the lakebottom.

So,d = 4.3*tan70

d = 11.8m

B) Since the light is moving from air (n1=1.00) to water (n2=1.33), we can use Snell's law to find the angle of refraction(θ_r)

So,

n1*sinθ_i = n2*sinθ_r

Thus; sinθ_r = (n1*sinθ_i)/n2

sinθ_r = (1 * sin70)/1.33

sinθ_r = 0.7065

θ_r = sin^(-1)0.7065

θ_r = 44.95°

Thus; xonsidering refraction, distance from point B at which the laser beam strikes the lake-bottom is calculated from;

d = 4.3 tan44.95

d = 4.293 m

4 0
3 years ago
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