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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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A 1600kg cannon fires a 5kg cannonball horizontally. The exit velocity of the cannonball is 80m/s and the barrel length is 2m. W
evablogger [386]

Answer:

 a = 1600 m / s²

Explanation:

For this exercise we use the kinematics relations,

         v² = v₀² + 2 a x

where v₀ is the initial velocity of the bullet, which as part of rest is zero, for the distance (x) we can assume that the gases accelerate along the entire trajectory of the cannon x = 2m

         a = \frac{v^2}{2x}

let's calculate

         a = \frac{80^2}{2 \ 2}

         a = 1600 m / s²

7 0
3 years ago
Does any one know the answer
valina [46]
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6 0
3 years ago
Find the intensity of the electromagnetic wave described in each case. (a) an electromagnetic wave with a wavelength of 630 nm a
qaws [65]

Find the intensity of the electromagnetic wave described in each case.

(a) an electromagnetic wave with a wavelength of 645 nm and a peak electric field magnitude of 8.5 V/m.
  
(b) an electromagnetic wave with an angular frequency of 6.3 ✕ 1018 rad/s and a peak magnetic field magnitude of 10−10 T.

3 0
4 years ago
This question is related to inertia:
luda_lava [24]
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3 0
3 years ago
Cynthia forgot to put the fabric softener in the wash. As her socks tumbled in the dryer, they became charged. If a small piece
polet [3.4K]

Answer:

E = 24000 N/C = 24 KN/C

Explanation:

The electric field experienced by a test charge is given by the following formula:

E = \frac{F}{q}\\\\

where,

E = Electric Field = ?

F = Force of attraction = 3 x 10⁻⁶ N

q = Charge on piece of lint = 1.25 x 10⁻¹⁰ C

Therefore, using these values in the equation, we get:

E = \frac{3\ x\ 10^{-6}\ N}{1.25\ x\ ^{-10}\ C}\\\\

<u>E = 24000 N/C = 24 KN/C</u>

6 0
3 years ago
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