Can someone help me with this

A parallel plate capacitor is charged up by a battery. The battery is then disconnected, but the charge remains on the plates. T

larisa [96]

Answer:

a. Decreases

b. Increases

c. Remains the same

d. Increases

Explanation:

a. Capacitance is given by c= Ak€/d

where A is conductivity plate with Area

K is a constant

€ is dielectric with permittivity.

d is the distance

b. Potential difference is given by

V = Ed, since, the electric field remains the

same, the potential diterence also increases with increase in distance.

Since the capacitance depends upon the distance, and all the other factors are kept constant, the capacitance decreases.

c. Electric field remains the same because charge on the

plate remains the same.

d. since electric field remains the same and capacitance decreases, the energy increases.

E= 1/2c * Q^2

7 0

3 years ago

A wooden object (conically shaped) has a diameter of 8cm and height of 14cm. It floats in oil with 6cm of its height above oil l

baherus [9]

Answer:

(a) The density of the object is 316/343 × the density of the oil

(b) The fraction of oil displaced after immersing the object is 0.461 of the oil volume

Explanation:

(a) The volume, V of a cone of height, h and base diameter, D = 2×r is given by the following equation;

$V = \dfrac{\pi r^{2} h}{3}$

The volume of the object is therefore;

$\dfrac{\pi \times 4^{2} \times 14}{3} = 74\tfrac{2}{3}\pi \, cm^3$

Where 6 cm is above the oil level we have;

$\dfrac{\pi \times \left (6 \times \dfrac{4}{14} \right )^{2} \times 6}{3} = 5\tfrac{43}{49}\pi \, cm^3$ above the oil level

Therefore, volume of the oil displaced = $68\tfrac{116}{147}\pi$ cm³ = 216.11 cm³

The density of the object is thus;

$\dfrac{68\tfrac{116}{147}\pi}{ 74\tfrac{2}{3}\pi} \times Density \ of \ the \ oil = \dfrac{316}{343} \right ) \times Density \ of \ the \ oil$

The density of the object = 316/343 × the density of the oil.

(b) The volume of the oil = 2 × Volume of the object = $2 \times 74\tfrac{2}{3}\pi \, cm^3 = 149\tfrac{1}{3}\pi \, cm^3$

The fraction of the volume displaced, x, after immersing the object is given as follows;

$x = \dfrac{68\tfrac{116}{147}\pi}{ 149\tfrac{1}{3}\pi} = \dfrac{158}{343} = 0.461$

The fraction of oil displaced after immersing the object = 0.461 of the volume of the oil

8 0

3 years ago

When the notebook is sitting on the table, what force prevents it from falling into the space below the table? If you don’t know

Xelga [282]

Answer:

reaction force/normal force (its when the tables kinda pushing from under so it doesnt fall)

and the notebook wont float bc of gravity

Explanation:

theres a quizlet on this if you wanna check it out lol

4 0

2 years ago

An object on a number line moved from x = 15 cm to x = 165 cm and

podryga [215]

Answer:

v_avg = 2.9 cm/s

Explanation:

The average velocity of the object is the sum of the distance of all its trajectories divided the time:

x_all is the total distance traveled by the object. In this case you have that the object traveled in the first trajectory 165cm-15cm = 150cm, and in the second one, 165cm - 25cm = 140cm

Then, x_all = 150cm + 140cm = 290cm

The average velocity is, for t = 100s

hence, the average velocity of the object in the total trajectory traveled is 2.9 cm/s

5 0

3 years ago

Which type of energy is stored in molecular bonds?

EastWind [94]

Answer:

Energy, potential energy, is stored in the covalent bonds holding atoms together in the form of molecules. This is often called chemical energy

5 0

3 years ago

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