PHYSICS HIGH SCHOOL

A parallel-plate air capacitor is made from two plates 0.210 m square, spaced 0.815 cm apart. it is connected to a 120 v battery. if the plates are pulled apart to a separation of 1.63 cm , suppose the battery remains connected while the plates are pulled apart. part a what is the capacitance?

Answers

Answer 1

Answer:

Answer:

at the beginning: $2.3\cdot 10^(-10) F$

when the plates are pulled apart: $1.1\cdot 10^(-10) F$

Explanation:

The capacitance of a parallel-plate capacitor is given by

$C=k \epsilon_0 (A)/(d)$

where

k is the relative permittivity of the medium (for air, k=1, so we can omit it)

$\epsilon_0 = 8.85\cdot 10^(-12) F/m$ is the permittivity of free space

A is the area of the plates of the capacitor

d is the separation between the plates

In this problem, we have:

$A=0.210 m^2$ is the area of the plates

$d=0.815 cm=8.15\cdot 10^(-3) m$ is the separation between the plates at the beginning

Substituting into the formula, we find

$C=(1)(8.85\cdot 10^(-12)F/m)(0.210 m^2)/(8.15\cdot 10^(-3) m)=2.3\cdot 10^(-10) F$

Later, the plates are pulled apart to $d=1.63 cm=0.0163 m$ , so the capacitance becomes

$C=(1)(8.85\cdot 10^(-12)F/m)(0.210 m^2)/(0.0163 m)=1.1\cdot 10^(-10) F$

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Julie finds a snail on the sidewalk and wants to know whether or not the snail moves throughout the day. She places a single mark on the sidewalk next to the snail.What will Julie use the mark for initially?

Answers

To make a base line mark to know where she started measuring the snails movement, at the end of the day she can measure from her first line to her last to see how far the snail went.
Hope this helped!

Answer:

To check the movement of the snail.

Explanation:

the answer was right :)

A grasshopper jumps and accelerates at 4.5 m/s2. If its legs exert a force of 0.062 N during the jump, what is the mass of the grasshopper?

Answers

Answer:

0.014 kg

Explanation:

trust me its right i took the assessment and got 100%. :)

The force of repulsion between two like-charged particles will increase ifA) the distance between the charged particles is increased.

B) the distance between the charged particles is decreased.

C) the magnitude of the charge on both the particles is decreased.

D) the magnitude of the charge on one of the particles is decreased.

Answers

The force of repulsion between two like-charged particles will increase if the distance between the charged particles is decreased.

What is coulomb's law?

According to Coulomb's law: The magnitude of each of the electric forces with which two point-at-rest charges interact is directly proportional to the product of the magnitude of both charges

And inversely proportional to the square of the distance that separates them and has the direction of the line that joins them.

The formula will be given as:

$F=(kq_1q_2)/(r^2)$

We can see that the force is inversely proportional to the distance of the two charges.

To know more about coulomb's law follow

brainly.com/question/66110

Answer:

B). the distance between the charged particles is decreased.

Explanation:

Two objects are connected by a light string that passes over a frictionless pulley as shown in the figure below. Assume the incline is frictionless and take m1 = 2.00 kg, m2 = 7.35 kg, and θ = 50.0°.1) Find the magnitude of the accelerations of the objects. (m/s^2)
2)Find the tension in the string. (N)
3)Find the speed of each object 2.50 s after being released from rest. (m/s)

Answers

Answer:

Part a)

$a = 3.81 m/s^2$

Part b)

$T = 27.2 N$

Part c)

$v_f = 9.53 m/s$

Explanation:

For m2 mass along the inclined plane we can write

$m_2g sin\theta - T = m_2 a$

for m1 mass we can write equation in vertical direction

$T - m_1 g = m_1 a$

so we will have

$m_2g sin\theta - m_1g = (m_1 + m_2) a$

so we have

$a = ((m_2 sin\theta - m_1)g)/(m_1 + m_2)$

now plug in all data in it

$a = ((7.35 sin50 - 2)9.81)/(7.35 + 2)$

$a = 3.81 m/s^2$

Part b)

from above equation

$T = m_1g + m_1 a$

$T = 2(9.81 + 3.81)$

$T = 27.2 N$

Part c)

Speed of the object after t = 2.50 s

$v_f = v_i + at$

$v_f = 0 + 3.81(2.50)$

$v_f = 9.53 m/s$

assume m1 moving down, so m2 is moving up and accelleration direction of each box is the same as its movement.

check forces act on m1
1. tension of the rope T up
2. m1g down
3. moving down with a
f = ma --> m1g - T = m1a ...(1)
or. T = m1g - m1a ... (1.1)

check forces act on m2
1. tension of the rope T up alone the incline. it must be equal to previous one because it is the same segment of the rope.
2. m2gsine(θ) down along the incline
3. moving up with a
f = ma --> T - m2gsine(θ) = m2a ...(2)

replace T from (1.1) to (2)
(m1g - m1a) - m2gsin(θ) = m2a
m1g - m2gsin(θ) = m2a + m1a
a = g(m1 - m2sin(θ))/(m1 + m2)

plug in value you will get the answer. if a is negative, it means direction on our assumption is not correct. it's the opposite (direction).

after get a, you will get T
and velocity use v = u + at

... i cannot post pict. draw the diagram will make you understand this better.

The purpose of a fuse or circuit breaker is _____.to speed up the current
to slow down the current
to stop current that is flowing too fast
to stop current that is flowing too slow

Answers

Answer:

to stop current that is flowing too fast

Explanation:

What lines the inferior surface of the diaphragm?

Answers

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