he electric field is dependent on the distance between the plates. false: The voltage of a connected charged capacitor decreases when the plate area is increased. false: The voltage of a disconnected charged capacitor increases when the plates are brought closer together.

Answers

Answer 1

Answer:

False.

Explanation:

The voltage of a disconnected charged capacitor decreases when the plates are brought closer together because the capacitance is inversely proportional to the area. If the area between plates decreases, its capacitance increases and vice versa. There is direct relationship between voltage of a disconnected charged capacitor and plates. If the distance between plates decreases, the voltage of a disconnected charged capacitor is also decreases while on the other hand, if the distance between plates increases, the voltage of a disconnected charged capacitor is also increases.


Related Questions

Which element would have properties most like helium (He)?

A. Ar
B. Hg
C. H
D. O

Answers

i believe the answer is d

Answer: A. Ar

Explanation: not anything else besides Ar

Suppose that an electron and a positron collide head-on. Both have kinetic energy of 1.20 MeV and rest energy of 0.511 MeV. They produce two photons, which by conservation of momentum must have equal energy and move in opposite directions. What is the energy Ephoton of one of these photons

Answers

Answer:

E = 1.711 MeV

Explanation:

From the law of the conservation of energy:

[tex]K.E_{e}+K.E_p + E_{e}+E_{p} = 2 E[/tex]

where,

[tex]K.E_e=K.Ep=[/tex] the kinetic energy of positron and electron = 1.2 MeV

[tex]E_e=E_p =[/tex] Rest energy of the electron and the positron = 0.511 MeV

E = Energy of Photon = ?

Therefore,

[tex]1.2\ MeV + 1.2\ MeV + 0.511\ MeV + 0.511\ MeV = 2E\\\\E = \frac{3.422\ MeV}{2}\\\\[/tex]

E = 1.711 MeV

A spinning disc with a mass of 2.5kg and a radius of 0.80m is rotating with an angular velocity of 1.5 rad/s. A ball of clay with unknown mass is dropped onto the disk and sticks to the very edge causing the angular velocity of the disk to slow to 1.13 rad/s. What is the mass of the ball of clay

Answers

Answer:

M = 1.90 Kg

Explanation:

Given data: mass = 2.5 Kg

radius R = 0.8 m

angular velocity ω = 1.5 rad/s

Angular momentum L =0.5×Iω^2

Where, I  is the moment of inertia of the spinning disc.

I = 0.5MR^2

I = 0.5×2.5×0.8^2

I = 0.8 Kg/m^2

Then L = 0.5×0.8×1.5^2 = 0.8×2.25 = 0.9 Kg-m^2/sec

Let unknown mass be M

New mass of disc = (2.5+M) Kg, R = 0.8 m

New I = 0.5(2.5+M)(0.8)^2

Since, angular momentum is conserved

Angular momentum before = angular momentum after

0.5×0.5(2.5+M)(0.8)^2×(1.13)^2 = 0.9

Solving for M we get

0.204304(2.5+M)=0.9

M = 1.90 Kg

In any given wave, when the frequency of the wave doubles (f = 2f), which of the following other changes would also take place?
A The wavelength would double (λ = 2λ).
B The velocity would double (v = 2v).
C The wavelength would be half (λ = λ/2).
D The velocity would be half (v = v/2).

Answers

Answer:

The correct answer is -

B. The velocity would double (v = 2v).

C. The wavelength would be half (λ = λ/2).

Explanation:

A wave has a speed or velocity that is related to the wavelength of the wave and the frequency of the wave and this relationship can be represented by the following equation-

Wave velocity V = Wavelength (λ) * Frequency (f)

Frequency (f) = Velocity (V) / Wavelength(λ).

The frequency and wavelength are inversely proportional and frequency and velocity are directly proportional to each other.

So, if f = 2f then,

putting value in the formula,

2f = 2v/λ, which means, f = 2v and f = λ/2

when the frequency is doubled, the wavelength will be halved and velocity will be doubled.

Please help ASAP with questions

Answers

What do you need what question

A 7.5 cm tall Aragorn action figure is placed 12.4 cm in front of a double convex lens. If the image of Aragorn is located 7.5 from the lens, what is the focal length of the lens?

Answers

Answer:

f = 4.67 cm

Explanation:

Here, we can use the thin lens formula, as follows:

[tex]\frac{1}{f}= \frac{1}{p}+ \frac{1}{q}\\\\[/tex]

where,

f = focal length of lens = ?

p = distance of object from lens = 12.4 cm

q = distance of image from lens = 7.5 cm

Therefore,

[tex]\frac{1}{f} =\frac{1}{12.4\ cm} +\frac{1}{7.5\ cm}\\\\\frac{1}{f} = \frac{1}{4.67\ cm}[/tex]

f = 4.67 cm

You are studying a population of flowering plants for several years. When you present your research findings you make the statement that, "Increased allocation of resources to reproduction relative to growth diminished future fecundity." Which of the following graph descriptions could accurately present your data?
a) With seeds in the current year on the y-axis and seeds in the previous year on the x-axis, you would see a line that increased from left to right
b) With survivorship on the y-axis and number of seeds produced on the x-axis, you would see a line that decreased left to right.
c) With leaf area on the y-axis and number of seeds produced on the x-axis, you would see a line that increased left to right
d) With survivorship on the y-axis and number of seeds produced on the x-axis, you would see a line that increased left to right.
e) With seeds in the current year on the y-axis and seeds in the previous year on the x-axis, you would see a line that decreased from left to right

Answers

Answer:

Option A

Explanation:

The graph for this problem must depict the following ""Increased allocation of resources to reproduction relative to growth diminished future fecundity."

Hence, the survivor ship must be on the Y axis and the resources on the X axis.

Here the resources include the number of seeds produced.

hence, the higher is the number of seeds (resource), the lower is the survivorship (future fecundity)

Hence, option A is correct

What are the biotic factors in this image?
Someone help thank you!!

Answers

animal plants bacteria fungi ! biotic mean living

What is Hooke's law? what is meant by elastic limit?
please answer me​

Answers

Answer:

Explanation:

The law is about how object will be deformed when certain amount force is  applied. Elastic limit is a point where object will not return its previous shape after being deformed

A 10- kg ball starting from rest rolls down a 5 m tall smooth hill from one person to another person who is standing at the bottom of the hill with a big spring whose constant is 100 N/m. How far does the spring compress in order to stop the ball

Answers

Answer: 3.13 m

Explanation:

Given

mas of the ball is m=10 kg

The ball rolls down a vertical distance of 5 m

Spring constant of spring is [tex]k=100\ N/m[/tex]

Here, the potential energy of the ball converted into kinetic energy which in turn converts into elastic potential energy

[tex]\Rightarrow mgh=\frac{1}{2}kx^2\quad [\text{x=compression in the spring}]\\\\\Rightarrow 10\times 9.8\times 5=\frac{1}{2}\cdot 100\cdot x^2\\\Rightarrow x=\sqrt{9.8}\\\Rightarrow x=3.13\ m[/tex]

Thus, the spring compresses by 3.13 m.

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