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olga2289 [7]
2 years ago
13

Sphere 1 with radius R_1 has positive charge q, Sphere 2 with radius 4.50 R_1 is far from sphere 1 and initially uncharged. The

separated spheres are then connected with a wire then uncouth to retain only negligible charge. (a) What is the ratio V_1/V_2 of the final potentials of the spheres? (b) what fraction of q ends up on sphere? (c) What fraction of q ends up on sphere 2? (d) What is the ratio q_1/q_2 of the surface charge densities of the spheres?
Physics
1 answer:
tia_tia [17]2 years ago
6 0

Answer:

Explanation:

capacitance of sphere 2 will be 4.5 times sphere 1

a ) when spheres are in contact they will have same potential finally . So

V_1 / V_2 = 1

b )

Charge will be distributed in the ratio of their capacity

charge on sphere1 = q  x 1 / ( 1 + 4.5 )

= q / 5.5

fraction = 1 / 5.5

c ) charge on sphere 2

= q x 4.5 / 5.5

fraction = 4.5 / 5.5

d ) surface charge density of sphere 1

= q /( 5.5 x A ) where A is surface area

surface charge density of sphere 2

= q x 4.5 /( 5.5 x 4.5² A ) where A is surface area

= q  /( 5.5 x 4.5 A )

q_1/q_2 = 4.5

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What have psychologists learned about perception from optical illusions ?
ruslelena [56]

Answer:

Explanation:

That an optical illusion somehow interferes with the way we see things. Even simple illusions can completely fool us. If you search out the term, you'll see all kinds of them.

Most critically we see one thing and know another to be true. But knowing the truth doesn't help us. We still see and believe the truth of the illusion.

5 0
3 years ago
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Two identical loudspeakers 2.00 m apart are emitting sound waves into a room where the speed of sound is 340 m/s. Abby is standi
ki77a [65]

Answer:

The lowest possible frequency of sound for which this is possible is 1307.69 Hz

Explanation:

From the question, Abby is standing 5.00m in front of one of the speakers, perpendicular to the line joining the speakers.

First, we will determine his distance from the second speaker using the Pythagorean theorem

l₂ = √(2.00²+5.00²)

l₂ = √4+25

l₂ = √29

l₂ = 5.39 m

Hence, the path difference is

ΔL = l₂ - l₁

ΔL = 5.39 m - 5.00 m

ΔL = 0.39 m

From the formula for destructive interference

ΔL = (n+1/2)λ

where n is any integer and λ is the wavelength

n = 1 in this case, the lowest possible frequency corresponds to the largest wavelength, which corresponds to the smallest value of n.

Then,

0.39 = (1+ 1/2)λ

0.39 = (3/2)λ

0.39 = 1.5λ

∴ λ = 0.39/1.5

λ = 0.26 m

From

v = fλ

f = v/λ

f = 340 / 0.26

f = 1307.69 Hz

Hence, the lowest possible frequency of sound for which this is possible is 1307.69 Hz.

5 0
3 years ago
how much gravitational potential energy do you give a 70 kg person when you lift him up 3 m in the air?
SCORPION-xisa [38]

Given gravitational potential energy when he's lifted is 2058 J.

Kinetic energy is transferred to the person.

Amount of kinetic energy the person has is -2058 J

velocity of person = 7.67 m/s².

<h3>Explanation:</h3>

Given:

Weight of person = 70 kg

Lifted height = 3 m

1. Gravitational potential energy of a lifted person is equal to the work done.

PE_g=W=m\times g\times h\\Acceleration due to gravity = g = 9.8 \ m/s^2 \\PE_g= m = m\times g\times h= 70\times 9.8 \times 3 = 2058\ kg.m/s^2 = 2058\ J

Gravitational potential energy is equal to 2058 Joules.

2. The Gravitational potential energy is converted into kinetic energy. Kinetic energy is being transferred to the person.

3. Kinetic energy gained = Potential energy lost = -PE_g = -2058\ kg.m/s^2

Kinetic energy gained by the person = (-2058 kg.m/s²)

4. Velocity = ?

Kinetic energy magnitude= \frac{1}{2} m\times v^2 = m\times g \times h

Solving for v, we get

v=\sqrt{2gh} =\sqrt{2\times 9.8 \times 3} = \sqrt{58.8} = 7.67 m/s^2

The person will be going at a speed of 7.67 m/s².

4 0
3 years ago
What minimum distance should you separate two sources emitting the same waves with wavelength 5mm in phase such that you obtain
Makovka662 [10]

To solve this problem we will apply the concept related to destructive interference (from the principle of superposition). This concept is understood as a superposition of two or more waves of identical or similar frequency that, when interfering, create a new wave pattern of less intensity (amplitude) at a point called a node. Mathematically it can be described as

d = n \frac{\lambda}{2}

Where,

d = Path difference

\lambda= wavelength

n = Any integer which represent the number of repetition of the spectrum

In this question the distance between the two source will be minimum for the case of minimum path difference, then n= 1

d = \frac{\lambda}{2}

d = \frac{5*10^{-3}}{2}

d = 2.5mm

Therefore the minimum distance that should you separate two sources emitting the same waves is 2.5mm

8 0
3 years ago
Calculate the energy, wavelength, and frequency of the emitted photon when an electron moves from an energy level of -3.40 eV to
jasenka [17]

Answer:

(a) The energy of the photon is 1.632 x 10^{-8} J.

(b) The wavelength of the photon is 1.2 x 10^{-17} m.

(c) The frequency of the photon is 2.47 x 10^{25} Hz.

Explanation:

Let;

E_{1} = -13.60 ev

E_{2} = -3.40 ev

(a) Energy of the emitted photon can be determined as;

E_{2} - E_{1} = -3.40 - (-13.60)

           = -3.40 + 13.60

           = 10.20 eV

           = 10.20(1.6 x 10^{-9})

E_{2} - E_{1} = 1.632 x 10^{-8} Joules

The energy of the emitted photon is 10.20 eV (or 1.632 x 10^{-8} Joules).

(b) The wavelength, λ, can be determined as;

E = (hc)/ λ

where: E is the energy of the photon, h is the Planck's constant (6.6 x 10^{-34} Js), c is the speed of light (3 x 10^{8} m/s) and λ is the wavelength.

10.20(1.6 x 10^{-9}) = (6.6 x 10^{-34} * 3 x 10^{8})/ λ

λ = \frac{1.98*10^{-25} }{1.632*10^{-8} }

  = 1.213 x 10^{-17}

Wavelength of the photon is 1.2 x 10^{-17} m.

(c) The frequency can be determined by;

E = hf

where f is the frequency of the photon.

1.632 x 10^{-8}  = 6.6 x 10^{-34} x f

f = \frac{1.632*10^{-8} }{6.6*10^{-34} }

 = 2.47 x 10^{25} Hz

Frequency of the emitted photon is 2.47 x 10^{25} Hz.

6 0
2 years ago
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