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masya89 [10]
3 years ago
13

What is the escape speed for an electron initially at rest on the surface of a sphere with a radius of 1.3 cm and a uniformly di

stributed charge of 2.3 ✕ 10−15 C. That is, what initial speed must the electron have to reach an infinite distance from the sphere and have zero kinetic energy when it gets there?
Physics
1 answer:
Vlad [161]3 years ago
4 0

Answer:

2.37 * 10^4 m/s

Explanation:

Constants :

Mass of electron = 9.11 * 10^(-31) kg

Electric charge of an electron = 1.602 * 10^(-19) C

Parameters given:

Radius of sphere = 1.3cm = 0.013m

Charge of sphere = 2.3 * 10^(−15) C

Using the law of conservation of energy, we have that:

K. E.(initial) + P. E.(initial) = K. E.(final) + P. E.(final)

K. E.(final) = 0, since final velocity is zero and P. E.(final) = 0 since the electron reaches a final distance of infinity.

Hence,

K. E.(initial) = P. E.(initial)

0.5mv^2 = (kqQ)/r

Where k = Coulumbs constant

Q = charge of the sphere.

r = radius of the sphere.

=> 0.5*m*v^2 = (kqQ)/r

0.5 * 9.11 * 10^(-31) * v^2 = (9 * 10^9 * 1.602 * 10^(-19) * 2.3 * 10^(-15))/0.013

4.555 * 10^(-31) * v^2 = 2550.88 * 10^(-25)

=> v^2 = 2550.88 * 10^(-25) / 4.555 * 10^(-31)

v^2 = 560 * 10^6 = 5.60 * 10^8

=> v = 2.37 * 10^4 m/s

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To solve this question we will use ideal gas equation:
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p = pressure
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n = number of moles
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We can rearrange formula to get:
\frac{p*V}{T} =n*R
We are working woth same gas so we can write following formula. Index 1 stands for conditions before change and index 2 stands for conditions after change.
\frac{ p_{1}*V_{1} }{T_{1}} = \frac{ p_{2}*V_{2} }{T_{2}}

We are given:
p1=92.1kPa = 92100Pa
V1=200mL = 0.2L
T1=275K
p2= 101325Pa
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We start by rearranging formula for V2. After that we can insert numbers:
{ p_{1}*V_{1} *T_{2}} = { p_{2}*V_{2}*T_{1} } \\  \\ V_{2}= \frac{p_{1}*V_{1} *T_{2}}{ p_{2}*T_{1}}  \\  \\ V_{2}=  \frac{92100*0.2*273}{101325*275}  \\  \\ V_{2}= 0.18L=180mL
7 0
4 years ago
Find the average velocity of the object from point B to C.
ExtremeBDS [4]
The average velocity of the object in moving from point-B to point-C is

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2. While standing near a bus stop, a student hears a distant horn beeping. The frequency emitted by the horn is 440 Hz. The bus
jarptica [38.1K]

Given Information:

Frequency of horn = f₀ = 440 Hz

Speed of sound = v = 330 m/s

Speed of bus = v₀ = 20 m/s

Answer:

Case 1. When the bus is crossing the student = 440 Hz

Case 2. When the bus is approaching the student = 414.9 Hz

Case 3. When the bus is moving away from the student = 468.4 Hz

Explanation:

There are 3 cases in this scenario:

Case 1. When the bus is crossing the student

Case 2. When the bus is approaching the student

Case 3. When the bus is moving away from the student

Let us explore each case:

Case 1. When the bus is crossing the student:

Student will hear the same frequency emitted by the horn that is 440 Hz.

f = 440 Hz

Case 2. When the bus is approaching the student

f = f₀ ( v / v+v₀ )

f = 440 ( 330/ 330+20 )

f = 440 ( 330/ 350 )

f = 440 ( 0.943 )

f = 414.9 Hz

Case 3. When the bus is moving away from the student

f = f₀ ( v / v+v₀ )

f = 440 ( 330/ 330-20 )

f = 440 ( 330/ 310 )

f = 440 ( 1.0645 )

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Answer:

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electrical field strength = potential diff / distance

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potential difference = electric field x distance

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= 562.5 V

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A bucket weighing 5 lbs is lifted at a constant rate from the bottom of a 100 ft well by a rope which weighs 5 lbs. The bucket h
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W_{bucket} = 24934.85\,lbf\cdot ft

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