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GaryK [48]
3 years ago
8

What is the charge on 1.0 kg of protons? (e = 1.60 × 10-19 c, mproton = 1.67 × 10-27 kg)?

Physics
2 answers:
Andru [333]3 years ago
8 0
First, we need to find the number of protons, which is the total mass divided by the mass of one proton:
N= \frac{m}{m_p}= \frac{1.0 kg}{1.67 \cdot 10^{-27} kg}=6.0 \cdot 10^{26} protons

Then, the total charge is the number of protons times the charge of a  single proton:
Q=Ne = 6.0 \cdot 10^{26}\cdot 1.60 \cdot 10^{-19} C=9.6 \cdot 10^7 C
Effectus [21]3 years ago
6 0

9.6 x 10⁷ C.

<h3>Further explanation</h3>

<u>Given:</u>

  • The total mass of protons = 1.0 kg
  • The mass of one proton = 1.67 \times 10^{-27} \ kg
  • The elementary charge of proton = 1.602 \times 10^{-19} \ C

<u>Question:</u>

What is the charge on 1.0 kg of protons?

<u>The Process:</u>

Atoms consist of smaller or subatomic particles, namely protons (p), electrons (e), and neutrons (n).

  • Protons are positively charged subatomic particles.
  • Electrons are negatively charged subatomic particles.
  • Neutrons are subatomic particles with no charge.

<u>Step-1: find the number of protons</u>

\boxed{ \ N = \frac{total \ mass}{mass \ of \ one \ proton} \ }

\boxed{ \ N = \frac{1.0 \ kg}{1.67 \times 10^{-27} \ kg} \ }

Hence, \boxed{ \ N = 5.99 \times 10^{26} \ protons} \ }

<u>Step-2: find the charge</u>

By using this formula,

\boxed{ \ Q = Ne \ }

we calculate the total charge of the protons.

Recall e = 1.602 \times 10^{-19} \ C

\boxed{ \ Q = (5.99 \times 10^{26})(1.602 \times 10^{-19}) \ }

Thus, the charge on 1.0 kg of protons is \boxed{\boxed{ \ 9.6 \times 10^7 \ C \ }}

<u>Quick Steps</u>

\boxed{ \ Q = 1.0 \ kg \times \frac{1 \ proton}{1.67 \times 10^{-27} \ kg} \times \frac{1.60 \times 10^{-19} \ C}{1 \ proton} \ } \rightarrow \boxed{ \ Q = 9.6 \times 10^7 \ C \ }

<h3>Learn more</h3>
  1. How many atoms thick was Rutherford's foil?  brainly.com/question/4929060  
  2. Find out the fraction of the space within the atom is occupied by the nucleus brainly.com/question/10818405
  3. Determine the shortest wavelength in electron transition brainly.com/question/4986277

Keywords: the charge, protons, the mass, the number, elementary

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Question:

Augment the rectifier circuit of Problem 4.68 with a  capacitor chosen to provide a peak-to-peak ripple voltage of  (i) 10% of the peak output and (ii) 1% of the peak output. In  each case:

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Problem 4.68:

A half-wave rectifier circuit with a 1-kΩ load operates from a 120-V (rms) 60-Hz household supply through  a 10-to-1 step-down transformer. It uses a silicon diode  that can be modeled to have a 0.7-V drop for any current.

Given Information:

Input voltage = 120 Vrms

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Load resistance = R = 1 kΩ

Required Information:

 (i) 10% of the peak output and (ii) 1% of the peak output. In  each case:

(a) What average output voltage results?

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Answer:

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Case (ii)

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Explanation:

Voltage at the secondary side of the transformer is

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Vrms = 120/10 = 12 V

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Vp = 12√2 = 16.97 V

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First we will calculate all the required parameters for the 10% ripple voltage and then for 1% ripple voltage.

case (i) 10% of the peak output:

(a) What average output voltage results?

Average output voltage = Vavg = Vp - Vd - 0.5Vr

Where Vp is the peak output voltage Vd is the voltage drop of diode and Vr is the ripple voltage which is given as a percentage of Vp

Vavg = Vp - Vd - 0.5Vr

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Vavg = 15.45 V

(b) What fraction of the cycle does the diode conduct?

ω = √2Vr/Vp - Vd

ω = √2*0.1(Vp-Vd)/Vp - Vd

ω = √2*0.1(16.97-0.7)/16.97 - 0.7

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Conduction of diode = (ω/2π)*100

Conduction of diode = (0.447/2π)*100

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(c) What is the average diode current?

Average current = Iavg = Vavg/R[ 1 + π( √2(Vp - Vd)/0.1(Vp-Vd))]

Average current = Iavg = 15.45/1000[ 1 + π( √2(16.97 - 0.7)/0.1(16.97-0.7))]

Average current = Iavg = 0.232 A

(d) What is the peak diode current?

Peak current = Ip = Vavg/R[ 1 + 2π( √2(Vp - Vd)/0.1(Vp-Vd))]

Peak current = Ip = 15.45/1000[ 1 + 2π( √2(16.97 - 0.7)/0.1(16.97-0.7))]

Peak current = Ip = 0.449 A

case (ii) 1% of the peak output:

(a) What average output voltage results?

Vavg = 16.97 - 0.7 - 0.5[0.01(16.97 - 0.7)]

Vavg = 16.18 V

(b) What fraction of the cycle does the diode conduct?

ω = √2*0.01(Vp-Vd)/Vp - Vd

ω = √2*0.01(16.97-0.7)/16.97 - 0.7

ω = 0.1417 rad

Conduction of diode = (0.1417/2π)*100

Conduction of diode = 2.25 %

(c) What is the average diode current?

Average current = Iavg = 16.18/1000[ 1 + π( √2(16.97 - 0.7)/0.01(16.97-0.7))]

Average current = Iavg = 0.735 A

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Peak current = Ip = 16.18/1000[ 1 + 2π( √2(16.97 - 0.7)/0.01(16.97-0.7))]

Peak current = Ip = 1.453 A

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