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rewona [7]
3 years ago
11

A technician is troubleshooting a problem. The technician tests the theory and determines the theory is confirmed. Which of the

following should be the technician’s NEXT step?
A. Implement the solution.
B. Document lessons learned.
C. Establish a plan of action.
D. Verify full system functionality.
Physics
1 answer:
vladimir1956 [14]3 years ago
3 0

Answer:

b)  Document lessons learned.

Explanation:

First he should do documentation

then C

then D

then A

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They can fight the infection but not the disease
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3 years ago
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A hollow conducting sphere with an outer radius of 0.295 m and an inner radius of 0.200 m has a uniform surface charge density o
IrinaK [193]

Answer:

a. 6.032\times10^{-6}C/m^2

b.6.816\times10^5N/C

Explanation:

#Apply  surface charge density, electric field, and Gauss law to solve:

a. Surface charge density is defined as charge per area denoted as \sigma

\sigma=\frac{Q}{4\pi r_{out}^2}, and the strength of the electric field outside the sphere E=\frac{\sigma _{new}}{\epsilon _o}

Using Gauss Law, total electric flux out of a closed surface is equal to the total charge enclosed divided by the permittivity.

\phi=\frac{Q_{enclosed}}{\epsilon_o}\\\\\sigma=\frac{Q}{4\pi r_{out}^2}\\\\\sigma=\frac{0.370\times 10^{-6}}{4\pi \times (0.295m)^2}\\\\=3.383\times10^{-7}C/m^2  #surface charge outside sphere.

\sigma_{new}=\sigma_{s}-\sigma\\\\\sigma_{new}=6.37\times10^{-6}C/m^2-3.383\times10^{-7}C/m^2\\\\\sigma_{new}=6.032\times10^{-6}C/m^2

Hence, the new charge density on the outside of the sphere is 6.032\times10^{-6}C/m^2

b. The strength of the electric field just outside the sphere is calculated as:

From a above, we know the new surface charge to be 6.032\times10^{-6}C/m^2,

E=\frac{\sigma _{new}}{\epsilon _o}\\\\=\frac{6.032\times10^{-6}C/m^2}{\epsilon _o}\\\\\epsilon _o=8.85\times10^{-12}C^2/N.m^2\\\\E=\frac{6.032\times10^{-6}C/m^2}{8.85\times10^{-12}C^2/N.m^2}\\\\E=6.816\times10^5N/C

Hence, the strength of the electric field just outside the sphere is 6.816\times10^5N/C

5 0
3 years ago
The Ha line of the Balmer series is emitted in the transition from n-3 to n 2. Compute the wavelength of this line for H and 2H.
Citrus2011 [14]

Explanation:

According to Rydberg's formula, the wavelength of the balmer series is given by:

\frac{1}{\lambda}=R(\frac{1}{2^2}-\frac{1}{3^2})

R is Rydberg constant for an especific hydrogen-like atom, we may calculate R for hydrogen and deuterium atoms from:

R=\frac{R_{\infty}}{(1+\frac{m_e}{M})}

Here, R_{\infty} is the "general" Rydberg constant, m_e is electron's mass and M is the mass of the atom nucleus

For hydrogen, we have, M=1.67*10^{-27}kg:

R_H=\frac{1.09737*10^7m^{-1}}{(1+\frac{9.11*10^{-31}kg}{1.67*10^{-27}kg})}\\R_H=1.09677*10^7m^{-1}

Now, we calculate the wavelength for hydrogen:

\frac{1}{\lambda}=R_H(\frac{1}{2^2}-\frac{1}{3^2})\\\lambda=[R_H(\frac{1}{2^2}-\frac{1}{3^2})]^{-1}\\\lambda=[1.0967*10^7m^{-1}(\frac{1}{2^2}-\frac{1}{3^2})]^{-1}\\\lambda=6.5646*10^{-7}m=656.46nm

For deuterium, we have M=2(1.67*10^{-27}kg):

R_D=\frac{1.09737*10^7m^{-1}}{(1+\frac{9.11*10^{-31}kg}{2*1.67*10^{-27}kg})}\\R_D=1.09707*10^7m^{-1}\\\\\lambda=[R_D(\frac{1}{2^2}-\frac{1}{3^2})]^{-1}\\\lambda=[1.09707*10^7m^{-1}(\frac{1}{2^2}-\frac{1}{3^2})]^{-1}\\\lambda=6.5629*10^{-7}=656.29nm

5 0
3 years ago
Plz answer it......<br>.<br>plz​
Lunna [17]

Answer:

Definitely C

Hope it helped!

(And I hope it's not wrong...lol)

7 0
3 years ago
a 0.11kg moving hockey puck is caught by a 80kg goalie who then slides on the frictionless ice at 0.076 m/s. What was the initia
aleksandrvk [35]

Answer:

55mph

Explanation:

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4 years ago
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