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marissa [1.9K]
3 years ago
15

A teacher tells her​ students, "When you do your math homework​ assignments, you must use white lined paper.​ Please, no​ tear-o

uts from spiral notebookslong dashI ​won't accept them under any​ circumstances! Divide each page with a vertical​ line, always putting the problem on the left side and your solution on the right side. Use a number two pencillong dashnothing ​lighter, and definitely no​ ink! I expect all erasures to be thorough and​ clean." Which one of the following basic needs is the teacher most likely to be putting in jeopardy​ here?
Engineering
1 answer:
Sindrei [870]3 years ago
8 0

Answer:

The white lined paper

Explanation:

The teacher is most likely putting the while line paper in jeopardy because of the detail process involved in taking care of the paper prior to the submission of the home work.

The fact that a mistake must not be visible due to the instruction of every erasures being thorough and clean.  this can cause jeopardy to the paper.

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Compute the volume percent of graphite, VGr, in a 3.2 wt% C cast iron, assuming that all the carbon exists as the graphite phase
Yanka [14]

Answer:

The volume percentage of graphite is 10.197 per cent.

Explanation:

The volume percent of graphite is the ratio of the volume occupied by the graphite phase to the volume occupied by the graphite and ferrite phases. The weight percent in the cast iron is 3.2 wt% (graphite) and 96.8 wt% (ferrite). The volume percentage of graphite is:

\%V_{gr} = \frac{V_{gr}}{V_{gr}+V_{fe}} \times 100\,\%

Where:

V_{gr} - Volume occupied by the graphite phase, measured in cubic centimeters.

V_{fe} - Volume occupied by the graphite phase, measured in cubic centimeters.

The expression is expanded by using the definition of density and subsequently simplified:

\%V_{gr} = \frac{\frac{m_{gr}}{\rho_{gr}} }{\frac{m_{gr}}{\rho_{gr}}+\frac{m_{fe}}{\rho_{fe}}}\times 100\,\%

Where:

m_{fe}, m_{gr} - Masses of the ferrite and graphite phases, measured in grams.

\rho_{fe}, \rho_{gr} - Densities of the ferrite and graphite phases, measured in grams per cubic centimeter.

\%V_{gr} = \frac{1}{1+\frac{\frac{m_{fe}}{\rho_{fe}} }{\frac{m_{gr}}{\rho_{gr}} } }\times 100\,\%

\%V_{gr} = \frac{1}{1 + \left(\frac{\rho_{gr}}{\rho_{fe}} \right)\cdot\left(\frac{m_{fe}}{m_{gr}} \right)} \times 100\,\%

If \rho_{gr} = 2.3\,\frac{g}{cm^{3}}, \rho_{fe} = 7.9\,\frac{g}{cm^{3}}, m_{gr} = 3.2\,g and m_{fe} = 96.8\,g, the volume percentage of graphite is:

\%V_{gr} = \frac{1}{1+\left(\frac{2.3\,\frac{g}{cm^{3}} }{7.9\,\frac{g}{cm^{3}} } \right)\cdot \left(\frac{96.8\,g}{3.2\,g} \right)} \times 100\,\%

\%V_{gr} = 10.197\,\%V

The volume percentage of graphite is 10.197 per cent.

5 0
3 years ago
Leland wants to work in a Production career operating heavy machinery. Which type of education or training should Leland seek?
zhenek [66]

Answer:

it is indeed C

Explanation:

4 0
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The electron beam in a TV picture tube carries 1015 electrons per second. As a design engineer, determine the voltage needed to
leonid [27]

Answer:

The voltage needed to accelerate the electron beam is 2.46 x 10^16 Volts

Explanation:

The rate of electron flow is given as:

q = 1015 electrons per second

The total current is given by:

Total Current = (Rate of electron flow)(Charge on one electron)

Total Current = I = (1015 electrons/s)(1.6 x 10^-19 C/electron)

I = 1.624 x 10^-16 A

Now, we know that electric power is given as:

Electric Power = Current x Voltage

P = IV

V = P/I

V = 4 W/1.624 X 10^-16 A

<u>V = 2.46 x 10^16 Volts</u>

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