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4 years ago

14

a digital multimeter is set to read dc volts on the 4 volt scale the meter leads are connected to a 12 volt battery what will th

e display read

2 answers:

Rainbow [258]4 years ago

8 0

Answer:

ssusscribanse mi ca na ljun a

Explanation:

IceJOKER [234]4 years ago

5 0

OR. Overange, or whatever icon the manufacturer chooses

Auto ranging meter saves rotating to a range scale higher than tha 12 volts.

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According to Moore and Marra's (2005) case study, which observed two online courses, students in the first course implemented a

Archy [21]

Your question is incomplete, please let me assume this to be your complete question;

ORIGINAL SOURCE:

When instructors are creating discussion board activities for online courses, at least two questions must be answered. First, what is the objective of the discussions? Different objectives might be to create a "social presence" among students so that they do not feel isolated, to ask questions regarding assignments or topics, or to determine if students understand a topic by having them analyze and evaluate contextual situations. Based on the response to this question, different rules might be implemented to focus on the quality of the interaction more so than the quantity. The second question is, how important is online discussions in comparison to the other activities that students will perform? This question alludes to the amount of participation that instructors expect from students in online discussions along with the other required activities for the course. If a small percentage of student effort is designated for class participation, our results show that it can affect the quality and quantity of interactions.

References:

Moore, J. L., & Marra, R. M. (2005) A comparative analysis of online discussion participation protocols.Journal of Research on Technology in Education, 38(2), 191-212.

STUDENT VERSION:

According to Moore and Marra's (2005) case study, which observed two online courses, students in the first course implemented a constructive argumentation approach while students in second course had less structure for their postings. As they stated, when instructors create online discussion board activities, they must answer at least two questions. These questions are: "What is the objective of the discussions?" And "How important are online discussions in comparison to the other activities that students will perform?". According to their findings, the discussion activities that were designed based on the answers to these questions can influence the quality and quantity of interactions (Moore & Marra, 2005).

References:

Moore, J. L., & Marra, R. M. (2005) A comparative analysis of online discussion participation protocols.Journal of Research on Technology in Education, 38(2), 191-212.

Which of the following is true for the students work;

Word-for-word plagiarism

Paraphrasing plagiarism

Not Plagiarism

ANSWER: IT IS NOT PLAGIARISM

Explanation: plagiarism is the act of extracting knowledge from someone's literature work, without acknowledging the literature work. In other words, this can be called a theft of knowledge, because when you failed to acknowledge the literature source that helped you to produce your paper work, it means you have claimed to be the original owner of that knowledge.

This is not a Plagiarism because the student has acknowledged the source of the knowledge in it's literature work. The original source and the student has cited the same literature work, this why their work looks similar but not exactly the same. So the student has not committed Plagiarism

4 0

4 years ago

A rigid canister with a radius of 5 in and a height of 10 in is filled with air. The initial pressure and temperature of air in

Elza [17]

Answer:1.458 Btu

Explanation:

Given

radius of canister $\left ( r\right )=5in$

Height of canister $\left ( h\right )=10 in.$

Initial pressure $\left ( P_i\right )=14.7 Psi$

Initia ltemprature $\left ( T_i\right )=70^{\circ}F$

Final pressure $\left ( P_f\right )=30Psi$

as canister is rigid therefore change in volume is zero

therefore

$\frac{P_i}{T_i}$ = $\frac{P_f}{T_f}$

$\frac{14.7}{70}$ = $\frac{30}{T_f}$

$T_f=142.85^{\circ}F$

volume of canister= $\pi \times r^{2}\times h$

= $\pi \times 5^{2}\times 10=250\pi$

volume of canister=12872,038.8 $mm^3$

now calculating mass of air

PV=mRT

substituting values

$\left ( 14.7Psi\right )\left ( 12872,038.8 mm^3\right )=m\left ( 0.287\right )\left ( 70^{\circ}F\right )$

m=21.0192 gm

Therefore heat transferred = $mc_p\left ( T_f-T_i\right )$

= $21.0192\times 10^{-3}\times \left ( 142.857-70\right )$

=1539.052J=1.458Btu

3 0

4 years ago

What are the four stages of BIM?

emmasim [6.3K]

Answer:A few short years ago, the use of building information modeling (BIM) was largely the domain of only the most sophisticated, forward-thinking and, some would say, well-financed project teams. Today, BIM has moved further into the mainstream and is being touted as a “transformative evolution” in the industry. For all the buzz, one thing is clear: Not all BIM is created equal. An early BIM adopter, DPR has always viewed technology as one key piece of a much larger puzzle when it comes to extracting the greatest cost savings and building the most value into a pro

Explanation:i hop this helps

3 0

3 years ago

1.00-L insulated bottle is full of tea at 90.08°C. You pour out one cup of tea and immediately screw the stopper back on the bot

liberstina [14]

Answer:

T_{f} = 90.07998 ° C

Explanation:

This is a calorimetry process where the heat given by the Te is absorbed by the air at room temperature (T₀ = 25ºC) with a specific heat of 1,009 J / kg ºC, we assume that the amount of Tea in the cup is V₀ = 100 ml. The bottle being thermally insulated does not intervene in the process

Qc = -Qb

M $c_{e_Te}$ (T₁ - $T_{f}$ ) = m $c_{e_air}$ (T_{f}-T₀)

Where M is the mass of Tea that remains after taking out the cup, the density of Te is the density of water plus the solids dissolved in them, the approximate values are from 1020 to 1200 kg / m³, for this calculation we use 1100 kg / m³

ρ = m / V

V = 1000 -100 = 900 ml

V = 0.900 l (1 m3 / 1000 l) = 0.900 10⁻³ m³

V_air = 0.100 l = 0.1 10⁻³ m³

Tea Mass

M = ρ V_te

M = 1100 0.9 10⁻³

M = 0.990 kg

Air mass

m = ρ _air V_air

m = 1.225 0.1 10⁻³

m = 0.1225 10⁻³ kg

(m c_{e_air} + M c_{e_Te}) T_{f}. = M c_{e_Te} T1 - m c_{e_air} T₀

T_{f} = (M c_{e_Te} T₁ - m c_{e_air} T₀) / (m c_{e_air} + M c_{e_Te})

Let's calculate

T_{f} = (0.990 1100 90.08– 0.1225 10⁻³ 1.225 25) / (0.1225 10⁻³ 1.225 + 0.990 1100)

T_{f} = (98097.12 -3.75 10⁻³) / (0.15 10⁻³ +1089)

T_{f} = 98097.11 / 1089.0002

T_{f} = 90.07998 ° C

This temperature decrease is very small and cannot be measured

3 0

3 years ago

50 points what shape is mars

Artyom0805 [142]

A sphere.............

5 0

3 years ago

Read 2 more answers