Answer:
performance bonuses
Explanation:
Create a simulation on multisim of over-voltage & under-voltage protection circuit.
Answer:protection circuit overload as usual
circuit overload
Explanation:
it is ery easy for wverone
CAN I GET ANSWERS PLEASE, I TRY THE FORMULA AND MY TEACHER HAS A FAMILY EMERGENCY SO HE CANT RESPOND BACK SO PLEASE HELP ME
Explanation:
I won't answer each of these, but will give you an explaination of how to solve for each.
You'll need to use Ohm's Law and Kirchhoff's Voltage Law.
Remember Ohm's Law as [tex]V = IR[/tex].
Kirchoff's Voltage Law says that the sum of voltages for a given circuit "loop" must equal zero. In the circuit shown, this means that the voltage provided by the battery (E_T) equals the voltage drop across each of the three resisters in the loop.
[tex]E_T = E_1 + E_2 + E_3[/tex]
A couple of other helpful notes:
These three resistors are in series which means that the current flowing through them is equal.
So it is easy to see that... [tex]V = IR \rightarrow E_1 = 4*2 = 8 V[/tex]
Solve for the voltage across E_2 and E_3. The sum of the three voltages equals the voltage of the battery (E_T).
Problem 3. The uniform beam is supported by two rods AB and CD that have cross-sectional areas of 10 mm2 and 15mm2, respectively. Determine the intensity w of the distributed load so that the average normal stress in each rod does not exceed 300 kPa.
Answer:
hello the diagram related to your question is missing attached below is the required diagram.
The value of AC = 6m
answer 2.25 N
Explanation:
From the diagram attached below
average intensity of loading = w/2
Total load due to linearly varying load = [tex]\frac{wL}{2}[/tex]
w = maximum intensity of load , L = span of varying load
since AC = 6m( L ) then
Total load due to linearly varying load = 3w ------ ( 1 )
attached below is the detailed solution of the given problem
In a certain chemical plant, a closed tank contains ethyl alcohol to a depth of 71 ft. Air at a pressure of 17 psi fills the gap at the top of the tank. Determine the pressure at a closed valve attached to the tank 10 ft above its bottom.
Answer:
the pressure at a closed valve attached to the tank 10 ft above its bottom is 37.88 psi
Explanation:
Given that;
depth 1 = 71 ft
depth 2 = 10 ft
pressure p = 17 psi = 2448 lb/ft²
depth h = 71 ft - 10 ft = 61 ft
we know that;
p = P_air + yh
where y is the specific weight of ethyl alcohol ( 49.3 lb/ft³ )
so we substitute;
p = 2448 + ( 49.3 × 61 )
= 2448 + 3007.3
= 5455.3 lb/ft³
= 37.88 psi
Therefore, the pressure at a closed valve attached to the tank 10 ft above its bottom is 37.88 psi
The device shown below contains 2 kg of water. The cylinder is allowed to fall 800 m during which the temperature of the water increases by 2.4°C. Some amount of water is added to the container and the experiment is repeated. All other values remain constant. This time the temperature of the water increases by 1.2°C. How much water was added to the container?
Answer:
m_added = 2 kg
Explanation:
From the question, we are told that the cylinder is allowed to fall 800 m in height. Thus, the potential energy will be converted into heat energy which will increase the temperature of water .
Now, let the mass of the falling cylinder be denoted by "m1" and let h be the height of fall.
Thus;
Formula for potential energy = mgh
Thus, as said earlier it's converted to heat generated. So heat generated = m1gh
Now let's calculate the heat absorbed;
heat absorbed = (m2)cΔt
Where;
ΔT is change in temperature
c is specific heat of water .
m2 is mass of water
Heat absorbed = heat generated
Thus;
(m2)cΔt = m1gh
Δt = m1gh/(m2•c)
Now, in both cases of the water and cylinder, m1, g , h and c are constant
Thus, we have;
Δt = (m1gh/m2) × 1/c
Where;
(m1gh/m2) is denoted as a constant k.
Thus;
Δt = K/m
For the first experiment, we have;
m = 2 kg
Δt = 2.4
Thus;
2.4 = K/2
Multiply both sides by 2 to get;
K = 4.8
For the second experiment, we have;
Δt = 1.2
Also,we have seen that K = 4.8
Thus;
Δt = K/m
Thus;
1.2 = 4.8/m
m = 1.2
m = 4 kg
Thus,mass added is;
m_added = 4 - 2
m_added = 2 kg
The production of carbon dioxide makes it unwise and unsafe to operate a tractor or any motor vehicle inside enclosed spaces such as a barn.
True
False
What is a height gage?
What is true about gaging?
How often do frequently used instruments need calibration?
What is a machinist rule?
Maych the unit of measurement to the vocabulary term
Explanation:
current- amperes
resistance- ohms
voltage- volts
electrical energy- kilowatt hours
electrical power- watts
hope it helps!
A sample of blue gray soilwas obtained from Paine Field in Everett, WA to determine its Atterberg Limits and classify itssoil type according to the Unified Soil Classification system.84% of the sample passed through the #200 sieve with 16% retained on the #4 sieve. For one of the PL determinations, the wet weight + dish = 11.53 grams and the dry weight + dish = 10.49 grams. The dish only weighed 4.15 grams. Compute the plastic limit. Three determinations of the liquid limit were made. For 17 blows, the water content was 49.8%; for 26 blows, the water content was 47.5%; and for 36 blows, the water content was 46.3%.
Required:
Evaluate the soil type, indicate the information on a plasticity chart,and give the Unified Soil Classification System (USCS) symboland the proper narrative description.
Solution :
1. PL [tex]$= \frac{11.53 - 10.49}{10.49 - 4.5} \times 100$[/tex]
= 17.36 %
2. Liquid limit at 25 number of blows,
[tex]$LL = \frac{(26-25)\times 49.8 + (25 - 2.17) \times 47.5}{26-17}$[/tex]
[tex]$= 47.75 $[/tex] %
Since 84 % of the passes the number 200 sieve the soil is immediately fine grained and MI, OI or CI because LL = 47.75 % > 35%
Therefore, the Plasticity index, PI [tex]$=47.75 - 17.36 = 30.39$[/tex] %
Equation of A-line [tex]$= 0.73 \times [47.75 - 20]$[/tex]
[tex]$=20.25 $[/tex] %
Thus the soil lies above the A-line, that is soil is medium compressible clay ---- CI
Find the value of v [v] that maximizes the power dissipated in the voltage source. You may use the Thevenin Equivalent Circuit Theorem. {Answer Format: answer in voltage (V) }
Answer:
Hello the circuit related to this question is missing attached below is the missing circuit
answer : 1.5 V
Explanation:
attached below is the detailed solution to the question
Applying Thevenin equivalent Circuit Theorem
First we perform Nodal analysis at Vth and draw the equivalent Thevenin circuits
Hence for a maximum power triangle
[tex]V_{RL} = \frac{V_{th} }{2} = \frac{-3}{2}[/tex] = -1.5
therefore the value of v that maximizes the power dissipated in the voltage source
V = [tex]-V_{RL}[/tex] = 1.5 V
List 3 specific things you can do with a drill press (not just cut holes in wood)
Answer:
tapping, horizontal dusting, slash drilling
Explanation:
University administrators have developed a Markov model to simulate graduation rates at their school. Students might drop out, repeat a year, or move on to the next year. Students have a 3% chance of repeating the year. First-years and sophomores have a 6% chance of dropping out. For juniors and seniors, the drop-out rate is 4%. Compute the probability that a student who starts as a 1st-year eventually graduates the university.
Solution :
The percentage of the students who have a chance of repeating their current year = 3%
The drop out students for the first year and the sophomores = 6%
Drop out rate of first year and the seniors = 4%
Now for the state space :
S = { first year(1), sophomores(2), juniors(3), seniors(4), graduates(G), Dropouts(D) }
Therefore
the first year students are indicated as '1'
Sophomores are indicated as '2'
Juniors are indicated as '3'
Seniors are indicated as '4
Graduates are indicated as 'G'
Dropouts are indicated as 'D'
The transition diagram is attached below.
The probability of the students who have the chance of repeating their current year = 3/100 = 0.03
Probability of first year dropouts and sophomores = 6/100 = 0.06
Probability of dropout rate of juniors and seniors = 4/100 = 0.04
Therefore, the probability matrix can be made as :
1 2 3 4 G D
[tex]\begin{matrix}1\\ 2\\ 3\\ 4\\ G\\ D\end{matrix}[/tex] [tex]\begin{bmatrix}0.03 & 0.91 & 0 & 0 & 0 & 0.06\\ 0& 0.03 & 0.91 & 0 & 0 & 0.06\\ 0& 0 & 0.03 & 0.93 & 0 & 0.04\\ 0& 0 & 0 & 0.03 & 0.93 & 0.04\\ 0& 0 & 0 & 0 & 1 & 0\\ 0& 0 & 0 & 0 & 0 & 1\end{bmatrix}[/tex]
Here, G represents 'graduates' and D represents 'Dropouts.'
Why are Gas cars Bad?(cons) give me reasons why gasoline cars are bad
Thx if u help
Answer: Gasoline cars are bad because
Explanation: They give off carbon emissions and pollute the ozone layer causing the greenhouse effect which leads to global warming
The pacific ocean surrounded by extremely high earthquake and volcanic activity the Atlantic ocean by contrast has little activity of this nature based on the Maps what attributes of the Atlantic ocean make it likely for earthquakes to occur or volcanoes to form
Answer: There are no major convergent boundaries in the Atlantic, which is where these events are more likely to occur
Explanation:
Water flows through this orifice meter by gravity. The orifice diameter is 50 cm and water height difference is 10 m. If the contraction coefficient is 0.62 and velocity coefficient is 0.90, the flow rate through the orifice is most nearly:
Answer:
Q ≅ 1.53 m³/s
Explanation:
From the given information:
The flow rate of the orifice is:
[tex]v = c_v \sqrt{2gh}[/tex]
[tex]v = 0.90 \times \sqrt{2*9.81 * 10}[/tex]
where;
[tex]Q = c_d \times \sqrt{2gh} \times A[/tex]; &
[tex]c_d = c_c \times c_v[/tex]
∴
[tex]Q = c_c \times c_v \sqrt{2gh} \times \dfrac{\pi}{4}\times d^2[/tex]
[tex]Q = 0.90 \times 0.62 \sqrt{2*9.81*10} \times \dfrac{\pi}{4}\times 0.5^2[/tex]
[tex]Q = 0.558 \times 14.00714104 \times 0.1963495408[/tex]
Q ≅ 1.53 m³/s
When the mass of an object decreases, the force of gravity does what?
Answer:
decreases too
Explanation:
Since the gravitational force is directly proportional to the mass of both interacting objects, more massive objects will attract each other with a greater gravitational force. So as the mass of either object increases, the force of gravitational attraction between them also increases.
Answer:
decreases
Explanation:
Since the gravitational force is directly proportional to the mass of interacting objects, more massive objects will attract each other with a greater gravitational force. So as the mass of either object increases (decreases), the force of gravitational attraction between them also increases (decreases).
When a pungent odor is detected during a sealed system recovery and/or repair which of the following is most likely
Answer: A leak in the system
Explanation:
For the velocity fields given below, determine:
a) whether the flow field is one-, two-, or three-dimensional,
b) whether the flow is steady or unsteady, and why. (The quantities a and b are constants.)
1. V = [ay2e−bt]i
2. V = ax^2i + bxj + ck
3. V = axyi - bytj
4. V = axi - byj + ctk
Answer and Explanation: A fluid flow is steady when its properties (velocity, pressure and others) do not change over time: P = P(x,y,z), with P being any of the properties.
So, an unsteady flow does depend upon time: P = P(x,y,z,t).
1. V = [[tex]ay2e^{-bt}[/tex]]i : This velocity field is one-dimension, because there is only the component in the x-direction and unsteady, because it is dependent upon the variable time (t);
2. V = [tex]ax^{2}i+bxj+ck[/tex] : is a three-dimensional field, because there is one component for each direction (x, y and z) and is steady, since it's independent of time;
3. V = axyi - bytj : is a 2-dimensional field, and since it changes with time, it is unsteady;
4. V = axi - byj + ctk : is a 3-dimensional field and the flow is unsteady;
if a current of 5 amps flows through a resistance of 40 ohms, what is the voltage across that resistor
Answer:
200V
Explanation:
I = 5A
R = 40Ω
V = IR = (5)(40) = 200V
Acertain foundation will experience a bearing capacity failure when it is subjected to a downward load of 2200 kN. Using ASD with a factor of safety of 3, determine the maximum allowable loadthat will satisfy geotechnical ULS requirements.
Answer:
550kN
Explanation:
We are required to find the maximum allowable load that will satisfy geotechnical ULS requirements
P' = pu/FOS ---1
Pu = design load
FOS = factor of safety
We get the design of safety
Pu = 0.75p
= 0.75 x 2200
= 1650kN
Factor of safety = 3
Design load = 1650
P' = 1650/3
= 550kN
So we have the maximum allowable load to be 550kN
how does a motion sensor work?
Answer:
A motion sensor uses one or multiple technologies to detect movement in an area. When a sensor detects motion, it sends a signal to your security systems control panel, which connects to your monitoring panel system. This alerts you and the monitoring center to a potential threat in your home.
Hope It Helps You................
discuss referencing techniques
Answer:Referencing is a standardised method of acknowledging the sources of information and ideas you have used.
Explanation:Depending on the way in which they record sources, scholarly reference styles can be divided into three main categories: documentary notes styles, parenthetical (or author-date) styles, and numbered styles. Within each category there are several, slightly different reference styles.
Parenthetical citations give a short reference in parentheses directly in the text.Numerical citations give only a number that corresponds to a footnote, endnote, or reference list entry.Referencing allows you to acknowledge the contribution of other writers and researchers in your work. Any university assignments that draw on the ideas, words or research of other writers must contain citations. Referencing is also a way to give credit to the writers from whom you have borrowed words and ideas.
What is the normal balance side of an asset?
