Explanation:
current- amperes
resistance- ohms
voltage- volts
electrical energy- kilowatt hours
electrical power- watts
hope it helps!
What is the normal balance side of an asset?
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.
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
What is a height gage?
What is true about gaging?
How often do frequently used instruments need calibration?
What is a machinist rule?
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
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.'
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;
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).
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
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
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
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
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
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).
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
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