The FDA! (food and drug administration)
The federal agency issues the model food code is FDA
How to determine the agencyThe U. S. Food and Drug Administration (FDA) publishes the Food Code, a model that assists food control jurisdictions at all levels of government by providing them with a scientifically sound technical and legal basis for regulating the retail and food service segment of the industry (restaurants and grocery stores.
The Food Code is a model for safeguarding public health and ensuring food is unadulterated and honestly presented when offered to the consumer. It represents FDA's best advice for a uniform system of provisions that address the safety and protection of food offered at retail and in food service.
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Three single-phase, 10 kVA, 460/120 V, 60 Hz transformers are connected to form a three-phase 460/208 V transformer bank. The equivalent impedance of each transformer referred to HV side is 1.0 +j2.0 Ω. The transformer delivers 20 kW at 0.8 pf leading. Answer the following questions:
(a) Draw a schematic diagram showing the transformer connection.
(b) Determine the magnitude of transformer primary and secondary winding currents.
(c) Determine the primary voltage magnitude for this operating condition. Determine the voltage regulation
Answer:
A) attached below
B) I₁ = 18.1 A , I₂ = 69.39 A
C) V( magnitude) = 454.5 ∠ 5.04° V , Voltage regulation = ≈ -1.2%
Explanation:
A) Schematic diagram attached below
attached below
B) magnitude of primary and secondary winding currents
I₂ ( secondary current ) = P / √3 * VL * cos∅ ---------- ( 1 )
VL = Line voltage = 208
cos∅ ( power factor ) = 0.8
P = 20 * 10^3 watts
insert values into equation 1
I₂ = 69.39 A
I₁ ( primary current ) = I₂V2 / V1
I₁ = ( 69.39 * 120 ) / 460 = 18.1 A
C ) Calculate the Primary voltage magnitude and the Voltage regulation
V(magnitude ) = Vp + ( I₁ ∠∅ ) Req ( 1 + j2 = 2.24 ∠63.43° )
= 460 + ( 18.1 * cos^-1 (0.8) ) ( 1 + j2 )
= 460 + 40.544 ∠ 100.3°
∴ V( magnitude) = 454.5 ∠ 5.04° V
Voltage regulation
= ((Vmag - V1) / V1 )) * 100
= (( 454.5 - 460 / 460 )) * 100
= -1.195 % ≈ -1.2%
Suppose you have two arrays: Arr1 and Arr2. Arr1 will be sorted values. For each element v in Arr2, you need to write a pseudo code that will print the number of elements in Arr1 that is less than or equal to v. For example: suppose you are given two arrays of size 5 and 3 respectively. 5 3 [size of the arrays] Arr1 = 1 3 5 7 9 Arr2 = 6 4 8 The output should be 3 2 4 Explanation: Firstly, you should search how many numbers are there in Arr1 which are less than 6. There are 1, 3, 5 which are less than 6 (total 3 numbers). Therefore, the answer for 6 will be 3. After that, you will do the same thing for 4 and 8 and output the corresponding answers which are 2 and 4. Your searching method should not take more than O (log n) time. Sample Input Sample Output 5 5 1 1 2 2 5 3 1 4 1 5 4 2 4 2 5
Answer:
The algorithm is as follows:
1. Declare Arr1 and Arr2
2. Get Input for Arr1 and Arr2
3. Initialize count to 0
4. For i in Arr2
4.1 For j in Arr1:
4.1.1 If i > j Then
4.1.1.1 count = count + 1
4.2 End j loop
4.3 Print count
4.4 count = 0
4.5 End i loop
5. End
Explanation:
This declares both arrays
1. Declare Arr1 and Arr2
This gets input for both arrays
2. Get Input for Arr1 and Arr2
This initializes count to 0
3. Initialize count to 0
This iterates through Arr2
4. For i in Arr2
This iterates through Arr1 (An inner loop)
4.1 For j in Arr1:
This checks if current element is greater than current element in Arr1
4.1.1 If i > j Then
If yes, count is incremented by 1
4.1.1.1 count = count + 1
This ends the inner loop
4.2 End j loop
Print count and set count to 0
4.3 Print count
4.4 count = 0
End the outer loop
4.5 End i loop
End the algorithm
5. End
What type of sensor is a crankshaft position sensor?
In addition to passing an ASE certification test, automotive technicians must have __________ year(s) of on the job training or __________ year(s) of on the job training and a two-year degree in automotive repair to qualify for certification.
4. Which of the following requires an endorsement on your CDL?
A. Air brakes
B. Double/triple trailers
C. Manual transmission
D. All of the above
The components which requires an endorsement on your Commercial Driver License (CDL) are: D. all of the above.
What is CDL?CDL is an acronym for Commercial Driver License and it can be defined as a category of driver's license that is issued to an individual, in order to indicate that he or she is qualified to operate and drive certain types of automobile vehicles or to use them in specific ways.
In the United States of America, the components which requires an endorsement on your Commercial Driver License (CDL) are:
Air brakesDouble/triple trailersManual transmissionRead more on CDL here: https://brainly.com/question/14326814
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Blood pressure is conventionally measured in the dimensions of millimeters in a column of mercury, and the readings are expressed as two numbers, for example, 120 and 80. The first number is called the systolic value, and it is the maximum pressure developed as the heart contracts. The second number (called the diastolic reading) is the pressure when the heart is at rest. In the units of kPa and psi, what is the difference in pressure between the given systolic and diastolic readings? The density of mercury is 13.54 Mg/m3.
Answer:
- the difference in pressure between the given systolic and diastolic readings in KPa is 5.313 KPa
- the difference in pressure between the given systolic and diastolic readings in psi is 0.77 psi
Explanation:
Given the data in the question;
blood pressure reading = 120 and 80 { systolic and diastolic }
To determine the difference in pressure between the two readings, we use the equation as follows;
change in pressure ΔP = p × g × h
where p is mercury density, g is acceleration due to gravity and h is difference of height in mercury column.
Frist,
difference of height in mercury column h = 120 - 80 = 40 mm = 0.04 m
given that; The density of mercury is 13.54 Mg/m³ = 13.54 × 10³ kg/m³
Not that Mg is Megagrams not Milligrams }
we know that g = 9.81 m/s²
so we substitute into our equation;
change in pressure ΔP = (13.54 × 10³) × 9.81 × 0.04
ΔP = 5313.096 kg/m-s² ≈ 5313.096 N/m²
ΔP = 5.313 KPa
Therefore, the difference in pressure between the given systolic and diastolic readings in KPa is 5.313 KPa.
In psi,
ΔP = 5.313 KPa
ΔP = 5313 Pa
ΔP = 5313 pa × ( 1.45 × 10⁻⁴ psi / 1 Pa )
ΔP = 0.770385 psi ≈ 0.77 psi
Therefore, the difference in pressure between the given systolic and diastolic readings in psi is 0.77 psi