Based on the engineering analysis, the relationship between the voltages of the output and adjust pins of LM1117 when in normal operation is "the LM1117 adjustable version establishes a 1.25V reference voltage, (VREF) between the output and the adjust terminal."
The LM1117 generally has a 1.2V at 800mA of load current.
The LM1117 is known to be a progression of low dropout linear voltage regulators. It is adjustable and be assigned between the output voltage of 1.25 to 13.8 V using just two external resistors.
Hence, in this case, it is concluded that the correct answer is "the LM1117 adjustable version establishes a 1.25V reference voltage, (VREF), between the output and the adjust terminal.
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How to calculate total photon flux incident on the photodiode
Answer:
Use this PDF file it should help you calculate total photon flux incident on the photodiode.
Explanation:
It's a file on lecture notes that help people learn about engineering and maybe in the lecture notes it might have information about how to calculate total photon flux incident on the photodiode
true/False
1.
Around 50 percent of all sales of farm commodities are from dairy
products.
Answer:true
Explanation:
It is true that around 50 percent of all sales of farm commodities are from dairy products.
What are dairy products?Dairy products or milk products, also known as lacticinia, are milk-based foods.
Cows, water buffalo, nanny goats, and ewes are the most common dairy animals. Dairy products include common grocery store foods such as yogurt, cheese, and butter in the Western world.
Milk, yogurt, cheese, lactose-free milk, and fortified soy milk and yogurt are all part of the Dairy Group. The Dairy Group excludes milk-based foods with low calcium and a high fat content. Cream cheese, sour cream, cream, and butter are all examples of this.
Dairy products account for roughly half of all farm commodity sales.
Thus, the given statement is true.
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Consider a 25-mm-diameter and 15-m-long smooth tube that is maintained at a constant surface temperature. Fluids enter the tube at 50°C with a mass flow rate of 0.01 kg/s. Determine the tube surface temperatures necessary to heat water, engine oil, and liquid mercury to the desired outlet temperature of 150°C
Answer: ⚠this is not my answer it from hamzaahmeds⚠
Water: h = 35.53 W/m².k
Engine oil: h = 18.84 W/m².k
Mercury: h = 1.19 W/m².k
Explanation:
Assuming the steady state, one-dimensional heat flow, it is clear that the added to the fluid by tube heat will be equal to the heat transfer through convection outside the tube.
Therefore,
mCΔT = hAΔT
mC = hA
h = mC/A
where,
h = convection coefficient
m = mass flow rate = 0.01 kg/s
C = specific heat capacity of fluid
A = surface area of tube = 2πrL = 2π(0.0125 m)(15 m) = 1.178 m²
FOR WATER:
C = 4186 J/Kg.k
Therefore,
h = (0.01 kg/s)(4186 J/Kg.k)/(1.178 m²)
h = 35.53 W/m².k
FOR ENGINE OIL:
C = 2220 J/Kg.k
Therefore,
h = (0.01 kg/s)(2220 J/Kg.k)/(1.178 m²)
h = 18.84 W/m².k
FOR LIQUID MERCURY:
C = 140 J/Kg.k
Therefore,
h = (0.01 kg/s)(140 J/Kg.k)/(1.178 m²)
h = 1.19 W/m².k
