What type of circuit measurement is made by placing a meters test leads in parallel with a deenergized component

Answers

Answer 1

Answer: If the component is DE-energized, meaning the whole device is
powered down, then the only thing you can measure with the meter-
probes on both ends of the component is its resistance.

If you have a fancy, expensive meter, then maybe you could measure
the component's capacitance or inductance. I never had one of those.

The normal meter measures volts, amps, and ohms. If you touch
the probes to both ends of the component and the circuit is energized,
then you measure the voltage across the component. If the circuit is
DE-energized, then you're measuring the component's resistance.

(Note: You have to know which one you're measuring, and set up the
meter properly. For example, if the circuit is energized and you try to
measure resistance, it's possible that you could fry your meter.)

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Nitrogen makes up about what percent of a human's body weight?

Answers

Answer:

the answer is 3.3 %

Explanation:

The wind-chill index is modeled by the function W = 13.12 + 0.6215T − 11.37v0.16 + 0.3965Tv0.16 where T is the temperature (°C) and v is the wind speed (km/h). When T = 12°C and v = 18 km/h, by how much would you expect the apparent temperature W to drop if the actual temperature decreases by 1°C? (Round your answers to two decimal places.)

Answers

This question involves the concepts of derivative, apparent temperature, actual temperature,and wind speed.

The drop in apparent temperature will be "1.25°C".

The apparent temperature (W) is given in terms of actual temperature (T) and wind speed (v) is given by the following function:

$W = 13.12 + 0.6215\ T-11.37\ v^(0.16)+0.3965\ Tv^(0.16)$

Taking the derivative with respect to actual temperature, we get:

$(dW)/(dT)=0.6215+0.3965\ v^(0.16)\n\n$

where,

dW = drop in apparent temperatures = ?

dT = drop in actual temperature = - 1°C

v = wind speed = 18 km/h

Therefore,

$dW=(-1)(0.6215-0.3965(18)^(0.16))$

dW = - 1.25°C

Learn more about derivatives here:

brainly.com/question/9964510?referrer=searchResults

Answer:

Δw=1.25°C

Explanation:

Given that

$w=13.12 +0.6215 T-11.37 v^(0.16)+0.3965 T v^(0.16)$

Given that T= 12°C and v=19 km/h

Now to find the drop in the apparent temperature w

$(dw)/(dT)=0.6215 +0.3965v^(0.16)$

$(\Delta w)/(\Delta T)= 0.6215 +0.3965 v^(0.16)$

Now by putting the values v=19 km/hr and ΔT=1

$(\Delta w)/(1)=0.6215 +0.3965* 18^(0.16)$

Δw=1.25°C

So we can say that when temperature is decrease by 1°C then apparent temperature will decrease by 1.25°C at given velocity.

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Answers

Answer:

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Explanation:

2. Annealed low-carbon steel has a flow curve with strength coefficient of 75000 psi and strain-hardening exponent of 0.25. A tensile test specimen with a gauge length of 2 in. is stretched to a length of 3.3 in. Determine the flow stress and the average flow stress that the metal experienced during this deformation.

Answers

Answer:

Flow stress= 9390Psi

Average flow stress= 4173.33Psi

Explanation:

Given:

Strength Coefficient = 75000psi

Strain hardening Exponent = 0.25

Gauge length = 2inches

Stretch length = 3.3 inches

Flow stress,Yf = 75000 × ln(3.3/2) × 0.25

Yf = 75000× ln(1.65) × 0.25

Yf = 75000× 0.5008 × 0.25

Flow stress = 9390Psi

Average flow stress = 75000× 0.5008 × (0.25/2.25)

Average flow stress= 4173.33psi

A force of 240.0 N causes an object to accelerate at 3.2 m/s2. What is the mass of the object?

Answers

the mass would be 75kg

2. An electrical heater 200mm long and 15mm in diameter is inserted into a drilled hole normal to the surface of a large block of material having a thermal conductivity of 5W/m·K. Estimate the temperature reached by the heater when dissipating 25 W with the surface of the block at a temperature of 35 °C.

Answers

Answer:

The final temperature is 50.8degrees celcius

Explanation:

Pls refer to attached handwritten document

Answer: 50.63° C

Explanation:

Given

Length of heater, L = 200 mm = 0.2 m

Diameter of heater, D = 15 mm = 0.015 m

Thermal conductivity, k = 5 W/m.K

Power of the heater, q = 25 W

Temperature of the block, = 35° C

T1 = T2 + (q/kS)

S can be gotten from the relationship

S = 2πL/In(4L/D)

On substituting we have

S = (2 * 3.142 * 0.2) / In (4 * 0.2 / 0.015)

S = 1.2568 / In 53.33

S = 1.2568 / 3.98

S = 0.32 m

Proceeding to substitute into the main equation, we have

T1 = T2 + (q/kS)

T1 = 35 + (25 / 5 * 0.32)

T1 = 35 + (25 / 1.6)

T1 = 35 + 15.625

T1 = 50.63° C

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