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Alexandra [31]
3 years ago
6

Technician A says amperage cannot exist without both voltage and resistance. Technician B says if amperage is high, then you kno

w that one of two conditions is present low voltage or high resistance. Who is correct? Select one: O a. Technician A O b. Technician B c. Both Technicians A and B d. Neither Technician A nor B​
NEED HELP ASAP PLZ!!!
Engineering
1 answer:
Ivan3 years ago
8 0

Answer:

Technician A

Explanation:

Ohms law:  I= E/R so rest resistance must be present along with E/potential difference.  Even if just wire shorted together there is resistance but very little.

Tech B: Again ohms law.  Current flow is directly proportional to the voltage and inversely  proportional to R (resistance or impedance).

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How is the difference between science and engineering Best stated?
stiv31 [10]

Answer:Science is the body of knowledge that explores the physical and natural world. Engineering is the application of knowledge in order to design, build and maintain a product or a process

Explanation:

8 0
2 years ago
A safety interlock module operates by monitoring the voltage from the
In-s [12.5K]

Answer: its an Ignition coil

8 0
3 years ago
Two different fuels are being considered for a 2.5 MW (net output) heat engine which can operate between the highest temperature
sveta [45]

Answer:

If the heat engine operates for one hour:

a) the fuel cost at Carnot efficiency for fuel 1 is $409.09 while fuel 2 is $421.88.

b) the fuel cost at 40% of Carnot efficiency for fuel 1 is $1022.73 while fuel 2 is $1054.68.

In both cases the total cost of using fuel 1 is minor, therefore it is recommended to use this fuel over fuel 2. The final observation is that fuel 1 is cheaper.

Explanation:

The Carnot efficiency is obtained as:

\epsilon_{car}=1-\frac{T_c}{T_H}

Where T_c is the atmospheric temperature and T_H is the maximum burn temperature.

For the case (B), the efficiency we will use is:

\epsilon_{b}=0.4\epsilon_{car}

The work done by the engine can be calculated as:

W=\epsilon Q=\epsilon H_v\cdot m_{fuel} where Hv is the heat value.

If the average net power of the engine is work over time, considering a net power of 2.5MW for 1 hour (3600s), we can calculate the mass of fuel used in each case.

m=\frac{P\cdot t}{\epsilon H_v}

If we want to calculate the total fuel cost, we only have to multiply the fuel mass with the cost per kilogram.

TC=m\cdot c

8 0
3 years ago
A cone penetration test was carried out in normally consolidated sand, for which the results are summarized below: Depth (m) Con
Cerrena [4.2K]

Answer:

hello your question is incomplete attached below is the missing equation related to the question  

answer : 40.389° , 38.987° , 38° , 39.869° , 40.265°

Explanation:

<u>Determine the friction angle at each depth</u>

attached below is the detailed solution

To calculate the vertical stress = depth * unit weight of sand

also inverse of Tan = Tan^-1

also qc is in Mpa while σ0 is in kPa

Friction angle at each depth

2 meters = 40.389°

3.5 meters  = 38.987°

5 meters = 38.022°

6.5 meters = 39.869°

8 meters = 40.265°

6 0
3 years ago
1. Fatigue equations are based solely on theoretical assumptions. Experimental data is only used to verify the theory. a. True.b
Rainbow [258]

Answer:

1.  b. False

2. b. False

3.  b. False

4.  b. False

5. a. True

6. a. True

7.  b. False

8.  b. False

9. a. True

Explanation:

1. The fatigue properties of a material  are determined by series of test.

2. For most steels there is a level of fatigue limit below which a component will survive an infinite number of cycles, for aluminum and titanium a fatigue limit can not be defined, as failure will eventually occur after enough experienced cycles.

3. Although there is a cyclic stress, there are also stresses complex circumstances involving tensile to compresive and constant stress, where the solution is given into the mean stress and the stress amplitude or stress range, which is double the stress amplitude.

4. Low‐cycle fatigue is defined as few thousand cycles and high cycle fatigue is around more than 10,000 cycles.

5. The number of cycles for failure on brittle materials are less and determined compared with the ductile materials.

6.  The bending fatigue could be handled with specific load requirements  for uniform bending or axial fatigue of the same section size where the material near the surface is subjected to the  maximum stress, as in torsional fatigue, which can be performed on  axial-type specially designed machines also, using the proper fixtures if  the maximum twist required is small, in which linear motion is changed to rotational motion.

7.  A SN-Curve for a given material, is a plot displayed on logarithmic scales of the magnitude of an alternating stress in relation to the number of cycles to failure

8. The strain life method measures the strain resistance of local stresses and strains around stress concentration that controls the fatigue life of the material. It is more accurate than determining fatigue performance as the stress-life method is for long life millions of cycles in elastic stresses, but an it gets an effective stress concentration in fatigue loading.

9. Linear Elastic Fracture Mechanics (LEFM) states that the material is isotropic and linear elastic so, when the stresses near the crack surpasses the material fracture toughness, the crack grows.

7 0
3 years ago
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