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algol [13]
2 years ago
10

Technician A says that a radio may be able to receive AM signals, but not FM signals if the antenna is defective. Technician B s

ays that a good antenna should give a reading of about 500 ohms when
tested with an ohmmeter between the center antenna wire and ground. Which technician is correct?
A. Technician A only
B. Technician B only
C. Both Technician A and Technician B
D. Neither Technician A nor Technician B
Engineering
1 answer:
DIA [1.3K]2 years ago
3 0

The response to whether the statements made by both technicians are correct is that;

D: Neither Technician A nor Technician B are correct.

<h3>Radio Antennas</h3>

In radios, antennas are the means by which signals to the sought frequency be it AM or FM are received.

Now, if the antenna is bad, it means it cannot pick any radio frequency at all and so Technician A is wrong.

Now, most commercial antennas usually come around a resistance of 60 ohms and so it is not required for a good antenna to have as much as 500 ohms resistance and so Technician B is wrong.

Read more about Antennas at; brainly.com/question/25789224

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Which examples demonstrate tasks commonly performed in Maintenance/Operations jobs? Check all that apply.
Verdich [7]

1.Ross fixes a dishwasher for a homeowner.

3.Cassandra fixes holes in an old road.

4 0
3 years ago
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At 45° latitude, the gravitational acceleration as a function of elevation z above sea level is given by g = a − bz , where a =
Ahat [919]

Answer:

8861.75 m approximately 8862 m

Explanation:

We need to remember Newton's 2nd Law which says that the force experienced by an object is proportional to his acceleration and that the constant of proportionality between those two vectors correspond to the mass of the object.

F=ma for the weight of an object (which is a force) we have that the acceleration experienced by that object is equal to the gravitational acceleration, obtaining that  W = mg

For simplicity we work with g =9.807 \frac{m}{s^{2}} despiting the effect of the height above sea level. In this problem, we've been asked by the height above sea level that makes the weight of an object 0.30% more lighter.

In accord with the formula g = a-bz the "normal" or "standard" weight of an object is given by W = mg = ma when z = 0, so we need to find the value of z that makes W = m(a-bz) = 0.997ma meaning that the original weight decrease by a 0.30%, so now we operate...

m(a-bz) = 0.997ma now we group like terms on the same sides ma(1-0.997) = mbz we cancel equal tems on both sides and obtain that z = \frac{a}{b} (0.003) = \frac{9.807 \frac{m}{s^{2} } }{3.32*10^{-6} s^{-2} } (0.003) = 8861.75 m

7 0
3 years ago
Air at 400kPa, 970 K enters a turbine operating at steady state and exits at 100 kPa, 670 K. Heat transfer from the turbine occu
Sonja [21]

Answer:

a

The rate of work developed is \frac{\r W}{\r m}= 300kJ/kg

b

The rate of entropy produced within the turbine is   \frac{\sigma}{\r m}=  0.0861kJ/kg \cdot K

Explanation:

     From  the question we are told

          The rate at which heat is transferred is \frac{\r Q}{\r m } = -  30KJ/kg

the negative sign because the heat is transferred from the turbine

          The specific heat capacity of air is c_p = 1.1KJ/kg \cdot K

          The inlet temperature is  T_1 = 970K

          The outlet temperature is T_2 = 670K

           The pressure at the inlet of the turbine is p_1 = 400 kPa

          The pressure at the exist of the turbine is p_2 = 100kPa

           The temperature at outer surface is T_s = 315K

         The individual gas constant of air  R with a constant value R = 0.287kJ/kg \cdot K

The general equation for the turbine operating at steady state is \

               \r Q - \r W + \r m (h_1 - h_2) = 0

h is the enthalpy of the turbine and it is mathematically represented as          

        h = c_p T

The above equation becomes

             \r Q - \r W + \r m c_p(T_1 - T_2) = 0

              \frac{\r W}{\r m}  = \frac{\r Q}{\r m} + c_p (T_1 -T_2)

Where \r Q is the heat transfer from the turbine

           \r W is the work output from the turbine

            \r m is the mass flow rate of air

             \frac{\r W}{\r m} is the rate of work developed

Substituting values

              \frac{\r W}{\r m} =  (-30)+1.1(970-670)

                   \frac{\r W}{\r m}= 300kJ/kg

The general balance  equation for an entropy rate is represented mathematically as

                       \frac{\r Q}{T_s} + \r m (s_1 -s_2) + \sigma  = 0

          =>          \frac{\sigma}{\r m} = - \frac{\r Q}{\r m T_s} + (s_1 -s_2)

    generally (s_1 -s_2) = \Delta s = c_p\ ln[\frac{T_2}{T_1} ] + R \ ln[\frac{v_2}{v_1} ]

substituting for (s_1 -s_2)

                      \frac{\sigma}{\r m} = \frac{-\r Q}{\r m} * \frac{1}{T_s} +  c_p\ ln[\frac{T_2}{T_1} ] - R \ ln[\frac{p_2}{p_1} ]

                      Where \frac{\sigma}{\r m} is the rate of entropy produced within the turbine

 substituting values

                \frac{\sigma}{\r m} = - (-30) * \frac{1}{315} + 1.1 * ln\frac{670}{970} - 0.287 * ln [\frac{100kPa}{400kPa} ]

                    \frac{\sigma}{\r m}=  0.0861kJ/kg \cdot K

           

 

                   

   

5 0
4 years ago
Only answer this if your name is riley
Sati [7]

Answer:

hey im like kinda riley

Explanation:

y u wanna talk to moi

3 0
3 years ago
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Is it possible to have a heat engine with efficiency of 100%?
Alex17521 [72]

Answer:

No

Explanation:

Heat engines are used for converting the heat into mechanical energy which is used for doing mechanical work.

The efficiency of heat engine is the fraction of mechanical energy to the thermal energy. The efficiency can not be 100% as some of the energy always loss due to friction and motion of the body parts of the heat engine.

7 0
3 years ago
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