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vova2212 [387]
3 years ago
10

A radio technician measures the frequency of an AM radio transmitter. The frequency is . What is the frequency in megahertz? Wri

te your answer as a decimal.
Physics
1 answer:
sukhopar [10]3 years ago
8 0

Complete Question

A radio technician measures the frequency of an AM radio transmitter. The frequency is 14603 kHz . What is the frequency in megahertz? Write your answer as a decimal.

Answer:

The value is  x =  14.6 \  MHz

Explanation:

From the question we are told that

  The  frequency is  f =  14603  \  kHz = 14603 *1000 = 14603000 \ Hz

Generally  

       1 Hz \to  1.0 *10^{-6} \  MHz

       14603000 \ Hz  \to x MHz

=>   x =  \frac{14603000 *  1.0*10^{-6}}{1 }

=>    x =  14.6 \  MHz

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A technician is checking refrigerant system pressures. Both high- and low-side service ports are located on the A/C compressor.
konstantin123 [22]

Answer:

Two major causes are outline bellow

1. The presence of air in the system

2. Clogged condenser

Explanation:

1. The presence of air in the system

One of the causes that have been established in relation to high compressor discharge pressure is the presence of air in the system. When this takes place, your best solution is to recharge the system.

2. Clogged condenser

Another is a clogged condenser in which case you will need to clean the condenser so that it will function properly. When you happen to spot that the discharge valve is closed and it is causing high discharge pressure on the compressor, you can solve that easily by opening the valve

5 0
3 years ago
Gasoline burns in the cylinder of an automobile engine. During the combustion reaction, the production of gas forces the piston
serg [7]

Answer:

\Delta U = 1640 J

Explanation:

As we know by first law of thermodynamics that for ideal gas system we have

Heat given = change in internal energy + Work done

so here we will have

Heat given to the system = 2.2 kJ

Q = 2200 J

also we know that work done by the system is given as

W = 560 J

so we have

\Delta U = Q - W

\Delta U = 2200 - 560

\Delta U = 1640 J

6 0
3 years ago
An airliner arrives at the terminal, and its engines are shut off. The rotor of one of the engines has an initial clockwise angu
Ilia_Sergeevich [38]

(a) 1200 rad/s

The angular acceleration of the rotor is given by:

\alpha = \frac{\omega_f - \omega_i}{t}

where we have

\alpha = -80.0 rad/s^2 is the angular acceleration (negative since the rotor is slowing down)

\omega_f is the final angular speed

\omega_i = 2000 rad/s is the initial angular speed

t = 10.0 s is the time interval

Solving for \omega_f, we find the final angular speed after 10.0 s:

\omega_f = \omega_i + \alpha t = 2000 rad/s + (-80.0 rad/s^2)(10.0 s)=1200 rad/s

(b) 25 s

We can calculate the time needed for the rotor to come to rest, by using again the same formula:

\alpha = \frac{\omega_f - \omega_i}{t}

If we re-arrange it for t, we get:

t = \frac{\omega_f - \omega_i}{\alpha}

where here we have

\omega_i = 2000 rad/s is the initial angular speed

\omega_f=0 is the final angular speed

\alpha = -80.0 rad/s^2 is the angular acceleration

Solving the equation,

t=\frac{0-2000 rad/s}{-80.0 rad/s^2}=25 s

6 0
3 years ago
Learning Goal:To understand and be able to use the rules for determining allowable orbital angular momentum states.Several numbe
Free_Kalibri [48]

Answer:

Explanation:

1) for a given n value the l value can be from 0 to n-1

So if n= 5 it can take 0,1,2,3,4

i.e it can take 5 values

2)for an electron with l =3

it can be from -3 -2 -1 0 1 2 3

i.e it can take 7 values

3) n = 3 !!

l = 0 , 1 , 2

for l=0 , m = 0 total = 1

for l= 1 ,m = -1,0,1 total = 3

for l = 2, m=-2,-1,0,1,2 total = 5

5+3+1 = 9

total possible states = 9 * 2 = 18

Answer is 168

4)given l=3 and n=3

orbital quantum number cannot be equal to principal quantum number

its max value is l-1 only

5)L = sqrt(l(l+1))x h'

for it to be max l should be max

for n = 3 max l value is 2

therfore it is sqrt(2(2+1)) x h'

\sqrt(6) \times h'

this is the answer

5 0
3 years ago
7. You probably had difficulty in measuring the period, yet its calculation should prove effortless at this point. Why do you su
Art [367]

Answer:

* First fight the reaction time of the person, which is approximately 1/10 s, remember that the stopwatch does not stop alone

* Problems to synchronize when starting to measure the oscillation and end point of the oscilloscope, this error needs a good control system to be decreased.

*Effect of friction with air

Explanation:

In the measurements of the oscillatory movement in general, the most difficult magnitude of the measurement is the period, by reasoning

* First fight the reaction time of the person, which is approximately 1/10 s, remember that the stopwatch does not stop alone

* Problems to synchronize when starting to measure the oscillation and end point of the oscilloscope, this error needs a good control system to be decreased.

* Effect of friction with air

5 0
3 years ago
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