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pychu [463]
3 years ago
12

Which of the following is NOT true of electromagnetic waves? A. An electric field is created in any region of space in which a m

agnetic field is changing with time. B. A magnetic field is created in any region of space in which an electric field is changing with time. C. Induction of electric fields by changing magnetic fields occurs only if a conducting material is present. D. Induction of magnetic fields by changing magnetic fields occurs even in the absence of a conducting material. Reset Selection
Physics
2 answers:
Iteru [2.4K]3 years ago
6 0
The incorrect statement about electromagnetic waves is C. induction of electric fields by changing magnetic fields only occurs if a conducting material is present.
Electromagnetic waves do not rely on any medium for propagation, which means that the generation of fields is irrespective of the presence of a conducting material. 
Savatey [412]3 years ago
5 0

. C. Induction of electric fields by changing magnetic fields occurs only if a conducting material is present.  is correct

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Steam in a heating system flows through tubes whose outer diameter is 5 cm and whose walls are maintained at a temperature of 13
svet-max [94.6K]

Answer:

5945.27 W per meter of tube length.

Explanation:

Let's assume that:

  • Steady operations exist;
  • The heat transfer coefficient (h) is uniform over the entire fin surfaces;
  • Thermal conductivity (k) is constant;
  • Heat transfer by radiation is negligible.

First, let's calculate the heat transfer (Q) that occurs when there's no fin in the tubes. The heat will be transferred by convection, so let's use Newton's law of cooling:

Q = A*h*(Tb - T∞)

A is the area of the section of the tube,

A = π*D*L, where D is the diameter (5 cm = 0.05 m), and L is the length. The question wants the heat by length, thus, L= 1m.

A = π*0.05*1 = 0.1571 m²

Q = 0.1571*40*(130 - 25)

Q = 659.73 W

Now, when the fin is added, the heat will be transferred by the fin by convection, and between the fin and the tube by convection, thus:

Qfin = nf*Afin*h*(Tb - T∞)

Afin = 2π*(r2² - r1²) + 2π*r2*t

r2 is the outer radius of the fin (3 cm = 0.03 m), r1 is the radius difference of the fin and the tube ( 0.03 - 0.025 = 0.005 m), and t is the thickness ( 0.001 m).

Afin = 0.006 m²

Qfin = 0.97*0.006*40*(130 - 25)

Qfin = 24.44 W

The heat transferred at the space between the fin and the tube will be:

Qspace = Aspace*h*(Tb - T∞)

Aspace = π*D*S, where D is the tube diameter and S is the space between then,

Aspace = π*0.05*0.003 = 0.0005

Qspace = 0.0005*40*(130 - 25) = 1.98 W

The total heat is the sum of them multiplied by the total number of fins,

Qtotal = 250*(24.44 + 1.98) = 6605 W

So, the increase in heat is 6605 - 659.73 = 5945.27 W per meter of tube length.

5 0
3 years ago
Please help I wasn’t here for this lesson
katovenus [111]

Answer:

27 cm squared

Explanation:

have a good rest of ur day

8 0
2 years ago
1.
Mashutka [201]

Answer:

a) acceleration

Explanation:

Acceleration is, by definition, the change of an object's velocity.

4 0
2 years ago
Assuming the diode is ideal but with a forward voltage drop of. 65 volts, what is the current in ma if v1=0v, and r1=490ω?
weeeeeb [17]

The current in the ideal diode with forward biased voltage drop of 65V is 132.6 mA.

To find the answer, we have to know more about the ideal diode.

<h3>What is an ideal diode?</h3>
  • A type of electronic component known as an ideal diode has two terminals, only permits the flow of current in one direction, and has less zero resistance in one direction and infinite resistance in another.
  • A semiconductor diode is the kind of diode that is used the most commonly.
  • It is a PN junction-containing crystalline semiconductor component that is wired to two electrical terminals.
<h3>How to find the current in ideal diode?</h3>
  • Here we have given with the values,

                       V_2=65V\\V_1=0V\\R_1=490Ohm.

  • We have the expression for current in mA of the ideal diode with forward biased voltage drop as,

                I=\frac{V_2-V_1}{R_1} =\frac{65}{490} =132.6mA

Thus, we can conclude that, the current in mA of the ideal diode with forward biased voltage drop of 65 V is 132.6.

Learn more about the ideal diode here:

brainly.com/question/14988926

#SPJ4

7 0
2 years ago
What is the unit for electrical power?
Vaselesa [24]

Answer:

The answer is Watt

Explanation:

Electricity is measured in Watts and kilowatts

Electricity is measured in units of power called Watts, named to honor James Watt, the inventor of the steam engine. A Watt is the unit of electrical power equal to one ampere under the pressure of one volt. One Watt is a small amount of power.

5 0
3 years ago
Read 2 more answers
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