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snow_tiger [21]
3 years ago
6

How is the radiating electric field (or electromagnetic signal) produced when radio stations broadcast

Physics
1 answer:
Natali5045456 [20]3 years ago
5 0

Answer:

Radio stations have dipole type antennas

this field increases in intensity and propagates outwards,

Explanation:

Radio stations have dipole type antennas, that is, all sides are isolated from each other, when the AC signal from the radio station arrives, the lcharge begins at times and by the Lens law a field appears that opposes this movement, this field increases in intensity and propagates outwards, when the voltage reaches a maximum, the generated wave also reaches the maximum, now the incident wave begins to decrease, an electric hand appears to oppose this prisoner, and in this way a cap is created. electric .

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Since the aluminum bar is not an isolated system, the second law of thermodynamics cannot be applied to the bar alone. Rather, i
max2010maxim [7]

Answer:

ΔS total ≥ 0 (ΔS total = 0 if the process is carried out reversibly in the surroundings)

Explanation:

Assuming that the entropy change in the aluminium bar is due to heat exchange with the surroundings ( the lake) , then the entropy change of the aluminium bar is, according to the second law of thermodynamics, :

ΔS al ≥ ∫dQ/T

if the heat transfer is carried out reversibly

ΔS al =∫dQ/T  

in the surroundings

ΔS surr ≥ -∫dQ/T = -ΔS al → ΔS surr ≥ -ΔS al = - (-1238 J/K) = 1238 J/K

the total entropy change will be

ΔS total = ΔS al + ΔS surr

ΔS total ≥ ΔS al + (-ΔS al) =

ΔS total ≥ 0

the total entropy change will be ΔS total = 0 if the process is carried out reversibly in the surroundings

4 0
3 years ago
Which of the following is a zone found in the open ocean
Svetlanka [38]
What are your answer choices?

5 0
3 years ago
Read 2 more answers
How is work calculated?
BartSMP [9]

Work is force times distance. If there's no distance, there's no work being done.

6 0
3 years ago
The mass of Object 2 is double the mass of Object 5. The mass of Object 4 is half of the mass of Object 5 and the mass of Object
SVETLANKA909090 [29]
This is a great problem if you like getting tied up in knots
and making smoke come out of your brain.

I found that it makes the problem a lot easier if I give the objects some
numbers. I'm going to say that the mass of Object 5 is 20 clods.

Let the mass of Mass of Object 5 be 20 clods .

Then . . .

-- The mass of Object 2 is double the mass of Object 5 = 40 clods.

-- The mass of Object 4 is half of the mass of Object 5 = 10 clods.
and
-- the mass of Object 3 is half of the mass of Object 4 = 5 clods.

So now, here are the masses:

Object #1 . . . . . unknown
Object #2 . . . . . 40 clods
Object #3 . . . . . 5 clods
Object #4 . . . . . 10 clods
Object #5 . . . . . 20 clods .

Now let's check out the statements, and see how they stack up:

Choice-A:
Object 3 and Object 5 exert the same gravitational force on Object 1.
Can't be.
Objects #3 and #5 have different masses, so they can't both
exert the same force on the same mass.

Choice-B.
Object 2 and Object 4 exert the same gravitational force on Object 1.
Can't be.
Objects #2 and #4 have different masses, so they can't both
exert the same force on the same mass.

Choice-C.
The gravitational force between Object 1 and Object 2 is greater than
the gravitational force between Object 1 and Object 4.
Yes ! Yay !
Object-2 has more mass than Object-4 has, so it must exert more force on
ANYTHING than Object-4 does, (as long as the distances are the same).

Choice-D.
The gravitational force between Object 1 and Object 3 is greater than the gravitational force between Object 1 and Object 5.
Can't be.
Object-3 has less mass than Object-5 has, so it must exert less force on
ANYTHING than Object-4 does, (as long as the distances are the same).

Conclusion:
If the DISTANCE is the same for all the tests, then Choice-C is
the only one that can be true.
8 0
3 years ago
Blood in a carotid artery carrying blood to the head is moving at 0.15 m/s when it reaches a section where plaque has narrowed t
sp2606 [1]

Answer:

26.9 Pa

Explanation:

We can answer this question by using the continuity equation, which states that the volume flow rate of a fluid in a pipe must be constant; mathematically:

A_1 v_1 = A_2 v_2 (1)

where

A_1 is the cross-sectional area of the 1st section of the pipe

A_2 is the cross-sectional area of the 2nd section of the pipe

v_1 is the velocity of the 1st section of the pipe

v_2 is the velocity of the 2nd section of the pipe

In this problem we have:

v_1=0.15 m/s is the velocity of blood in the 1st section

The diameter of the 2nd section is 74% of that of the 1st section, so

d_2=0.74d_1

The cross-sectional area is proportional to the square of the diameter, so:

A_2=(0.74)^2 A_1=0.548 A_1

And solving eq.(1) for v2, we find the final velocity:

v_2=\frac{A_1 v_1}{A_2}=\frac{A_1 (0.15)}{0.548 A_1}=0.274 m/s

Now we can use Bernoulli's equation to find the pressure drop:

p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho v_2^2

where

\rho=1025 kg/m^3 is the blood density

p_1,p_2 are the initial and final pressure

So the pressure drop is:

p_1 - p_2 = \frac{1}{2}\rho (v_2^2-v_1^2)=\frac{1}{2}(1025)(0.274^2-0.15^2)=26.9 Pa

8 0
3 years ago
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