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spin [16.1K]
3 years ago
10

what were the negative results with the american television from analog broadcast to digital broadcast?

Physics
2 answers:
alexandr1967 [171]3 years ago
8 0

Answer:

Explanation:

everything

Finger [1]3 years ago
7 0
Https://en.m.wikipedia.org/wiki/Digital_television
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A small charged sphere is attached to a thread and placed in an electric field. The other end of the thread is anchored so that
VMariaS [17]

Answer:

E = 9.66\times 10^{-6} N/C

direction is Horizontal

Explanation:

As we know that the string is horizontal here

so the tension force in the string is due to electrostatic force on it

now we will have

F = qE

so here the force is tension force on it

F = 6.57 \times 10^{-2} N

Q = 6.80 \times 10^3 C

now we have

6.57 \times 10^{-2} = (6.80 \times 10^3)E

E = 9.66\times 10^{-6} N/C

direction is Horizontal

4 0
3 years ago
An ocean thermal energy conversion system is being proposed for electric power generation. Such a system is based on the standar
defon

Answer:

Explanation:

Dear Student, this question is incomplete, and to attempt this question, we have attached the complete copy of the question in the image below. Please, Kindly refer to it when going through the solution to the question.

To objective is to find the:

(i) required heat exchanger area.

(ii) flow rate to be maintained in the evaporator.

Given that:

water temperature = 300 K

At a reasonable depth, the water is cold and its temperature = 280 K

The power output W = 2 MW

Efficiency \zeta = 3%

where;

\zeta = \dfrac{W_{out}}{Q_{supplied }}

Q_{supplied } = \dfrac{2}{0.03} \ MW

Q_{supplied } = 66.66 \ MW

However, from the evaporator, the heat transfer Q can be determined by using the formula:

Q = UA(L MTD)

where;

LMTD = \dfrac{\Delta T_1 - \Delta T_2}{In (\dfrac{\Delta T_1}{\Delta T_2} )}

Also;

\Delta T_1 = T_{h_{in}}- T_{c_{out}} \\ \\ \Delta T_1 = 300 -290 \\ \\ \Delta T_1 = 10 \ K

\Delta T_2 = T_{h_{in}}- T_{c_{out}} \\ \\ \Delta T_2 = 292 -290 \\ \\ \Delta T_2 = 2\ K

LMTD = \dfrac{10 -2}{In (\dfrac{10}{2} )}

LMTD = \dfrac{8}{In (5)}

LMTD = 4.97

Thus, the required heat exchanger area A is calculated by using the formula:

Q_H = UA (LMTD)

where;

U = overall heat coefficient given as 1200 W/m².K

66.667 \times 10^6 = 1200 \times A \times 4.97 \\ \\  A= \dfrac{66.667 \times 10^6}{1200 \times 4.97} \\ \\  \mathbf{A = 11178.236 \ m^2}

The mass flow rate:

Q_{H} = mC_p(T_{in} -T_{out} )  \\ \\  66.667 \times 10^6= m \times 4.18 (300 -292) \\ \\ m = \dfrac{  66.667 \times 10^6}{4.18 \times 8} \\ \\  \mathbf{m = 1993630.383 \ kg/s}

3 0
3 years ago
Different ____ of light through two separate mediums causes the bending of waves fronts associated with light rays.
Fittoniya [83]

Different speeds of light through two separate media ... and the difference in wavelength caused by the difference ... causes the bending of waves fronts associated with light rays.

AFTER the bend, since the light rays then travel in a different direction, we may also say that the 'velocity' has changed.

8 0
3 years ago
A 8.8 cm diameter circular loop of wire is in a 1.04 T magnetic field. The loop is removed from the field in 0.30 s . Assume tha
denis23 [38]

Answer:

0.021 V

Explanation:

The average induced emf (E) can be calculated usgin the Faraday's Law:

E = - \frac{N*\Delta \phi}{\Delta t}  

<u>Where:</u>

<em>N = is the number of turns = 1   </em>

<em>ΔΦ = ΔB*A                                            </em>

<em>Δt = is the time = 0.3 s   </em>

<em>A = is the loop of wire area = πr² = πd²/4 </em>

<em>ΔB: is the magnetic field = (0 - 1.04) T                     </em>

Hence the average induced emf is:

E = - \frac{\Delta B*A}{\Delta t} = - \frac{(0- 1.04 T) \pi (0.088 m)^{2}}{4*0.3 s} = 0.021 V                      

Therefore, the average induced emf is 0.021 V.

I hope it helps you!

8 0
3 years ago
An arch carries the thrust of weight to its _____(1)______. With a _____(2)______, the horizontal part of the structure supports
finlep [7]

Option D is correct. An arch carries the thrust of weight to its <u>sides </u>with a <u>post-and-lintel.</u>

<u></u>

<h3>What is an arch?</h3>

An arch is indeed a vertical curving construction that covers an elevated space that may or may not sustain the load above it or the pressure gradient against it

In the case of a horizontally arched, such as an embankment dam. While arches and vaults are often confused, A vault is defined as an ongoing arch forming a roof.

Option D satisfies the fill-in blanks option.

Hence option D is correct. An arch carries the thrust of weight to its <u>sides </u>with a <u>post-and-lintel.</u>

<u></u>

To learn more about the arch refer to the link;

brainly.com/question/18162421

6 0
2 years ago
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