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Pavel [41]
2 years ago
8

Despite the advent of digital television, some viewers still use "rabbit ears" atop their sets instead of purchasing cable telev

ision service or satellite dishes.
Certain orientations of the receiving antenna on a television set give better reception than others. Furthermore, the best orientation varies from station to station.
Explain.
Physics
1 answer:
Maslowich2 years ago
7 0

Answer:

This is due to difference in bandwidths of different stations, and the inability of the bunny ear to accommodate distant bandwidths most times at the same time.

Explanation:

Bunny ears are a simple half-wave dipole antenna used to receive VHF television bands. It is constructed of two telescoping rods attached to a base, these rods extend out to about 1 meter length, and are collapsible when not in use. For the best reception the rods should be adjusted to be a little less than 1/4 wavelength at the frequency of the television channel being received and bandwidth of different station differs one frone another, so adjustment has to be made sometimes in order to get reception from other stations. Thankfully, the dipole has a wide bandwidth, so most times, adequate reception is achieved without having to adjust the length of the bunny ear.

The half wave dipole of a bunny ear has a low gain of about 2.14 dBi; which means it is not as directional and sensitive to distant stations as a large rooftop antenna, but its wide angle reception pattern may allow it to receive several stations located in different directions without requiring readjustment when the channel is changed. Sometime an adjustment of the rods is necessary.

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A field measuring 12 meters by 16 meters is to have a brick paver walkway installed all around it, increasing the total area to
kolezko [41]

Answer:

1.5 m

Explanation:

Length. L = 12 m

Width, W = 16 m

Area, A = 12 x 16 = 192 m^2

Let the width of pavement be d.

The new length, L' = 12 + 2d

the new width, W' = 16 + 2d  

New Area, A' = L' x W' = (12 + 2d)(16 + 2d) = 192 + 56 d + 4d^2

Difference in area = A' - A

285 =  192 + 56 d + 4d^2 - 192

93 =  56 d + 4d^2

4d^2 + 56 d - 93 = 0

d = \frac{-56\pm \sqrt{56^{2}+4\times 4\times 93}}{8}

d=\frac{-56\pm 87.72}{8}\

d = 1.5 m

Thus, the width of the pavement is 1.5 m.

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Before he gets there, while he's coming at you,
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effort is the answer

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