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Evgen [1.6K]
4 years ago
9

A television is tuned to a station broadcasting at a frequency of 6.60×10^7 Hz . For best reception, the rabbit-ear antenna used

by the TV should be adjusted to have a tip-to-tip length equal to half a wavelength of the broadcast signal.
Find the optimum length of the antenna.
Physics
1 answer:
den301095 [7]4 years ago
4 0

Answer:

The optimum length of the antenna is 2.27 m

Explanation:

Given:

Broadcasting frequency f = 6.60 \times 10^{7} Hz

From the formula of wavelength in terms of frequency,

  v = f \lambda

Where v = speed of light = 3 \times 10^{8} \frac{m}{s}

  \lambda = \frac{v}{f}

  \lambda = \frac{3 \times 10^{8} }{6.60 \times 10^{7} }

  \lambda = 4.54 m

For finding the length of antenna,

   l = \frac{\lambda}{2}

   l = \frac{4.54}{2}

   l = 2.27 m

Therefore, the optimum length of the antenna is 2.27 m

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Four charges 7 × 10−9 C at (0 m, 0 m), −9 × 10−9 C at (3 m, 3 m), 7 × 10−9 C at (1 m, 3 m), and −8 × 10−9 C at (−3 m, 2 m), are
Ivanshal [37]

Answer:

Magnitude of the resulting force on the 7 nC charge at the origin:

Fn₁= 23.95*10⁻⁹ N

Explanation:

Look at the attached graphic:

Charges of positive signs exert repulsive forces on q₁ + and charges of negative signs exert attractive forces on q₁ +.

q₁ experiences three forces (F₂₁,F₃₁,F₄₁) and we calculate them with Coulomb's law:

F = (k*q₁*q)/(d)²

d_{12} = \sqrt{3^{2}+3^{2}  }  = \sqrt{18} m : distance from q₁ to q₂

(d₁₂)² = 18 m²

d_{13} =\sqrt{1^{2}+3^{2}  } = \sqrt{10} m  : distance from q₁ to q₃

(d₁₃)² = 10 m²

d_{14} =\sqrt{3^{2}+2^{2}  } = \sqrt{13} m  : distance from q₁ to q₄

(d₁₄)² = 13 m²

K=  8.98755 × 10⁹ N *m²/C²

q₁=  7*10⁻⁹C

k*q₁=8.98755*10⁹ *7*10⁻⁹= 62.9

F₂₁= (62.9)*(9* 10⁻⁹) /(18) = 31.45*10⁻⁹ C

F₃₁= (62.9)*(7* 10⁻⁹) /(10) = 44*10⁻⁹ C

F₄₁= (62.9)*(8* 10⁻⁹) /(13) = 38.7*10⁻⁹ C

x-y components of the net force on q₁ (Fn₁):

α= tan⁻¹(3/3)= 45°  ,  β= tan⁻¹(3/1)= 71.56° , θ= tan⁻¹(2/3)= 33.69°

Fn₁x = F₂₁x+ F₃₁x+F₄₁x

F₂₁x =+ F₂₁*cosα =+ (31.45*10⁻⁹)* (cos 45°) = +22.24 *10⁻⁹ N

F₃₁x= -F₃₁*cosβ = - ( 44*10⁻⁹)* (cos 71.56°) = -13.91 *10⁻⁹ N

F₄₁x= -F₄₁*cosθ = -(38.7*10⁻⁹)* (cos 33.69°) = -32.2*10⁻⁹ N

Fn₁x = (+22.24 - 13.91 - 32.2)*10⁻⁹ N

Fn₁x = -23.87 *10⁻⁹ N

Fn₁y = F₂₁y+ F₃₁y+F₄₁y

F₂₁x =+ F₂₁*sinα =+ (31.45*10⁻⁹)* (sin 45°) = +22.24 *10⁻⁹ N

F₃₁x= -F₃₁*sinβ = - ( 44*10⁻⁹)* (sin 71.56°) = -41.74 *10⁻⁹ N

F₄₁x= +F₄₁*sinθ = +(38.7*10⁻⁹)* (sin 33.69°) =+21.47*10⁻⁹ N

Fn₁y = (22.24 -41.74+21.47)*10⁻⁹ N  

Fn₁y = 1.97*10⁻⁹ N

Magnitude of the resulting force on the 7 nC charge at the origin (q₁):

F_{n1} =\sqrt{(Fn_{1x} )^{2}+(Fn_{1y} )^{2} }

F_{n1} =\sqrt{(23.87 )^{2}+(1.97 )^{2} }

Fn₁= 23.95*10⁻⁹ N

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What does the pupil of the eye controls
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The correct answer would be D
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if you're shopping for a rack switch, what component on the switch tells you it can be mounted to a rack?
Tatiana [17]

if you're shopping for a rack switch, the component on the switch that tells you it can be mounted to a rack is the:

  • Rack ears

<h3>What are rack ears?</h3>

Rack ears are L-shaped objects that can be used to hold a rack switch firmly to the support walls. Rack ears are often found at the front panel of the rack which is to be mounted.

When a person purchases a rack switch and finds the rack ears there, it is a signal that the item can be secured firmly to rails. Sometimes, the rack ears also appear as extensions.

Learn more about rack ears here:

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6 0
2 years ago
two forces whose magnitude are in ratio of 3:5 gives a resultant of 35N.if the angle of inclination is 60degree.calculate the ma
nadya68 [22]

Answer:

the magnitude of first force = 3 × 5= 15 N

ANd, the magnitude of second force = 5 × 5 = 25 N

Explanation:

The computation of the magnitude of the each force is shown below:

Provided that

Ratio of forces = 3: 5

Let us assume the common factor is x

Now

first force =  3x

And, the second force = 5x

Resultant force = 35 N

The Angle between the forces = 60 degrees  

Based on the above information

Resultant force i.e. F = √ F_1^2 +F_2^2 + 2 F_1F_2cos\theta

35 = √[(3x)²+ (5x)²+ 2 (3x)(5x) cos 60°]

 35 =√ 9x² + 25x² + 15x²    (cos 60° = 0.5)

35 = √49 x²

 x = 5

So, the magnitude of first force = 3 × 5= 15 N

ANd, the magnitude of second force = 5 × 5 = 25 N

7 0
3 years ago
4. How much time does it take for a student running at a speed of 5 m/s to cover a distance of 2,000 m?
Monica [59]

Answer:

6 Minutes 40 Seconds or 400 Seconds

Explanation:

Time to cover a distance of 5m = 1 Second

Time to cover a distance of 2000m = 2000÷5

= 400 Seconds

After converting 400 Seconds into minutes it will become 6 minutes 40 seconds.

Those who found this helpful please give me a Thanks to support me. So, I can explain other questions more clearly. If you don't want to mark me Brainliest don't mark. But, please give me a Thanks.

6 0
3 years ago
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