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eduard
3 years ago
12

The audio power of the human voice is concentrated at about 300 Hz. Antennas of the appropriate size for this frequency are impr

acticably large, so that to send voice by radio the voice signal must be used to modulate a higher (carrier) frequency for which the natural antenna size is smaller. a. What is the length of an antenna one-half wavelength long for sending radio at 300 Hz
Physics
1 answer:
earnstyle [38]3 years ago
6 0

Answer:

the length of an antenna one-half wavelength long for sending radio at 300 Hz is 500 km

Explanation:

Given the data in the question;

we know that wave length is;

λ = c/f

c is the speed of voice ( 3 × 10⁸  m/s )

frequency f = 300 Hz

so we substitute

λ = 3 × 10⁸ / 300

λ = 1000000 m

we know that; 1 km = 1000 m

so

λ = 1000000 m / 1000

λ = 1000 km

hence, an antenna one-half wavelength will be;

λ /2

=  1000 km / 2

= 500 km

Therefore, the length of an antenna one-half wavelength long for sending radio at 300 Hz is 500 km

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A 5.30kg block hangs from a spring with a spring constant 1700 N/m. The block is pulled down 4.50cm from the equilibrium positio
Andru [333]

To solve this problem it is necessary to apply the concepts related to the frequency in a spring, the conservation of energy and the total mechanical energy in the body (kinetic or potential as the case may be)

PART A) By definition the frequency in a spring is given by the equation

f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}

Where,

m = mass

k = spring constant

Our values are,

k=1700N/m

m=5.3 kg

Replacing,

f = \frac{1}{2\pi} \sqrt{\frac{1700}{5.3}}

f=2.85 Hz

PART B) To solve this section it is necessary to apply the concepts related to the conservation of energy both potential (simple harmonic) and kinetic in the spring.

\frac{1}{2}kA^2 = \frac{1}{2}mv^2 + \frac{1}{2} kY^2

Where,

k = Spring constant

m = mass

y = Vertical compression

v = Velocity

This expression is equivalent to,

kA^2 =mV^2 +ky^2

Our values are given as,

k=1700 N/m

V=1.70 m/s

y=0.045m

m=5.3 kg

Replacing we have,

1700*A^2=5.3*1.7^2 +1700*(0.045)^2

Solving for A,

A^2 = \frac{5.3*1.7^2 +1700*(0.045)^2}{1700}

A ^2 = 0.011035

A=0.105 m \approx 10.5 cm

PART C) Finally, the total mechanical energy is given by the equation

E = \frac{1}{2}kA^2

E=\frac{1}{2}1700*(0.105)^2

E= 9.3712 J

3 0
3 years ago
If you shout on the moon,will the sound travel faster or slower than on earth?why?
labwork [276]
As long as the sound is inside the helmet of your space suit, it will travel
at the same speed as it would on Earth, through the same mixture of gases
at the same pressure.  Once it passes through the visor of your space helmet,
its 'speed' has no meaning, since there's nothing for sound to travel through on
the moon, and it doesn't travel at all.
7 0
3 years ago
You attach a meter stick to an oak tree, such that the top of the meter stick is 1.87 meters above the ground. Later, an acorn f
Alborosie

Answer:

3.25 m

Explanation:

t = Time taken = 0.166 seconds

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration = 9.81 m/s²

s = 1 because meter stick is 1 meter in length

s=ut+\frac{1}{2}at^2\\\Rightarrow u=\frac{s-\frac{1}{2}at^2}{t}\\\Rightarrow u=\frac{1-\frac{1}{2}\times 9.81\times 0.166^2}{0.166}\\\Rightarrow u=5.21\ m/s

Here, the initial velocity of point B is calculated from the time which is given. This velocity will be the final velocity of the acorn which falling from point A.

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{5.21^2-0^2}{2\times 9.81}\\\Rightarrow s=1.38\ m

The distance of the acorn from the ground is 1.87+1.38 = 3.25 m

4 0
3 years ago
You push on a box with 100 N of force, causing it to accelerate at 5 m/s?.
Xelga [282]
<h3><u>Given</u><u>:</u><u>-</u></h3>

Force,F = 100 N

Acceleration,a = 5 m/s²

<h3><u>To</u><u> </u><u>be</u><u> </u><u>calculated:-</u><u> </u></h3>

Calculate the mass of the box .

<h3><u>Formula</u><u> </u><u>used:-</u><u> </u></h3>

\bf \: Force = Mass  \times  Acceleration

<h3><u>Solution:-</u><u> </u></h3>

\sf \: Force = Mass × Acceleration

★ Substituting the values in the above formula,we get:

\sf \implies \: 100 = Mass \times 5

\sf \implies \: Mass =   \cancel\dfrac{100}{5}

\sf \implies \: Mass = 20 \: kg

4 0
3 years ago
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Why can you plant plants, but not water water?
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Answer:

Because water is water and its a liquid meanwhile plants are plants living things that you can move around.

Explanation:

3 0
2 years ago
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