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worty [1.4K]
3 years ago
12

An 800-kHz radio signal is detected at a point 2.7 km distant from a transmitter tower. The electric field amplitude of the sign

al at that point is 0.36 V/m. Assume that the signal power is radiated uniformly in all directions and that radio waves incident upon the ground are completely absorbed. What is the intensity of the radio signal at that point?
a. 240 µW/m^2
b. 340 µW/m^2
c. 86 µW/m^2
d. 170 µW/m^2
e. 120 µW/m^2
Physics
1 answer:
dimaraw [331]3 years ago
7 0

Answer:

Option D is correct: 170 µW/m²

Explanation:

Given that,

Frequency f = 800kHz

Distance d = 2.7km = 2700m

Electric field Eo = 0.36V/m

Intensity of radio signal

The intensity of radial signal is given as

I = c•εo•Eo²/2

Where c is speed of light

c = 3×10^8m/s

εo = 8.85 × 10^-12 C²/Nm²

I = 3×10^8 × 8.85×10^-12 × 0.36²/2

I = 1.72 × 10^-4W/m²

I = 172 × 10^-6 W/m²

I = 172 µW/m²

Then, the intensity of the radio wave at that point is approximately 170 µW/m²

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Relative formula mass of CuCO3
Elodia [21]

AnMolar mass of CuCO3 = 123.5549 g/mol

This compound is also known as Copper(II) Carbonate.

Convert grams CuCO3 to moles  or  moles CuCO3 to grams

Molecular weight calculation:

63.546 + 12.0107 + 15.9994*3

Percent composition by element

Element   Symbol   Atomic Mass   # of Atoms   Mass Percent

Copper Cu 63.546 1 51.431%

Carbon C 12.0107 1 9.721%

Oxygen O 15.9994 3 38.848%

Explanation:

5 0
2 years ago
An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the
andrew-mc [135]

Answer:

50 W

Explanation:

Case 1

Power = V * I

100 = 220 * I

I = \frac{100}{220} A

Case 2

P = V * I

P = 110 * \frac{100}{220}

P = 50 W

I think the answer is 50 W

Hope it helps

8 0
3 years ago
Describe what happens to radiant energy once it enters the atmosphere? How does Radiant energy cause weather?
Triss [41]
Once energy from the Sun gets to Earth, several things can happen to it:

Energy can be scattered or absorbed by aerosols in the atmosphere. Aerosols are dust, soot, sulfates and nitric oxides. When aerosols absorb energy, the atmosphere becomes warmer. When aerosols scatter energy, the atmosphere is cooled.
Short wavelengths are absorbed by ozone in the stratosphere.
Clouds may act to either reflect energy out to space or absorb energy, trapping it in the atmosphere.
The land and water at Earth's surface may act to either reflect energy or absorb it. Light colored surfaces are more likely to reflect sunlight, while dark surfaces typically absorb the energy, warming the planet.
Albedo is the percentage of the Sun's energy that is reflected back by a surface. Light colored surfaces like ice have a high albedo, while dark colored surfaces tend to have a lower albedo. The buildings and pavement in cities have such a low albedo that cities have been called "heat islands" because they absorb so much energy that they warm up.
6 0
3 years ago
For Part A, Sebastian decided to use the copper cylinder. How would the magnitude of his q and ∆H compare if he were to redo Par
Vitek1552 [10]

The magnitudes of his q and ∆H for the copper trial would be lower than the aluminum trial.

The given parameters;

  • <em>initial temperature of metals, =  </em>T_m<em />
  • <em>initial temperature of water, = </em>T_i<em> </em>
  • <em>specific heat capacity of copper, </em>C_p<em> = 0.385 J/g.K</em>
  • <em>specific heat capacity of aluminum, </em>C_A = 0.9 J/g.K
  • <em>both metals have equal mass = m</em>

The quantity of heat transferred by each metal is calculated as follows;

Q = mcΔt

<em>For</em><em> copper metal</em><em>, the quantity of heat transferred is calculated as</em>;

Q_p = (m_wc_w + m_pc_p)(T_m - T_i)\\\\Q_p = (T_m - T_i)(m_wc_w ) + (T_m - T_i)(m_pc_p)\\\\Q_p = (T_m - T_i)(m_wc_w ) + 0.385m_p(T_m - T_i)\\\\m_p = m\\\\Q_p = (T_m - T_i)(m_wc_w ) + 0.385m(T_m - T_i)\\\\let \ (T_m - T_i)(m_wc_w )  = Q_i, \ \ \ and \ let \ (T_m- T_i) = \Delta t\\\\Q_p = Q_i + 0.385m\Delta t

<em>The </em><em>change</em><em> in </em><em>heat </em><em>energy for </em><em>copper metal</em>;

\Delta H = Q_p - Q_i\\\\\Delta H = (Q_i + 0.385m \Delta t) - Q_i\\\\\Delta H = 0.385 m \Delta t

<em>For </em><em>aluminum metal</em><em>, the quantity of heat transferred is calculated as</em>;

Q_A = (m_wc_w + m_Ac_A)(T_m - T_i)\\\\Q_A = (T_m -T_i)(m_wc_w) + (T_m -T_i) (m_Ac_A)\\\\let \ (T_m -T_i)(m_wc_w)  = Q_i, \ and \ let (T_m - T_i) = \Delta t\\\\Q_A = Q_i \ + \ m_Ac_A\Delta t\\\\m_A = m\\\\Q_A = Q_i \ + \ 0.9m\Delta t

<em>The </em><em>change</em><em> in </em><em>heat </em><em>energy for </em><em>aluminum metal </em><em>;</em>

\Delta H = Q_A - Q_i\\\\\Delta H = (Q_i + 0.9m\Delta t) - Q_i\\\\\Delta H = 0.9m\Delta t

Thus, we can conclude that the magnitudes of his q and ∆H for the copper trial would be lower than the aluminum trial.

Learn more here:brainly.com/question/15345295

6 0
3 years ago
A spool of thread has an average radius of 1.00 cm. If the spool contains 62.8 m of thread, how many turns of thread are on the
Simora [160]

To solve this problem we will start from the given concept in which the number of turns is equivalent to the length of the thread per circumference of spool. That is:

N = \frac{l}{\phi}

Where,

l = length of the thread

\phi= circumference of spool

For \phi we have that,

\phi = 2\pi r \rightarrow 2\pi (0.01)

For l  we have that

l = 62.8m

Finally the number of Turns would be,

N = \frac{l}{\phi}

N = \frac{62.8}{2\pi (0.01)}

N = 1000turns

Therefore the number of turns of thread on the spool are 1000turns.

7 0
3 years ago
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