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3 years ago

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An 800-kHz radio signal is detected at a point 4.5 km distant from a transmitter tower. The electric field amplitude of the sign

al at that point is 0.63 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 magnetic field amplitude of the signal at that point

1 answer:

Hitman42 [59]3 years ago

3 0

Answer:

$B_2=2.1nT$

Explanation:

From the question we are told that:

Frequency $F=800kHz$

Distance $d=4.5km$

Electric field amplitude $B_2=0.63V/m$

Generally the equation for momentum is mathematically given by

$B=\frac{E}{C}$

Therefore

$B_2=\frac{0.63}{3*10^8}$

$B_2=0.21*10^{-8}$

$B_2=2.1nT$

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The wavelength of a water wave is 54 m. It has a frequency of 0.03 Hz. What is the velocity of the wave?

dalvyx [7]

Answer:

1.62 m/s

Explanation:

Wavelength of the water wave= 54 m

The frequency is 0.03 Hz

Therefore the velocity can be calculated as follows

Velocity= frequency × wavelength

= 0.03 × 54

= 1.62 m/s

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Earth rotates on its axis once every 24 hours, so that objects on its surface execute uniform circular motion about the axis wit

cestrela7 [59]

Answer:

$v=1667.9km/h$

$a_{cp}=436.6km/h^2$

Explanation:

The speed is the distance traveled divided by the time taken. The distance traveled in 24hs while standing on the equator is the circumference of the Earth $C=2\pi R$ , where $R=6371km$ is the radius of the Earth.

We have then:

$v=\frac{C}{t}=\frac{2\pi R}{t}=\frac{2\pi (6371km)}{(24h)}=1667.9km/h$

And then we use the centripetal acceleration formula:

$a_{cp}=\frac{v^2}{R}=\frac{(1667.9km/h)^2}{(6371km)}=436.6km/h^2$

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3 years ago

A paleontologist estimates that when a particular rock formed, it contained 12 mg of the radioactive isotope potassium-40, which

leva [86]

Answer:

$t = 2.52 billion \:years$

Explanation:

As we know by radioactivity law

$N = N_o e^{-\lambda t}$

so here we will have

$N = 3 mg$

$N_o = 12 mg$

now we will have

$3 = 12 e^{-\lambda t}$

$\lambda t = ln 4$

now we also know that

$\lambda = \frac{ln2}{1.26 \times 10^6 yrs}$

$t = 1.26\times 10^6\times \frac{ln4}{ln2}$

$t = 2.52 billion \:years$

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What are the applied forces on a box that is at rest on a floor

lutik1710 [3]

The only force applied is gravity

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An insulated beaker with negligible mass contains 0.250 kg of water at 75.0C. How many kilograms of ice at -20.0C must be droppe

kkurt [141]

Answer:

The amount of kilograms of ice at -20.0°C that must be dropped into the water to make the final temperature of the system 40.0°C = 0.0674 kg

Explanation:

Heat gained by ice in taking the total temperature to 40°C = Heat lost by the water

Total Heat gained by ice = Heat used by ice to move from -20°C to 0°C + Heat used to melt at 0°C + Heat used to reach 40°C from 0°C

To do this, we require the specific heat capacity of ice, latent heat of ice and the specific heat capacity of water. All will be obtained from literature.

Specific heat capacity of ice = Cᵢ = 2108 J/kg.°C

Latent heat of ice = L = 334000 J/kg

Specific heat capacity of water = C = 4186 J/kg.°C

Heat gained by ice in taking the total temperature to 40°C = mCᵢ ΔT + mL + mC ΔT = m(2108)(0 - (-20)) + m(334000) + m(4186)(40 - 0) = 42160m + 334000m + 167440m = 543600 m

Heat lost by water = mC ΔT = 0.25 (4186)(75 - 40) = 36627.5 J

543600 m = 36627.5

m = 0.0674 kg = 67.4 g of ice.

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3 years ago