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Darina [25.2K]
3 years ago
7

An 800-kHz radio signal is detected at a point 8.5 km distant from a transmitter tower. The electric field amplitude of the sign

al at that point is 0.90 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 average electromagnetic energy density at that point
Physics
2 answers:
Temka [501]3 years ago
6 0

Answer:

Explanation:

Given that,

Frequency of radio signal is

f = 800kHz = 800,000 Hz.

Distance from transmitter

d = 8.5km = 8500m

Electric field amplitude

E = 0.9 V/m

The average energy density can be calculated using

U_E = ½•ϵo•E²

Where ϵo = 8.85 × 10^-12 F/m

Then,

U_E = ½ × 8.85 × 10^-12 × 0.9²

U_E = 3.58 × 10^-12 J/m²

The average electromagnetic energy density is 3.58 × 10^-12 J/m²

sveta [45]3 years ago
4 0

Answer:

4.96*10^-20 J

Explanation:

To find the average electromagnetic energy of the wave you use the fact that the wave can be taken as a spherical electromagnetic wave. IN this case, the amplitude of the wave change in space according to:

|E|=\frac{E_o}{r}

for a distance of 8.5km = 5.5*10^3m you have:

|E|=\frac{0.90\ V/m}{8.5*10^3m}=1.058*10^{-4}\ V/m

Next, to find the average energy at this point you use the following formula:

u_E=\frac{1}{2}\epsilon_0|E|^2

εo : dielectric permittivity of vacuum = 8.85*10^-12C^2/Nm^2

u_E=\frac{1}{2}(8.85*10^{-12})|1.058*10^{-4}|^2\ J=4.96*10^{-20}J

hence, the average energy of the wave is 4.96*10^-20 J

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Electrons: negative charge
Protons: positive charge
Neutrons: negative charge

The atom would have to have more electrons than protons

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Mirrors reflect light waves.
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Block A, with a mass of 4.0 kg, is moving with a speed of 2.0 m/s while block B , with a mass of 8.0 kg, is moving in the opposi
Anon25 [30]
Consider velocity to the right as positive.

First mass:
m₁ = 4.0 kg
v₁ = 2.0 m/s to the right

Second mass:
m₂ = 8.0 kg
v₂ = -3.0 m/s to the left

Total momentum of the system is
P = m₁v₁ + m₂v₂
   = 4*2 + 8*(-3)
  = -16 (kg-m)/s

Let v (m/s) be the velocity of the center of mass of the 2-block system.

Because momentum of the system is preserved, therefore
(m₁+m₂)v= -16
(4+8 kg)*(v m/s) = -16 (kg-m)/s
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Answer:
The center of mass is moving at 1.33 m/s to the left.
5 0
3 years ago
To push a 26.0 kg crate up a frictionless incline, angled at 25.0° to the horizontal, a worker exerts a force of 209 N parallel
alukav5142 [94]

Answer:

(a) W = +397.1 J

(b) W = -204.6 J

(c) W = 0

(d) W= + 192.5 J

Explanation:

Work (W) is defined as the product of force (F) by the distance (d)the body travels due to this force. :

W= F*d Formula ( 1)

The forces that perform work on an object must be parallel to its displacement.

The forces perpendicular to the displacement of an object do not perform work on it.

The work is positive (W+) if the force has the same direction of movement of the object.  

The work is negative (W-) if the force has the opposite direction of the movement of the object.

Problem development

(a) Work performed by the worker's applied force on the box .

W= 209 N * 1.9 m = +397.1 J

(b) Work performed by the gravitational force on the crate

We calculate the weight component parallel to the displacement of the box:

We define the x-axis in the direction of the inclined plane ,25.0° to the horizontal.

We define the y-axis and in the direction of the plane perpendicular to the inclined plane.

W= m*g=26*9.8= 254.8N : total box weight

Wx= W*sen25.0°= 254.8*sen25.0°= 107.68 N

W = -Wx *d =107.68 N *1.9 m= -204.6 J

(c) Work performed by normal force (N) exerted by the incline on the crate

The force N is perpendicular to the displacement, then:

W=0

(d) Total work done on the crate

W = 397.1 J -204.6 J

W = 192.5 J

4 0
2 years ago
What is the answer be fast
svetlana [45]

Answer:

Explanation:

Same numbers of protons but different number of neutron so i would go for A same atomic number different number of neutrons

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