PHYSICS COLLEGE

An 800 kHz radio signal is detected at a point 2.1 km distant from a transmitter tower. The electric field amplitude of the signal at that point is 800 mV/m. Assume that the signal power is radiated uniformly in all directions and that radio waves incident upon the ground are completely absorbed. The intensity of the radio signal at that point is closest to

Answers

Answer 1

Answer:

Answer:

$I=8.48* 10^(-4)\ W/m^2$

Explanation:

Given that,

Frequency of the radio signal, $f=800\ kHz=8* 10^5\ Hz$

It is detected at a pint 2.1 km from the transmitter tower, x = 2.1 km

The amplitude of the electric field is, E = 800 mV/m

Let I is the intensity of the radio signal at that point. Mathematically, it is given by :

$I=(E^2_(rms))/(c\mu_o)$

$E_(rms)$ is the rms value of electric field, $E_(rms)=(E)/(√(2) )$

$I=(E^2)/(2c\mu_o)$

$I=((800* 10^(-3))^2)/(2* 3* 10^8* 4\pi * 10^(-7))$

$I=8.48* 10^(-4)\ W/m^2$

So, the intensity of the radio signal at that point is $8.48* 10^(-4)\ W/m^2$ . Hence, this is the required solution.

Related Questions

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It's a little hard can u help

Answers

I can't really tell what (I) is, but F) blender def converts electric into motion of spinning blades

You're driving along at 25 m/s with your aunt's valuable antiques in the back of your pickup truck when suddenly you see a giant hole in the road 55 m ahead of you. Fortunately, your foot is right beside the brake and your reaction time is zero! Can you stop without the antiques sliding and being damaged? Their coefficients of friction are μs=0.6 and μk=0.3. Hint: You're not trying to stop in the shortest possible distance. What's your best strategy for avoiding damage to the antiques?

Answers

Answer:

Explanation:

initial velocity, u = 25 m/s

distance, s = 55 m

coefficient of static friction = 0.6

coefficient of kinetic friction = 0.3

Let the acceleration is a.

Use third equation of motion

v² = u² + 2as

0 = 25 x 25 - 2 x a x 55

a = 5.68

a = μg

μ = 5.68 / 9.8 = 0.58

so, the coefficient of friction is less then the coefficient of static friction so the antiques are safe.

Which type of diffraction occurs when the point source and the screen are at finite distances from the obstacle forming the diffraction pattern?A. fraunhofer
B. fresnel
C. far-field
D. single slit

Answers

Fresnel diffraction

A skateboarder with mass ms = 54 kg is standing at the top of a ramp which is hy = 3.3 m above the ground. The skateboarder then jumps on his skateboard and descends down the ramp. His speed at the bottom of the ramp is vf = 6.2 m/s.

Answers

The work ( $W_f$ ) done by the friction force between the ramp and the skateboarder is given by $-\mu_k \cdot m_s \cdot g \cdot h_y$ .

The workdone by the friction force ( $W_f$ ) can be calculated using the formula for work, which is the product of the force applied ( $F_f$ ) and the displacement (d) over which the force is applied:

$W_f = F_f \cdot d$

In this scenario, the frictionforce works against the skateboarder's momentum down the ramp, therefore it does no good.

Given:

Mass of skateboarder ( $m_s$ ) = 54 kg

Height of the ramp ( $h_y$ ) = 3.3 m

Final velocity ( $v_f$ ) = 6.2 m/s

Coefficient of kineticfriction ( $\mu_k$ ) between skateboarder and ramp

Acceleration due to gravity (g) = $9.81 \, \text{m/s}^2$

The normal force ( $F_{\text{normal}}$ ) is equal to the weight of the skateboarder:

$F_{\text{normal}} = m_s \cdot g$

The displacement (d) is the vertical distance ( $h_y$ ) that the skateboarder descends down the ramp.

Now we can write the expression for the work done by the friction force ( $W_f$ ):

$W_f = -\mu_k \cdot F_{\text{normal}} \cdot d$

Substitute the expression for the normal force:

$W_f = -\mu_k \cdot (m_s \cdot g) \cdot h_y$

Thus, this expression represents the work done by the friction force between the ramp and the skateboarder in terms of the given variables.

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Your question seems incomplete, the probable complete question is:

Write an expression for the work, Wf, done by the friction force between the ramp and the skateboarder in terms of the variables given in the problem statement.

Final answer:

The momentum of the box with respect to the floor can be found by multiplying its mass by its velocity. When the box is put down on the frictionless skating surface, its velocity becomes zero and its momentum with respect to the floor is also zero.

Explanation:

To find the momentum of the box, we can use the formula:

Momentum = mass x velocity

a. The momentum of the box with respect to the floor is: 5 kg x 5 m/s = 25 kg·m/s

b. When the box is put down on the frictionless skating surface, its velocity becomes zero. So, the momentum of the box with respect to the floor is also zero.

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An implanted pacemaker supplies the heart with 72 pulses per minute, each pulse providing 6.0 V for 0.65 ms. The resistance of the heart muscle between the pacemaker’s electrodes is 550 Ω. Find (a) the current that flows during a pulse, (b) the energy delivered in one pulse, and (c) the average power supplied by the pacemaker.

Answers

Answer:

a) Current = 11 mA

b) Energy = 66 mJ

c) Power = 101.54 W

Explanation:

a) Voltage, V = IR

Voltage, V = 6 V, Resistance, R = 550 Ω

Current, I $=(6)/(550)=0.011A=11mA$

b) Energy = Current x Voltage = 6 x 0.011 = 0.066 J = 66 mJ

c) $\texttt{Power=}(Energy)/(Time)=(0.066)/(0.65* 10^(-3))=101.54W$

A space vehicle of mass m has a speed v. At some instant, it separates into two pieces, each of mass 0.5m. One of the pieces is at rest just after the separation. How much work was done by the internal forces that caused the separation

Answers

Answer:

W = ½ m v²

Explanation:

In this exercise we must solve it in parts, in a first part we use the conservation of the moment to find the speed after the separation

We define the system formed by the two parts of the rocket, therefore the forces during internal separation and the moment are conserved

initial instant. before separation

p₀ = m v

final attempt. after separation

$p_(f)$ = m /2 0 + m /2 v_{f}

p₀ = p_{f}

m v = m /2 $v_(f)$

v_{f}= 2 v

this is the speed of the second part of the ship

now we can use the relation of work and energy, which establishes that the work is initial to the variation of the kinetic energy of the body

initial energy

K₀ = ½ m v²

final energy

$K_(f)$ = ½ m/2 0 + ½ m/2 v_{f}²

K_{f} = ¼ m (2v)²

K_{f} = m v²

the expression for work is

W = ΔK = K_{f} - K₀

W = m v² - ½ m v²

W = ½ m v²

Final answer:

The principle of conservation of momentum implies that no work is performed by the internal forces during the separation of the space vehicle. This is granted that external forces are ignored and the total momentum and kinetic energy of the closed system remain constant.

Explanation:

The subject you're asking about centers around the principle of conservation of momentum. In the case of this space vehicle, before separation, the momentum of the whole system is given by the product of the mass and velocity, mv. After separation, one piece is at rest, leaving the other piece with momentum mv. As there is no external force, the total momentum does not change, so no work is performed by the internal forces causing the separation.

In more detail, the principle of conservation of momentum states that the total linear momentum of a closed system remains constant, regardless of any interactions happening within the system. The system is 'closed' meaning that no external forces are acting upon it. In this case, the space vehicle and the two smaller pieces it separates into form a closed system. This is consistent with your question's stipulation to ignore external forces, such as gravitational forces.

This can also be understood from the work-energy theorem, which states that the work done on an object is equal to the change in its kinetic energy. If we consider the vehicle before and after the separation, the kinetic energy of the system remains the same: initially all the energy is concentrated in the moving vehicle, and after the separation, all the kinetic energy is transferred to the moving piece while the at-rest piece has none. Therefore, the work done by the internal forces - which would change the kinetic energy - must be zero.

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