An oscillating dipole antenna 1.73 m long with a maximum 36.0 mV potential creates a 500 Hz electromagnetic wave. (a) What is the maximum electric field strength created

Answers

Answer 1

Answer:

[tex]E_0=0.021v/m[/tex]

Explanation:

From the question we are told that:

Length [tex]l=1.73m[/tex]

Voltage [tex]V=36.0mV[/tex]

Frequency [tex]F=500Hz[/tex]

Generally the equation for maximum Electric Field  is mathematically given by

 [tex]E_0=\frac{\triangle}{r}[/tex]

 [tex]E_0=\frac{36*10^{-3}}{1.73}[/tex]

 [tex]E_0=0.021v/m[/tex]


Related Questions

given A=4i-10j and B= 7i+5j find b such that A+bB is a vector pointing along the x-axis (i.e has no y component)​

Answers

Answer:

-4/7

Explanation:

Given the following

A=4i-10j and B= 7i+5j

A+ bB = 4i-10j + (7i+5j)b

A+ bB =  4i-10j + 7ib+5jb

A+ bB =

The vector along the x-axis is expressed as i + 0j

If the vector A+ bB is pointing in the direction of the x-axis then;

[tex]A+ bB * \frac{i+0j}{|i+0j|} = 0 \\ (4+7b)i-(10-5b)j* \frac{i+0j}{\sqrt{1^2+0^2} } = 0\\(4+7b)i-(10-5b)j *(i+0j) = 0\\4+7b-0 =0\\7b=-4\\b = -4/7[/tex]

Hence the value of b is -4/7

The value of [tex]\beta[/tex] such that [tex]\vec C = \vec A + \beta \cdot \vec B = c\,\hat{i}[/tex] is 2.

According to the statement, we have following system of vectorial equations:

[tex]\vec A = 4\,\hat {i} - 10\,\hat{j}[/tex] (1)

[tex]\vec {B} = 7\,\hat{i} + 5\,\hat{j}[/tex] (2)

[tex]\vec C = \vec A + \beta \cdot \vec B = c\,\hat{i}[/tex] (3)

By applying (1) and (2) in (3):

[tex](4\,\hat{i}-10\,\hat{j}) + \beta\cdot (7\,\hat{i}+5\,\hat{j}) = c\,\hat{i}[/tex]

[tex](4+7\cdot \beta)\,\hat{i} +(-10+5\cdot \beta)\,\hat{j} = c\,\hat{i}[/tex]

And we get two scalar equations after analyzing each component:

[tex]4+7\cdot \beta = c[/tex] (4)

[tex]-10+5\cdot \beta = 0[/tex] (5)

We solve for [tex]\beta[/tex] in (5):

[tex]\beta = 2[/tex]

And for [tex]c[/tex] in (4):

[tex]c = 4+7\cdot (2)[/tex]

[tex]c = 18[/tex]

The value of [tex]\beta[/tex] such that [tex]\vec C = \vec A + \beta \cdot \vec B = c\,\hat{i}[/tex] is 2.

Please see this question related to Sum of Vectors for further details: https://brainly.com/question/11881720

Why are hydraulic brakes used?​

Answers

Answer:

Hydraulic brake systems are used as the main braking system on almost all passenger vehicles and light trucks. Hydraulic brakes use brake fluid to transmit force when the brakes are applied.

Explanation:

If car A passes car B, then car A must be
A. accelerating at a greater rate than car B.
B. moving faster than car B, but not necessarily accelerating
C. accelerating
D. moving faster than car B and accelerating more than car B

Answers

Answer:

B. moving faster than car B, but not necessarily accelerating

Explanation:

Velocity is the speed of something. So car A's velocity is greater than car B but does not mean car A is accelerating.

A physical pendulum in the form of a planar object moves in simple harmonic motion with a frequency of 0.680 Hz. The pendulum has a mass of 2.00 kg, and the pivot is located 0.340 m from the center of mass. Determine the moment of inertia of the pendulum about the pivot point.

Answers

Answer:

Therefore, the moment of inertia is:

[tex]I=0.37 \: kgm^{2} [/tex]

Explanation:

The period of an oscillation equation of a solid pendulum is given by:

[tex]T=2\pi \sqrt{\frac{I}{Mgd}}[/tex] (1)

Where:

I is the moment of inertiaM is the mass of the pendulumd is the distance from the center of mass to the pivotg is the gravity

Let's solve the equation (1) for I

[tex]T=2\pi \sqrt{\frac{I}{Mgd}}[/tex]

[tex]I=Mgd(\frac{T}{2\pi})^{2}[/tex]

Before find I, we need to remember that

[tex]T = \frac{1}{f}=\frac{1}{0.680}=1.47\: s[/tex]

Now, the moment of inertia will be:

[tex]I=2*9.81*0.340(\frac{1.47}{2\pi})^{2}[/tex]  

Therefore, the moment of inertia is:

[tex]I=0.37 \: kgm^{2} [/tex]

I hope it helps you!

A gymnast weighs 450 N. She stands on a balance beam of uniform construction which weighs 250 N. The balance beam is 3.0 m long and is supported at each end. If the support force at the right end is four times the force at the left end, how far from the right end is the gymnast

Answers

Answer:

   x = 9.32 cm

Explanation:

For this exercise we have an applied torque and the bar is in equilibrium, which is why we use the endowment equilibrium equation

Suppose the counterclockwise turn is positive, let's set our reference frame at the left end of the bar

          - W l / 2 - W_{child} x + N₂ l = 0

             x = [tex]\frac{-W l/2 + n_2 l}{W_{child}}[/tex]             1)

now let's use the expression for translational equilibrium

         N₁ - W - W_(child) + N₂ = 0

indicate that N₂ = 4 N₁

we substitute

           N₁ - W - W_child + 4 N₁ = 0

           5 N₁ -W - W_{child} = 0

           N₁ = ( W + W_{child}) / 5

         

we calculate

           N₁ = (450 + 250) / 5

          N₁ = 140 N

           

we calculate with equation 1

           x = -250 1.50 + 4 140 3) / 140

           x = 9.32 cm

a method of reducing friction​

Answers

Answer:

Lubrication

Explanation:

People oil/lubricate bicycle chains because the chain turns around the cogs and rub together so this help with friction.

Hope this helps :)

Answer:

The method of reducing friction are :

i) In moving parts of machine friction can be reduced by using a ball bearing between the moving surfaces

ii) The bodies of aeroplane ,ship ,boat etc are made streamlined to reduce friction.

iii) Friction can be reduced by polishing rough surfaces. For example : carrom boards are highly polished to reduce friction.

I hope this help you:)

Which of the following is not true about Triton, the large moon of Neptune? It is more reflective than Earth's Moon. It is larger than Earth's Moon. It is in a retrograde orbit. It has a thin atmosphere. It has nitrogen geysers.

Answers

Answer:

Triton is the largest of Neptune's 13 moons. It is unusual because it is the only large moon in our solar system that orbits in the opposite direction of its planet's rotation―a retrograde orbit. ... Like our own moon, Triton is locked in synchronous rotation with Neptune―one side faces the planet at all times.

In the diagram, disk 1 has a moment of inertia of 3.4 kg · m2 and is rotating in the counterclockwise direction with an angular velocity of 6.1 rad/s about a frictionless rod passing through its center. A second disk rotating clockwise with an angular velocity of 9.3 rad/s falls from above onto disk 1. The two then rotate as one in the clockwise direction with an angular velocity of 1.8 rad/s. Determine the moment of inertia, in kg · m2, of disk 2.

Answers

Answer:

I = 3.6 kg•m²

Explanation:

Conservation of angular momentum

Let's assume CW is the positive direction

3.4(-6.1) + I(9.3) = 3.4(1.8) + I(1.8)

I(9.3 - 1.8) = 3.4(1.8 + 6.1)

I(7.5) = 3.4(7.9)

I = 3.4(7.9)/(7.5) = 3.5813333333...

The moment of inertia of the second disk will be  [tex]I=3.58\ kg-m^2[/tex]

What is moment of inertia?

The moment of inertia is defined as the product of mass of section and the square of the distance between the reference axis and the centroid of the section.

here it is given that

MOI of disk one   [tex]I_1=3.4\ kg-m^2[/tex]

Angular velocity  [tex]w_1=6.1\ \frac{rad}{s}[/tex]

Angular velocity of disk two  [tex]w=1.8\ \frac{rad}{s}[/tex]

MOI of the disk two [tex]I=?[/tex]

The final angular velocity [tex]w_f= 1.8\ \frac{rad}{sec}[/tex]

Now from the conservation of the momentum the angular momentum before collision will be equal to the angular momentum after collision.

[tex]I_1w_1+I_2w_2=(I_1+I_2)w_f[/tex]

Now put the values in the formula

[tex](3.4\times 6.10)+(I_2\times 9.3)=(3.4+I_2)\times 1.8[/tex]

[tex]I_2=3.58\ kg-m^2[/tex]

Thus the moment of inertia of the second disk will be  [tex]I=3.58\ kg-m^2[/tex]

To know more about moment of inertia follow

https://brainly.com/question/18721435


How is the sun used to make food?

Answers

Answer:

Plants use a process called photosynthesis to make food. During photosynthesis, plants trap light energy with their leaves. Plants use the energy of the sun to change water and carbon dioxide into a sugar called glucose. Glucose is used by plants for energy and to make other substances like cellulose and starch.

                                             Thank you

Answer:

Plants use a process called photosynthesis to make food. During photosynthesis, plants trap light energy with their leaves. Plants use the energy of the sun to change water and carbon dioxide into a sugar called glucose.

Describe how you could whether sound travels best through wood, plastic, or metal.

Answers

Answer:

metal

Explanation:

sound can travel best in materials with higher elastic properties like metal than it can through other solids like plastic or rubber which have lower elastic properties

I hope this helps

You have a 1 W light bulb in your lab. It puts out light of only 1 frequency. The wavelength of this light is 500nm. you set up a detector with a surface area of 1 square centimeter facing the light source at a distance of 100m.

Required:
a. Find the number of photons hitting the detector every second.
b. What is the maximum E field of the E M wave hitting the detector?
c. What is the maximum value of the B field of this E M wave?
d. How far away would you have to place the detector to only receive 1 photon per second from the light bulb?

Answers

Answer:

a)   # _photon = 2.5 10¹⁸ photons / s,   b) E = 10⁻² N / C,  c)     B = 3 10⁻¹¹ T

d)  r=  2 10⁹ m

Explanation:

a) Let's solve this exercise in part, let's start by finding the energy of each photon using the Planck relation

          E₀ = h f

          c = λ f

          E₀ = h c /λ

          E₀ = 6.63 10⁻³³⁴   3 10⁸/500 10⁻⁹

          E₀ = 3.978 10⁻⁻¹⁹ J

Let's use a direct ratio rule to find the number of photons

         #_foton = E / Eo

         #_fototn = 1 / 3.978 10⁻¹⁹

         # _photon = 2.5 10¹⁸ photons / s

b) The intensity received by the detector is related to the electric field

          I = E²

Let's look for the intensity that the detector receives, suppose that the emission is shapeless throughout the space

          I = P / A

          P = I A

Let's use index 1 for the point on the bulb and index 2 for the point on the detector.

The area of ​​a sphere is

          A = 4π r²

         P = I₁ A₁ = I₂ A₂

         I₁ r₁² = I₂ r₂²

         I₂ = I₁  r₁²/r₂²

         I₂ = I₁    1 / 100²

         I₂ = I₁ 10⁻⁴

we must know the intensity at the output of the bulb suppose that I₁ = 1 J

          I₂ = 10⁻⁴ J

let's look for the electric field

         E =√I

         E = √10⁻⁴

         E = 10⁻² N / C

c) for the calculation of the magnetic field we use that the field is in phase

               E / B = c

               B = E / c

               B = 10⁻² / 3 10⁸

               B = 3 10⁻¹¹ T

d) Let's use a direct proportions rule if we fear 2.5 10¹⁸ photons in an area  A = 4π R² where R = 100 m how many photons are there in the area of ​​the detector r = 1 cm,   A’= 10⁻⁴ m²

             #_photons = 2.5 10¹⁸ A_detector / A_sphere

             #_photons = 2.5 1018 10-4 / 4π 10⁴

             #_photons = 2 10⁹ photons in the detector area

for the number of photons to decrease to 1, the radius of the sphere must be 2 10⁹ m

A 55kg bungee jumper has fallen far enough that her bungee cord is beginning to stretch and resist her downward motion . Find the ( magnitude and direction ) exerted on her by the bungee cord at an instant when her downward acceleration has a magnitude of 7.1m/s2​

Answers

Answer:

148.5 N

Explanation:

Given that,

The mass of a bungee jumper, m = 55 kg

The downward acceleration, a = 7.1 m/s²

We need to find the net force acting on the jumper. As it is moving in downward direction, net force is :

T = m(g-a)

Put all the values,

T = 55(9.8 - 7.1)

= 148.5 N

So, the force exerted on the bungee cord is 148.5  N.

Answer:

The downward force is 148.5 N.

Explanation:

mass, m = 55 kg

downwards acceleration, a = 7.1 m/s^2

Let the force is F.

According to the newton's second law

m g - F = m a

F = m (g - a)

F = 55 (9.8 - 7.1)

F = 148.5 N

A mysterious constant force of 10 N acts horizontally on everything. The direction of the force is found to be always pointed toward a wall in a big hall. Find the potential energy of a particle due to this force when it is at a distance x from the wall, assuming the potential energy at the wall to be zero.

Answers

Answer:

it will be 10x

Explanation:

workdone(potential energy before it hits the wall)= horizontal force × distance

=10× x = 10x joules

A mysterious constant force of 10 N acts horizontally on everything. The direction of the force is found to be always pointed toward a wall in a big hall.The potential energy of a particle due to this force is  10x.

What is force?

A force is an effect that can alter an object's motion according to physics. An object with mass can change its velocity, or accelerate, as a result of a force. An obvious way to describe force is as a push or a pull. A force is a vector quantity since it has both magnitude and direction.

Given in the question a mysterious constant force of 10 N acts horizontally on everything. The direction of the force is found to be always pointed toward a wall in a big hall the potential energy,

Work done (potential energy before it hits the wall)

= horizontal force × distance

=10× x = 10x joules

The potential energy of a particle due to this force is  10x.

To learn more about force refer to the link:

brainly.com/question/13191643

#SPJ2

Observe: Air pressure is equal to the weight of a column of air on a particular location. Air pressure is measured in hectopascals (hPa). Note how the air pressure changes as you move Station B towards the center of the high-pressure system.

a. What do you notice?
b. Why do you think this is called a high-pressure system?

Answers

Answer:

A. When moving towards a high pressure center the pressure values ​​increase in the equipment

B. This area is called high prison since the weight of the atmosphere on top is maximum

Explanation:

A) A high atmospheric pressure system is an area where the pressure is increasing the maximum value is close to 107 Kpa, the other side as low pressure can have small values ​​85.5 kPa.

When moving towards a high pressure center the pressure values ​​increase in the equipment

B) This area is called high prison since the weight of the atmosphere on top is maximum

in general they are areas of good weather

What is the energy of a photon with a frequency of 3.6 × 1015 Hz? Planck’s constant is 6.63 × 10–34 J•s.

1.8 × 10–49 J
2.4 × 10–19 J
1.8 × 10–18 J
2.4 × 10–18 J

Answers

Frequency =v=3.6×10^15Hz

We know

[tex]\boxed{\sf E=hv}[/tex]

[tex]\\ \sf\longmapsto E=6.63\times 10^{-34}J\times 3.6\times 10^{15}s^{-1}[/tex]

[tex]\\ \sf\longmapsto E=23.86\times 10^{-19}J[/tex]

[tex]\\ \sf\longmapsto E=2.38\times 10^{-18}J[/tex]

[tex]\\ \sf\longmapsto E=2.4\times 10^{-18}J[/tex]

Answer:

D!!!!!

Explanation:

Our system is a block attached to a horizontal spring on a frictionless table. The spring has a spring constant of 4.0 N/m and a rest length of 1.0 m, and the block has a mass of 0.25 kg.

Compute the PE when the spring is compressed by 0.50 m.

Answers

Answer

E - 1/2 K x^2      potential energy of compressed spring

E = 1/2 * 4 N / m * (.5 m)^2 = 2 * .5^2 N-m = .5 N-m

A magnifying glass produces a maximum angular magnification of 5.4 when used by a young person with a near point of 20 cm. What is the maximum angular magnification obtained by an older person with a near point of 65 cm

Answers

67 points to the 55th power add it then you get exactly 1500

If the mass of an object is 15 kg and the velocity is -4 m/s, what is the momentum?

Answers

momentum p= m x v = 15 x -4 = -60 N.s


The area around a charged object that can exert a force on other charged objects is an electric ___

Answers

Answer is:

Electric field.

Flag question
Consider the pressure and force acting on the
dam retaining a reservoir of water. Suppose the
dam is 500-m wide
and the water is 80.0-m
deep at the dam, as illustrated below. What is
the average pressure on the dam due to the
water?​

Answers

Answer:

P = density (p) * g * h

P = 1000  kg/m^3 * 9.8 m/s^2 * 40 m = 392,000 N/m^2

since kg m / s^2 = Newtons

The average pressure is 1/2 (pressure at 0m +  pressure 80 m) for liquid of uniform density

A car whose tire have radii 50cm travels at 20km/h. What is the angular velocity of the tires?

Answers

Radius=r=50cm=0.5mVelocity=20km/h=v

Convert to m/s

[tex]\\ \sf\longmapsto v=20\times 5/18=5.5m/s[/tex]

We know

[tex]\boxed{\sf \omega=\dfrac{rv}{|r|^2}}[/tex]

[tex]\\ \sf\longmapsto \omega=\dfrac{0.5(5.5)}{|0.5|^2}[/tex]

[tex]\\ \sf\longmapsto \omega=\dfrac{2.75}{0.25}[/tex]

[tex]\\ \sf\longmapsto \omega=11rad/s[/tex]

During normal beating, the heart creates a maximum 4.10-mV potential across 0.350 m of a person's chest, creating a 1.00-Hz electromagnetic wave. (a) What is the maximum electric field strength created? V/m (b) What is the corresponding maximum magnetic field strength in the electromagnetic wave? T (c) What is the wavelength of the electromagnetic wave?

Answers

Explanation:

Given that,

Maximum potential, V = 4. mV

Distance, d = 0.350 m

Frequency of the wave, f = 100 Hz

(a) The maximum electric field strength created is given by:

[tex]E=\dfrac{V}{d}\\\\E=\dfrac{4.1\times 10^{-3}}{0.350 }\\\\E=0.0117\ V/m[/tex]

(b) The corresponding maximum magnetic field strength in the electromagnetic wave is given by :

[tex]B=\dfrac{E}{c}\\\\B=\dfrac{0.0117}{3\times 10^8}\\\\B=3.9\times 10^{-11}\ T[/tex]

(c) The wavelength of the electromagnetic wave can be calculated as :

[tex]\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{100}\\\\=3\times 10^6\ m[/tex]

So, the wavelength of the electromagnetic wave is [tex]3\times 10^6\ m[/tex].

A 2.0 kg puck is at rest on a level table. It is pushed straight north with a constant force of 5N for 1.50 s and then let go. How far does the puck move from rest in 2.25 s?

Answers

Answer:

d = 6.32 m

Explanation:

Given that,

The mass of a puck, m = 2 kg

It is pushed straight north with a constant force of 5N for 1.50 s and then let go.

We need to find the distance covered by the puck when move from rest in 2.25 s.

We know that,

F = ma

[tex]a=\dfrac{F}{m}\\\\a=\dfrac{5}{2}\\\\a=2.5\ m/s^2[/tex]

Let d is the distance moved in 2.25 s. Using second equation of motion,

[tex]d=ut+\dfrac{1}{2}at^2\\\\d=0+\dfrac{1}{2}\times 2.5\times (2.25)^2\\\\d=6.32\ m[/tex]

So, it will move 6.32 m from rest in 2.25 seconds.

1 of 3 : please help got an extra day for a test and i don’t get this (must show work) points and brainliest!

Answers

Answer:

y = 1/2at^2

we could also write it as-

y = (at^2)/2

2y = at^2

2y/a = t^2

√2y/a = t

hope it helps

Two loudspeakers emit sound waves along the x-axis. The sound has maximum intensity when the speakers are 21 cm apart. The sound intensity decreases as the distance between the speakers is increased, reaching zero at a separation of 61 cm. a. What is the wavelength of the sound

Answers

Answer:

The answer is "80 cm".

Explanation:

The distance of 21 cm between the speaker's effect of high strength but a spacing of 61 cm corresponds to a zero to zero intensity, that also is, the waves are all in phase with others [tex]\Delta \ x_1 = 21 \ cm[/tex] this is out of phase [tex]\Delta\ x_2 = 61\ cm[/tex]

[tex]\therefore\\\\\Delta\ x_2 -\Delta\ x_1 = \frac{\lambda}{2}\\\\\lambda= 2( \Delta\ x_2 -\Delta\ x_1)[/tex]

   [tex]= 2 ( 61\ cm - 21\ cm)\\\\ = 2(40\ cm)\\\\= 80\ cm[/tex]

A linear accelerator can be used to accelerate which of the following?
Question 3 options:

protons and electrons

protons and neutrons

protons only

protons, electrons, and neutrons

Answers

Answer: protons and electrons

a student standing between two walls shouts once.he hears the first echo after 3 seconds and the next after 5 seconds. calculate the distance between the walls.​

Answers

Explanation:

It took [tex]t_1 =1.5\:\text{s}[/tex] for the sound to reach the 1st wall and at the same time time, the same sound took [tex]t_2 = 2.5\:\text{s}[/tex] to reach the 2nd wall. Assuming that the sound travels at 343 m/s, then let [tex]x_1[/tex] be the distance of the person to the 1st wall and [tex]x_2[/tex] be the distance to the 2nd wall. So the distance between the walls X is

[tex]X = x_1 + x_2 = v_st_1 + v_st_2 = v_s(t_1 + t_2)[/tex]

[tex]\:\:\:\:\:= (343\:\text{m/s})(4.0\:\text{s}) = 1372\:\text{m}[/tex]

the wave frequency is a measure of

Answers

Answer: hertz

Explanation:

Wavelength is also measured in metres ( ) - it is a length after all. The frequency ( ) of a wave is the number of waves passing a point in a certain time. We normally use a time of one second, so this gives frequency the unit hertz ( ), since one hertz is equal to one wave per second.

Answer:

The number of waves that pass a particular point in a unit of time.

Explanation:

The frequency of a wave describes the number of waves that pass a particular point in a unit of time. The frequency is inversely proportional to the period of the wave, which is the time required for one full wave cycle to complete.

PF

a person lifts 60kg on the surface of the earth, how much mass can he lift on the surface of the moon if he applies same magnitude of force​

Answers

Explanation:

Hey there!

According to the question;

A person can lift mass of 60 kg on earth.

mass(m1) = 60kg

acceleration due to gravity on earth (a) = 9.8m/s²

Now;

force (f) = m.a

= 60*9.8

= 588 N

Since, there is application of same magnitude of force on moon,

mass(m) =?

acceleration due to gravity on moon (a) = 1.67m/s²

Now;

force (f) = m.a

588 = m*1.67

m = 352.09 kg

Therefore, the person who can lift the mass of 60 kg on earth can lift mass of 352 kg on moon.

Hope it helps!

An observer on Earth sees rocket 1 leave Earth and travel toward Planet X at 0.3c. The observer on Earth also sees that Planet X is stationary. An observer on Planet X sees rocket 2 travel toward Earth at 0.4c. What is the speed of rocket 1 according to an observer on rocket 2?

Answers

Answer:

0.625 c

Explanation:

Relative speed of a body may be defined as the speed of one body with respect to some other or the speed of one body in comparison to the speed of second body.

In the context,

The relative speed of body 2 with respect to body 1 can be expressed as :

[tex]$u'=\frac{u-v}{1-\frac{uv}{c^2}}$[/tex]

Speed of rocket 1 with respect to rocket 2 :

[tex]$u' = \frac{0.4 c- (-0.3 c)}{1-\frac{(0.4 c)(-0.3 c)}{c^2}}$[/tex]

[tex]$u' = \frac{0.7 c}{1.12}$[/tex]

[tex]u'=0.625 c[/tex]

Therefore, the speed of rocket 1 according to an observer on rocket 2 is 0.625 c

Other Questions
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