Consider the interactions involved when you use a TV remote control to change the channel. Classify each interaction as long range or short range.

Answers

Answer 1
Answer:

Explanation:

Following are two interactions that are generally involved when we use a TV remote control to change the channel :

1. Figure touches remote buttons, and its a short range interaction.

2. Now remote sends signal to Television, this is a long range interaction.

Answer 2
Answer:

Final answer:

The interactions of a TV remote and the TV involve short-range infrared communication, while the TV receives signals from long-range electromagnetic waves broadcasted for channels in frequency ranges for VHF and UHF.

Explanation:

When you use a TV remote control to change the channel, two main interactions are involved. The first interaction is the infrared communication between the remote and the TV, which is a form of electromagnetic radiation. Infrared signals require a direct line of sight, operating over a relatively short range. On the other hand, the TV itself receives broadcast signals through antennas that capture electromagnetic waves broadcasted over a long range - these signals can be VHF or UHF TV channels.

Additionally, the TV channels are broadcasted on frequencies ranging from 54 to 88 MHz and 174 to 222 MHz for VHF, while UHF channels utilize frequencies from 470 to 1000 MHz. These signals are sent over a significant distance to your TV’s antenna, showing that television broadcast interaction is long range. These broadcast signals are part of electromagnetic spectrum and carry a large range of frequencies due to the variety of content (audio and visual information) that needs to be transmitted.


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The planet earth orbits around the sun and also spins around its own axis. 33% part (a) calculate the angular momentum of the earth in its orbit around the sun in kg • m2/s
Which equation represents inverse proportionality?Responses y=m/x y equals m divided by x y=mx2+b for nonzero b y equals m x squared plus b for nonzero b y=mx y equals m x y=mx2
Two atoms collide while moving in a dilute gas. The larger atom has a mass M1 = 6 Daltons and a speed v1 = 200 m/s, while the smaller has a mass M2 = 1 Daltons. During the collision both atoms simply bounce off each other. They do not change their speeds, but after the collision they each change their directions, bouncing in the indicated directions. (You may express your results using the mass unit "Daltons". 1 Dalton is approximately equal to the mass of a proton or neutron and is defined as one-twelfth the mass of a single neutral carbon-12 atom in its ground state.)A. What is the magnitude of the change in the momentum, Δp1, of mass M1? B. What is the change in the total momentum of the pair? C. What is the magnitude of the change in the momentum Δp2, of mass M2?
Consider three drinking glasses. All three have the same area base, and all three are filled to the same depth with water. Glass A is cylindrical. Glass B is wider at the top than at the bottom, and so holds more water than A. Glass C is narrower at the top than at the bottom, and so holds less water than A. In which glass is the pressure on the base greatest liquid pressure at the bottom?a. Glass Ab. Glass Bc. Glass Cd. All three have equal non-zero pressure at the bottom.e. All three have zero pressure at the bottom.

Determine whether or not each of the following statement is true. If a statement is true, prove it. If the statement is false, provide a counterexample and explain how it constitutes a counterexample. Diagrams can be useful in explaining such things. If the electric potential in a certain region of space is constant, then the charge enclosed by any closed surface completely contained within that region is zero.

Answers

Answer:

True

Explanation:

This is a representation of Gauss law.

Gauss’s law does hold for moving charges, and in this respect Gauss’s law is more general than Coulomb’s law. In words, Gauss’s law states that: The net outward normal electric flux through any closed surface is proportional to the total electric charge enclosed within that closed surface. The law can be expressed mathematically using vector calculus in integral form and differential form, both are equivalent since they are related by the divergence theorem, also called Gauss’s theorem.

Sound with frequency 1300 Hz leaves a room through a doorway with a width of 1.03 m . At what minimum angle relative to the centerline perpendicular to the doorway will someone outside the room hear no sound

Answers

Answer:

  about 14.7°

Explanation:

The formula for the angle of the first minimum is ...

  sin(θ) = λ/a

where θ is the angle relative to the door centerline, λ is the wavelength of the sound, and "a" is the width of the door.

The wavelength of the sound is the speed of sound divided by the frequency:

  λ = (340 m/s)/(1300 Hz) ≈ 0.261538 m

Then the angle of interest is ...

  θ = arcsin(0.261538/1.03) ≈ 14.7°

At an angle of about 14.7°, someone outside the room will hear no sound.

Four electrons are located at the corners of a square 10.0 nm on a side, with an alpha particle at its midpoint. Part A How much work is needed to move the alpha particle to the midpoint of one of the sides of the square?

Answers

Final answer:

The total work done in moving an alpha particle from the center to the side of a square with electrons at its corners involves finding the potential energy change, which can be calculated using the charges, distances, and Coulomb's constant.

Explanation:

The question deals with the fundamental concepts of electrostatics and the energy associated with moving charges in an electric field. Given the aforementioned question, we are required to find the work done moving an alpha particle (a helium nucleus, having a charge of +2e) from the center of a square to one of its sides, with electrons (each having a charge of -e) being situated at its corners.

To determine the work done, we must consider the potential energy changes resulting from moving the alpha particle. The potential energy associated with two point charges is given by the formula: U = k*q1*q2/r, where q1 and q2 are charges, r is the distance between them, and k is Coulomb's constant.

First, we calculate the potential energy at the center due to all four electrons then find the potential energy at the midpoint of the side. The work done is the difference between these two potential energies. As the electrons are all at an equal distance from the alpha particle (in the center and on the side), the calculations would involve plugging in the values for the charge of an electron, the charge of an alpha particle, the given distance values, and Coulomb's constant into the aforementioned formula.

Learn more about Electric Potential Energy here:

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Final answer:

The work required to move the alpha particle from the midpoint to the midpoint of one of the side of the square with four electrons at its corners would be zero as the net electric field at the midpoint due to the electrons is zero.

Explanation:

The subject of this question pertains to the concept of electrostatics and potential energy in physics. In this scenario, the alpha particle is initially at the midpoint of a square with four electrons at its corners. As per Coulomb's Law, the electrostatic force between two charges is inversely proportional to the square of the distance between them.

Since the alpha particle placed in the center of the square and four electrons at the corners form a symmetrical system, the net force and hence the net electric field at the midpoint due to the electrons is zero. Thus, no work would be required to move the alpha particle to the midpoint of one of the sides of the square as work done is calculated by the formula W = F x d x cos(θ), where F is force, d is the displacement, and θ is the angle between the force and displacement. Since F is equal to zero, the work done will also be zero.

Learn more about Work Done in Moving a Charge in an Electric Field here:

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While leaning out a window that is 6.0 m above the ground, you drop a 0.60-kg basketball to a friend at ground level. Your friend catches the ball at a height of 1.6 m above the ground. Determine the following.(a) the amount of work done by the force of gravity on the ball.(b) the gravitational potential energy of the ball-earth system, relative to the ground when it is released.(c) the gravitational potential energy of the ball-earth system, relative to the ground when it is caught.

Answers

Answer:

a) W = 25.872 J

b) - 35.28 J

c) - 9.408

Explanation:

a) The amount of work done by the force of gravity on the ball = Change in potential energy between the two vertical points = - mg (H₂ - H₁)

F = - mg (gravity is acting downwards)

F = - 0.6 × 9.8 = - 5.88 N

(H₂ - H₁) = (1.6 - 6) = - 4.4 m

W = (-5.88)(-4.4) = 25.872 J

b) Gravitational-potential energy of the ball when it was released relative to the ground = (- mg) H₁ = (- 0.6 × 9.8) × 6 = - 35.28 J

c) Gravitational-potential energy of the ball when it is caught relative to the ground = (-mg)(H₂) = -0.6 × 9.8 × 1.6 = - 9.408 J

Professional baseball pitchers deliver pitches that can reach the blazing speed of 100 mph (miles per hour). A local team has drafted an up-and-coming, left-handed pitcher who can consistently pitch at 42.91 m/s (96.00 mph) . Assuming a pitched ball has a mass of 0.1434 kg and has this speed just before a batter makes contact with it, how much kinetic energy does the ball have?

Answers

Answer: 132.02 J

Explanation:

By definition, the kinetic energy is written as follows:

KE = 1/2 m v²

In our question, we know from the question, the following information:

m = 0.1434 Kg

v= 42.91 m/s

Replacing in the equation for KE, we have:

KE = 1/2 . 0.1434 Kg. (42.91)² m²/s² ⇒ KE = 132.02 N. m = 132.02 J

HELP ASAP PLEASE!!!In which direction(s) does the ground shake during an earthquake?
A. sideways
B. up and down
C. back and forth
D. all of the above

Answers

Answer: D i am pretty sure

Explanation:

Answer:

all

Explanation: