when a ball is dropped off a cliff in free fall, it has an acceleration of 9.8 m/s^2. what is its acceleration as it gets closer to the ground

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.

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.

Answer:

Changing the battery's voltage will also change the flow of electrons through the circuit. An increase in the voltage will produce more electron movement, and a reduction in the voltage will produce less electron movement.

Explanation:

The voltage is the potential energy between two points in an electric circuit. It is also the work done per unit charge to move a charge between these two points, this work is done against the resistance (analogous to frictional forces in the wire) of the wire. The potential energy is like the push required to move an electron through an electric circuit, and negatively charged particle (electron in the case of the wire) are pulled towards the higher potential, which is conventionally at the positive terminal. Current (flow of electrons) will not occur without any potential between two points.

Changing the battery's voltage will change the push that is experienced by the electrons. If the potential is increased, the electrons will experience more push, which means there will be more movement or flow of electrons through the circuit. Reducing the battery's voltage reduces the push experienced by the electrons; meaning a reduced flow or movement of these electrons through the circuit.

Answer:

0.157 V

Explanation:

Parameters given:

Number of turns, N = 1207

Diameter of coil = 20 cm = 0.2 m

Radius of coil, r = 0.2/2 = 0.1 m

Magnetic field strength, B =

Time interval, t = 10 ms =

The average EMF induced in a coil due to a magnetic field is given as:

EMF =

where A = Area of coil

A = π

Therefore, EMF will be:

Answer:

speed when it reaches y = 4.00cm is

v = 14.9 g.m/s

Explanation:

given

q₁=q₂ =2.00 ×10⁻⁶

distance along x = 3.00cm= 3×10⁻²

q₃= 4×10⁻⁶C

mass= 10×10 ⁻³g

distance along y = 4×10⁻²m

r₁ = = = 3.61cm = 0.036m

r₂ = = = 5cm = 0.05m

electric potential V =

change in potential ΔV =

ΔV = , where 2.00μC

ΔV =

ΔV = 2 × 9×10⁹ × 2×10⁻⁶ ×

ΔV= 2.789×10⁵

= ΔV × q₃

ˣ 10×10⁻³ ×v² = 2.789×10⁵× 4 ×10⁻⁶

v² = 223.12 g.m/s

v = 14.9 g.m/s

The speed of the charge q₃ when it starts from rest at y = 2 cm and reaches y = 4 cm is; v = 14.89 m/s

We are given;

Charge 1; q₁ = 2.00 μC = 2 × 10⁻⁶ C

Charge 2; q₂ = 2.00 μC = 2 × 10⁻⁶ C

Distance of charge 1 along x = 3 cm = 3 × 10⁻² m

Distance of charge 2 along x = -3 cm = -3 × 10⁻² m

Charge 3; q₃ = +4.00 μC  = 4 × 10⁻⁶ C

mass; m = 0.01 g

distance of charge 3 along y = 4 cm = 4 × 10⁻² m

q₃ starts from rest at y = 2 × 10⁻² m and reaches y = 4 × 10⁻² m.

Thus;

Distance of charge 1 from the initial position of q₃;

r₁ = √((3 × 10⁻²)² + ((2 × 10⁻²)²)

r₁ = 0.0361 m

Distance of charge 2 from the final position of q₃;

r₂ = √((3 × 10⁻²)² + ((4 × 10⁻²)²)

r₂ = 0.05 m

Now, formula for electric potential is;

V = kq/r

Where k = 9 × 10⁹ N.m²/s²

Thus,change in potential is;

ΔV = V₁ - V₂

Now, Net V₁ = 2kq₁/r₁

Net V₂ = 2kq₂/r₂

Thus;

ΔV = 2kq₁/r₁ - 2kq₂/r₂

ΔV = (2 × 9 × 10⁹)[(2 × 10⁻⁶/0.0361) - (2 × 10⁻⁶/0.05)]

ΔV = 277229.92 V

Now, from conservation of energy;

½mv² = q₃ΔV

Thus;

½ × 0.01 × v² = 4 × 10⁻⁶ × 277229.92

v² = 2 × 4 × 10⁻⁶ × 277229.92/0.01

v = √(221.783936)

v = 14.89 m/s

Read more about point charges at;brainly.com/question/13914561