Answer:
Air resistance
Explanation:
Despite the law of conservation of energy stating that energy can neither be created nor destroyed but can only be transformed from one state to another, some energy is usually lost in the process of transformation and its majorly attributed to frictional loss. Friction opposes normal movement hence in air, air resistance tends to reduce the original energy compared to the initial. That is why the final energy in this case is slightly less than the original energy.
Answer:
Amplitude and Frequency
Explanation:
Analog signals are composed of continuous waves that can have any values for frequency and amplitude. These waves are smooth and curved.
Radio transmissions are a combination of two kinds of waves: audio frequency waves that represent the sounds being transmitted and radio frequency waves that "carry" the audio information. All waves have a wavelength, an amplitude and a frequency as shown in the figure. These properties of the wave allow it to be modified to carry sound information.
The two most common types of modulation used in radio are amplitude modulation (AM) and frequency modulation (FM). Frequency modulation minimizes noise and provides greater fidelity than amplitude modulation, which is the older method of broadcasting . Both AM and FM are analog transmission systems, that is, they process sounds into continuously varying patterns of electrical signals which resemble sound waves.
this is simple the word your looking for is 8 letters long and the definition is key to the answer
The resistance of a single light bulb is 220 ohms per bulb.
<h3>What is Ohm's Law?</h3>
Ohm's Law is a formula used to determine how voltage, current, and resistance in an electrical circuit relate to one another.
Ohm's Law (E = IR) is as basic to students of electronics as Einstein's Relativity equation (E = mc2) is to physicists.
E = I x R
The formula reads voltage = current x resistance, or V = A xΩ., or volts = amps x ohms.
110volts divided by .25amps = 440 ohms. 440 divided by 2 =220 ohms per bulb.
R = 110/(2*0.25) = 220 ohms
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