The answer 1 is
The best-known use of radio waves is for communication; television, cellphones and radios all receive radio waves and convert them to mechanical vibrations in the speaker to create sound waves that can be heard.Electromagnetic radiation is transmitted inwaves or particles at different wavelengths and frequencies.
the answer 2 is
Radio Waves: Instant Communication
Microwaves: Data and Heat
i think the answer number 3 is
It is all about wavelength versus tunnel diameter. The wavelength of GPS is about 20cm it would happily propagate in any normal tunnel if it could get in but the earth and other structures absorb it. AM radio (600kHz - 1500kHz) cannot propagate in any normal tunnel because the wavelength is too long (500m-200m) relative to the diameter, and thus gets reflected at the entrance. FM (100MHz ~ 3m) would propagate and it does for a while but then it suffers reflections
The Renaissance, was the rebirth to science as well as many other advancements
The sound level of the sound wave due to the ambulance is 140.
<h3>What do you mean by sound?</h3>
In terms of physics, the sound is a vibration that travels through a transmission medium like a gas, liquid, or solid as an acoustic wave. Sound is the reception of these waves and the brain's perception of them in terms of human physiology and psychology. Only acoustic waves with frequencies between roughly 20 Hz and 20 kHz, or the audio frequency range, can cause a human to have an auditory sensation. These are sound waves with wavelengths ranging from 17 meters (56 ft) to 1.7 millimeters in the air at atmospheric pressure (0.67 in). Ultrasounds are sound waves with a frequency higher than 20 kHz that are inaudible to humans. Infrasound refers to sound frequencies below 20 Hz. Animals of different species have different hearing ranges.
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Answer: The electrons flowing through the wire are referred to as a quantity of electricity, and the flow of electricity is referred to as “an electric current.”
Explanation: Hope it Helps have a blessed day
Answer:
1. Energy = 2880 Joules.
2. Energy = 60 Joules.
3. Quantity of charge = 120 Coulombs.
Explanation:
Given the following data;
1. Voltage = 12 Volts
Current = 0.5 Amps
Time, t = 8 mins to seconds = 8 * 60 = 480 seconds
To find the energy;
Power = current * voltage
Power = 12 * 0.5
Power = 6 Watts
Next, we find the energy transferred;
Energy = power * time
Energy = 6 * 480
Energy = 2880 Joules
2. Charge, Q = 4 coulombs
Potential difference, p.d = 15V
To find the total energy transferred;
Energy = Q * p.d
Energy = 4 * 15
Energy = 60 Joules
3. Voltage = 6 Volts
Current = 1 Amps
Time = 2 minutes to seconds = 2 * 60 = 120 seconds
To find the quantity of charge;
Quantity of charge = current * time
Quantity of charge = 1 * 120
Quantity of charge = 120 Coulombs
