Answer:
(a) The power wasted for 0.289 cm wire diameter is 15.93 W
(b) The power wasted for 0.417 cm wire diameter is 7.61 W
Explanation:
Given;
diameter of the wire, d = 0.289 cm = 0.00289 m
voltage of the wire, V = 120 V
Power drawn, P = 1850 W
The resistivity of the wire, ρ = 1.68 x 10⁻⁸ Ω⋅m
Area of the wire;
A = πd²/4
A = (π x 0.00289²) / 4
A = 6.561 x 10⁻⁶ m²
(a) At 26 m of this wire, the resistance of the is
R = ρL / A
R = (1.68 x 10⁻⁸ x 26) / 6.561 x 10⁻⁶
R = 0.067 Ω
Current in the wire is calculated as;
P = IV
I = P / V
I = 1850 / 120
I = 15.417 A
Power wasted = I²R
Power wasted = (15.417²)(0.067)
Power wasted = 15.93 W
(b) when a diameter of 0.417 cm is used instead;
d = 0.417 cm = 0.00417 m
A = πd²/4
A = (π x 0.00417²) / 4
A = 1.366 x 10⁻⁵ m²
Resistance of the wire at 26 m length of wire and 1.366 x 10⁻⁵ m² area;
R = ρL / A
R = (1.68 x 10⁻⁸ x 26) / 1.366 x 10⁻⁵
R = 0.032 Ω
Power wasted = I²R
Power wasted = (15.417²)(0.032)
Power wasted = 7.61 W
The answer is false! Hope that helps! :)
A message telephone messagemessage a communication usually brie that is written or spoken or signaled; "hesent a three-word <span>message"</span>
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.
Answer:
W = 0 J
Explanation:
The amount of work done by gas at constant pressure is given by the following formula:
where,
W = Work done by the gas
P = Pressure of the gas
ΔV = Change in the volume of the gas
Since the volume of the gas is constant. Therefore, there is no change in the volume of the gas:
<u>W = 0 J</u>
