The answer is A.
A positive charge’s electric field pushes out.
Hope this helps! -Avenging
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
Answer:
V = 0.248 L
Explanation:
To do this, use the following equation:
P1*V1/T1 = P2*V2/T2
This equation is used to find a relation between two differents conditions of a same gas, which is this case. From this equation we can solve for V2.
Solving for V2:
V2 = P1*V1*T2/T1*P2
Temperature must be at Kelvin, so, we have to sum the temperature 273 to convert it in K.
Replacing the data we have:
V2 = 1 * 4.91 * (-196+273) / 5.2 * (20+273)
V2 = 378.07 / 1523.6
V2 = 0.248 L
Answer: be a better person
Explanation:
I think transfers is the answer
