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:
2583.9 N/C
Explanation:
Parameters given:
Outer diameter = 14 cm
Outer radius, R = 7cm = 0.07m
Inner diameter = 7 cm
Inner radius, r = 3.5 cm = 0.035m
Charge of washer = 8 nC = 8 * 10^(-9)C
Distance from washer, z = 33 cm = 0.33m
The electric field due to a washer (hollow disk) is given as:
E = k * σ * 2π [ 1 - z/(√(z² + R²)]
Where σ = charge per unit area
σ = q/π(R² - r²)
σ = 8 * 10^(-9) /(π*(0.07 - 0.035)²)
σ = 2.077 * 10^(-6) C/m²
=> E = 9 * 10^9 * 2.077 * 10^(-6) * 2π * [1 - 0.33/(√(0.33² + 0.07²)]
E = 117.467 * 10^3 * (1 - 0.978)
E = 117.467 * 10^3 * 0.022
E = 2583.9 N/C
I think atoms and molecules in matter are always in motion because of kinetic energy.
Newton's first law of motion states that an object at rest tends to stay at rest, while an object in motion tends to stay in motion unless an external force acts upon it. This law appears in basketball when the player is shooting the ball. When the player is holding the ball, the ball is at rest but when a player shoots the ball, they use force to throw the ball in the hoop.
