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: - 45000 N.s
Explanation: Impulse is equal to the change in momentum
J = Δp
To solve for impulse we calculate the change in momentum
Δp = m ( Δv)
= 1500 kg ( 10 m/s - 40 m/s)
= - 45000 N.s
Answer:
C. 100
D.3
E. 33.3
Explanation:
C. Mechanical Advantage=Load / Effort
= 200N
--------
100N
Therefore,. = 100
D. I. Velocity Ratio= distance moved by the effort / distance moved by load
= 30cm/10cm
= 3
II. Efficiency= M.A / V.R
= 100/3
= 33.33
The statement is false. Vectors are used to solve projectile motion problems because they allow the analysis of one direction at a time for two-dimensional motion. Scalar quantities can be used to analyze linear motion problem, but not projectile motion.
Answer:
Limewater can be used to detect carbon dioxide. If carbon dioxide is bubbled through limewater then it turns from clear to cloudy/milky in colour. This is why limewater used in a simple respirometer can show that more carbon dioxide is present in exhaled air compared to inhaled air.
Explanation:
