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 basic structure would be:
Reactants → Products
Answer:
m = 28.7[kg]
Explanation:
To solve this problem we must use the definition of kinetic energy, which can be calculated by means of the following equation.
where:
Ek = kinetic energy = 1800 [J]
m = mass [kg]
v = 11.2 [m/s]
![1800=\frac{1}{2}*m*(11.2)^{2}\\m = 28.7[kg]](https://tex.z-dn.net/?f=1800%3D%5Cfrac%7B1%7D%7B2%7D%2Am%2A%2811.2%29%5E%7B2%7D%5C%5Cm%20%3D%2028.7%5Bkg%5D)
Answer: The trip takes 
Explanation:
Velocity 
is the variation of the position of a body (distance traveled 
) with time 
:
In this case, the car travels a distance 
at a velocity 
and we need to find the time it takes the trip.
Isolating :
Finally:
