how do you isolate t (time) in this equation? the question is asking me to derive an equation for time to hit the ground for a d
ropped object in terms of height of the drop, gravity, and initial velocity and that’s what i have so far.
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Answer:
a) The magnitude JJ of the current density for a copper cable is 5.91 × 10⁵A.m⁻²
b)The mass per unit length \lambdaλ for a copper cable is 0.757kg/m
c)The magnitude J of the current density for an aluminum cable is 3.5 × 10⁵A/m²
d)The mass per unit length \lambdaλ for an aluminum cable is 0.380kg/m
Explanation:
The expression for electric field of conductor is,
The general equation of voltage is,
V = iR
The expression for current density in term of electric field is,
Substitute (V/L) for E in the above equation of current density.
Substitute iR for V in equation (1)
Substitute 1.69 × 10⁸ Ω .m for p
50A for i
0.200Ω.km⁻¹ for (R/L) in eqn (2)
The magnitude JJ of the current density for a copper cable is 5.91 × 10⁵A.m⁻²
b) The expression for resistivity of the conductor is,
The expression for mass density of copper is,
m = dV
where, V is the density of the copper.
Substitute AL for V in equation of the mass density of copper.
m=d(AL)
m/L = dA
λ is use for (m/L)
substitute,
pL/R for A and λ is use for (m/L) in the eqn above
Substitute 0.200Ω.km⁻¹ for (R/L)
8960kgm⁻³ for d and 1.69 × 10⁸ Ω .m
c) Using the equation (2) current density for aluminum cable is,
p is the resistivity of the aluminum cable.
Substitute 2.82 × 10⁻⁸Ω.m for p ,
50A for i and 0.200Ω.km⁻¹ for (R/L)
The magnitude J of the current density for an aluminum cable is 3.5 × 10⁵A/m²
d) Using the equation (3) mass per unit length for aluminum cable is,
p is the resistivity and is the density of the aluminum cable.
Substitute 0.200Ω.km⁻¹ for (R/L), 2700 for d and 2.82 × 10⁻⁸Ω.m for p
The mass per unit length \lambdaλ for an aluminum cable is 0.380kg/m
Refer to the diagram shown below.
In 2.4 hours, the distance traveled by the first airplane heading a 51.3° at 750 mph is
a = 750*2.4 = 1800 miles.
The second airplane travels
b = 620*2.4 = 1488 mile
The angle between the two airplanes is
163° - 51.3° = 111.7°
Let c = the distance between the two airplanes after 2.4 hours.
From the Law of Cosines, obtain
c² = a² + b² - 2ab cos(111.7°)
= 3.24 x 10⁶ + 2.2141 x 10⁶
c = 2335.41 miles
Answer: 2335.4 miles
In 2.4 hours, the distance traveled by the first airplane heading a 51.3° at 750 mph is
a = 750*2.4 = 1800 miles.
The second airplane travels
b = 620*2.4 = 1488 mile
The angle between the two airplanes is
163° - 51.3° = 111.7°
Let c = the distance between the two airplanes after 2.4 hours.
From the Law of Cosines, obtain
c² = a² + b² - 2ab cos(111.7°)
= 3.24 x 10⁶ + 2.2141 x 10⁶
c = 2335.41 miles
Answer: 2335.4 miles