R = ρl/A
Where R = Resistance in Ohms, Ω, ρ = Resistivity in Ωm, l = Length in m.
Area in m²
ρ = Resistivity = 3.14 * 10⁻⁸ Ωm, Length l = 12m,
Area = πr² = π* (2*10⁻⁴)² m² ≈ 3.14 * (2*10⁻⁴)² m²
R = ρl/A
≈ 3.14 * 10⁻⁸ * 12 / (3.14 * (2*10⁻⁴)²)
≈ 3
Resistance, R ≈ 3 Ω
Answer:
Mass = 4152kg
Explanation:
Given
L = 208m
I = 154A
V = 0.245V
Density = 3610 kg/m3
ρ = 4.23 x 10-8Ω·m = resistivity of wire
Resistance R = ρL/ A
R = voltage / current = V/I = 0.245/154 = 1.59×10-³ohms
1.59×10-³ = 4.23 x 10-⁸×208/A
Rearranging,
A = 4.23 x 10-⁸×208/1.59×10-³
A = 5.53×10-³m²
Mass = density × volume
Volume = L×A = 208×5.53×10-³m³= 1.15m³
Mass = 3610×1.15 = 4152kg
Answer:
The time he can wait to pull the cord is 41.3 s
Explanation:
The equation for the height of the skydiver at a time "t" is as follows:
y = y0 + v0 · t + 1/2 · g · t²
Where:
y = height at time "t".
y0 = initial height.
v0 = initial velocity.
t = time.
g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).
First, let´s calculate how much time will it take for the skydiver to hit the ground if he doesn´t activate the parachute.
When he reaches the ground, the height will be 0 (placing the origin of the frame of reference on the ground). Then:
y = y0 + v0 · t + 1/2 · g · t²
0 m = 15000 m + 0 m/s · t - 1/2 · 9.8 m/s² · t²
0 m = 15000 m - 4.9 m/s² · t²
-15000 m / -4.9 m/s² = t²
t = 55.3 s
Then, if it takes 4.0 s for the parachute to be fully deployed and the parachute has to be fully deployed 10.0 s before reaching the ground, the skydiver has to pull the cord 14.0 s before reaching the ground. Then, the time he can wait before pulling the cord is (55.3 s - 14.0 s) 41.3 s.
Two factors influence the pressure of fluids. They are the depth of the fluid and its density.
