Answer:
B: precipitation seeping through pores and cracks in the ground
Explanation:
Man made wells aren't the main source because their isn't enough of them and morning dew doesn't go in to the ground in one main source.
Answer:
a. T₂ = 901.43 ° K
b. T₂ = 843.85 ° K
Explanation:
Given
η = 0.165 mol , Q = 690 J , V = 280 cm³ , P = 3.40 x 10 ⁶ Pa , T₁ = 700 ° K
Using the equation that describe the experiment of heat
Q = η * Cv * ΔT
a.
Nitrogen Cv = 20.76 J / mol ° K
ΔT = T₂ - T₁
T₂ = [ Q / ( η * Cv) ] + T₁
T₂ = [ 690 J / ( 0.165 mol * 20.76 J / mol ° K ) ] + 700 ° K
T₂ = 901.43 ° K
b.
Nitrogen Cp = 29.07 J / mol ° K
T₂ = [ Q / ( η * Cv) ] + T₁
T₂ = [ 690 J / ( 0.165 mol * 29.07 J / mol ° K ) ] + 700 ° K
T₂ = 843.85 ° K
Complete question is;
A copper wire has a diameter of 4.00 × 10^(-2) inches and is originally 10.0 ft long. What is the greatest load that can be supported by this wire without exceeding its elastic limit? Use the value of 2.30 × 10⁴ lb/in² for the elastic limit of copper.
Answer:
F_max = 28.9 lbf
Explanation:
Elastic limit is simply the maximum amount of stress that can be applied to the wire before it permanently deform.
Thus;
Elastic limit = Max stress
Formula for max stress is;
Max stress = F_max/A
Thus;
Elastic limit = F_max/A
F_max is maximum load
A is area = πr²
We have diameter; d = 4 × 10^(-2) inches = 0.04 in
Radius; r = d/2 = 0.04/2 = 0.02
Plugging in the relevant values into the elastic limit equation, we have;
2.30 × 10⁴ = F_max/(π × 0.02²)
F_max = 2.30 × 10⁴ × (π × 0.02²)
F_max = 28.9 lbf
