Answer:
I dont really know, I am sorry, but I am going to ask my teacher
Answer:
(a)
<em>d</em>Q = m<em>d</em>q
<em>d</em>q = 
<em>d</em>T
= (T₂ - T₁)
From the above equations, the underlying assumption is that remains constant with change in temperature.
(b)
Given;
V = 2L
T₁ = 300 K
Q₁ = 16.73 KJ , Q₂ = 6.14 KJ
ΔT = 3.10 K , ΔT₂ = 3.10 K for calorimeter
Let 
be heat constant of calorimeter
Q₂ = ΔT
Heat absorbed by n-C₆H₁₄ = Q₁ - Q₂
Q₁ - Q₂ = m ΔT
number of moles of n-C₆H₁₄, n = m/M
ρ = 650 kg/m³ at 300 K
M = 86.178 g/mol
m = ρv = 650 (2x10⁻³) = 1.3 kg
n = m/M => 1.3 / 0.086178 = 15.085 moles
Q₁ - Q₂ = m ' ΔT
= (16.73 - 6.14) / (15.085 x 3.10)
= 0.22646 KJ mol⁻¹ k⁻¹
Answer:
Recall the formula for the maximum stress, σₐ = 2σ₀ *√ (α/ρₓ)
where
σ₀ = tensile stress = 140 MPa = 1.40x 10⁸Pa
α = crack length = 3.8 × 10–2 mm = 3.8 x 10⁻⁵m
ρₓ = radius of curvature = 1.9 × 10⁻⁴mm = 1.9 × 10⁻⁷m
Substituting these values into the formula, we can calculate the max stress as
====== 2 x 1.40x 10⁸ x √(3.8 x 10⁻⁵/1.9 × 10⁻⁷)
σₐ = 24.4MPa
Answer:
A tsunami's trough, the low point beneath the wave's crest, often reaches shore first. When it does, it produces a vacuum effect that sucks coastal water seaward and exposes harbor and sea floors. As the tsunami approaches water is drawn back from the beach to effectively help feed the wave. In a tide the wave is so long that this happens slowly, over a few hours.
Explanation:
Answer:
Method B is the more efficient way of heating the water.
Explanation:
Method B is more efficient because by placing a heating element in the water as in described in method B, the heat that is lost to the surroundings is minimized which implies that more heat is supplied directly to the water. Therefore, more heating is achieved with a lesser amount of electrical energy input. Whereas placing the pan on a range means more heat losses to the surrounding and as such it will take a longer time for the water to heat up and also take more electrical energy.
