Answer:
Airplanes' wings are curved on top and flatter on the bottom. That shape makes air flow over the top faster than under the bottom. As a result, less air pressure is on top of the wing. This lower pressure makes the wing, and the airplane it's attached to, move up.
Explanation:
Answer:
no of unit is 17941
Explanation:
given data
fixed cost = $338,000
variable cost = $143 per unit
fixed cost = $1,244,000
variable cost = $92.50 per unit
solution
we consider here no of unit is = n
so here total cost of labor will be sum of fix and variable cost i.e
total cost of labor = $33800 + $143 n ..........1
and
total cost of capital intensive = $1,244,000 + $92.5 n ..........2
so here in both we prefer cost of capital if cost of capital intensive less than cost of labor
$1,244,000 + $92.5 n < $33800 + $143 n
solve we get
n > 
n > 17941
and
cost of producing less than selling cost so here
$1,244,000 + $92.5 n < 197 n
solve it we get
n >
n > 11904
so in both we get greatest no is 17941
so no of unit is 17941
Answer:
b) The null hypothesis should be rejected.
Explanation:
The null hypothesis is that the mean shear strength of spot welds is at least
3.1 MPa
H0: u ≥3.1 MPa against the claim Ha: u< 3.1 MPa
The alternate hypothesis is that the mean shear strength of spot welds is less than 3.1 MPa.
This is one tailed test
The critical region Z(0.05) < ± 1.645
The Sample mean= x`= 3.07
The number of welds= n= 15
Standard Deviation= s= 0.069
Applying z test
z= x`-u/s/√n
z= 3.07-3.1/0.069/√15
z= -0.03/0.0178
z= -1.68
As the calculated z= -1.68 falls in the critical region Z(0.05) < ± 1.645 the null hypothesis is rejected and the alternate hypothesis is accepted that the mean shear strength of spot welds is less than 3.1 MPa
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