Answer:
Modal analysis is performed to identify the modes, and the response in that mode can be picked from the response spectrum. These peak responses are then combined to estimate a total response. A typical combination method is the square root of the sum of the squares (SRSS) if the modal frequencies are not close.
Answer:
The solution to this question is 5.153×10⁻⁴(kmol)/(m²·s)
That is the rate of diffusion of ammonia through the layer is
5.153×10⁻⁴(kmol)/(m²·s)
Explanation:
The diffusion through a stagnant layer is given by
Where
= Diffusion coefficient or diffusivity
z = Thickness in layer of transfer
R = universal gas constant
= Pressure at first boundary
= Pressure at the destination boundary
T = System temperature
= System pressure
Where = 101.3 kPa 
, 
, 
0.5×101.3 = 50.65 kPa
Δz = z₂ - z₁ = 1 mm = 1 × 10⁻³ m
R = 
T = 298 K and = 1.18 
= 1.8×10⁻⁵
= 5.153×10⁻⁴
Hence the rate of diffusion of ammonia through the layer is
5.153×10⁻⁴(kmol)/(m²·s)
Graduated cylinder & beaker; both are used for science activities while a measuring cup is mostly used for baking.
Answer:
The answer is 960 kg
Explanation:
Solution
Given that:
Assume the initial dye concentration as A₀
We write the expression for the dye concentration for one hour as follows:
ln (C₁) = ln (A₀) -kt
Here
C₁ = is the concentration at 1 hour
t =time
Now
Substitute 480 g for C₁ and 1 hour for t
ln (480) = ln (A₀) -k(1) ------- (1)
6.173786 = ln (A₀) -k
Now
We write the expression for the dye concentration for three hours as follows:
ln (C₃) = ln (A₀) -k
Here
C₃ = is the concentration at 3 hour
t =time
Thus
Substitute 480 g for C₃ and 3 hour for t
ln (120) = ln (A₀) -k(3) ------- (2)
4.787492 = ln (A₀) -3k
Solve for the equation 1 and 2
k =0.693
Now
Calculate the amount of blue present initially using the expression:
Substitute 0.693 for k in equation (2)
4.787492 = ln (A₀) -3 (0.693)
ln (A₀) =6.866492
A₀ =e^6.866492
= 960 kg
Therefore, the amount of the blue dye present from the beginning is 960 kg
