All categories

2 years ago

Differentiate between "Threshold and Resolution" with suitable examples.

Engineering

1 answer:

9966 [12]2 years ago

6 0

Answer:

to make the bace of a building more sturdy

Explanation:

example: the bace of the empire state building is stone very sturdy

You might be interested in

At what times should you use your headlights?

SVETLANKA909090 [29]

Answer:

Headlights are required to be used 1/2 hour after sunset to 1/2 hour before sunrise, when windshield wipers are being used, when visibility is less than 1000 feet, or when there is insufficient light or adverse weather.

Explanation:

hope this helps

8 0

2 years ago

Read 2 more answers

For some transformation having kinetics that obey the Avrami equation (Equation 10.17), the parameter n is known to have a value

OleMash [197]

Answer:

t = 25.10 sec

Explanation:

we know that Avrami equation

$Y = 1 - e^{-kt^n}$

here Y is percentage of completion of reaction = 50%

t is duration of reaction = 146 sec

so,

$0.50 = 1 - e^{-k^146^2.1}$

$0.50 = e^{-k306.6}$

taking natural log on both side

ln(0.5) = -k(306.6)

$k = 2.26\times 10^{-3}$

for 86 % completion

$0.86 = 1 - e^{-2.26\times 10^{-3} \times t^{2.1}}$

$e^{-2.26\times 10^{-3} \times t^{2.1}} = 0.14$

$-2.26\times 10^{-3} \times t^{2.1} = ln(0.14)$

$t^{2.1} = 869.96$

t = 25.10 sec

5 0

3 years ago

The heat required to raise the temperature of m (kg) of a liquid from T1 to T2 at constant pressure is Z T2CpT dT (1) In high sc

a_sh-v [17]

Answer:

(a)

dQ = mdq

dq = $C_p$ dT

$q = \int\limits^{T_2}_{T_1} {C_p} \, dT$ = $C_p$ (T₂ - T₁)

From the above equations, the underlying assumption is that $C_p$ 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 $C_{cal}$ be heat constant of calorimeter

Q₂ = $C_{cal}$ ΔT

Heat absorbed by n-C₆H₁₄ = Q₁ - Q₂

Q₁ - Q₂ = m $C_p$ 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 $C_p$ ' ΔT

$C_p$ = (16.73 - 6.14) / (15.085 x 3.10)

$C_p$ = 0.22646 KJ mol⁻¹ k⁻¹

6 0

2 years ago

Question 1: Final Results = What are the values of the resistances such that the gain = -100, Rin = 1 MI2. Don't use resistances

lidiya [134]

Answer:

Explanation:

In a study of algebra, you will encounter many families of equations, or groups of

equations that share common characteristics. Of interest to us here is the family of

linear equations in one variable, a study that lays the foundation for understanding

more advanced families. In addition to solving linear equations, we’ll use the skills we

develop to solve for a specified variable in a formula, a practice widely used in science,

business, industry, and research.

A. Solving Linear Equations Using Properties of Equality

An equation is a statement that two expressions are

equal. From the expressions and

we can form the equation

which is a linear equation in one variable. To solve

an equation, we attempt to find a specific input or xvalue that will make the equation true, meaning the

left-hand expression will be equal to the right. Using

Table 1.1, we find that is a

true equation when x is replaced by 2, and is a false

equation otherwise. Replacement values that make

the equation true are called solutions or roots of the equation.

4 0

2 years ago

On-site oil storage containers must be marked "Used Oil." True. False.

Scrat [10]

Answer:

How must used oil storage containers be marked? Containers and aboveground tanks used to store used oil at generator facilities must be labeled or marked clearly with the words “Used Oil" (40 CFR Section 279.22(c)).

Explanation:

i think it will help you

7 0

3 years ago