Orthographic projection, common method of representing three-dimensional objects, usually by three two-dimensional drawings in each of which the object is viewed along parallel lines that are perpendicular to the plane of the drawling.
Answer:
Activation energy for creep in this temperature range is Q = 252.2 kJ/mol
Explanation:
To calculate the creep rate at a particular temperature
creep rate, 
Creep rate at 800⁰C, 

.........................(1)
Creep rate at 700⁰C


.................(2)
Divide equation (1) by equation (2)
![\frac{0.01}{5.5 * 10^{-4} } = \exp[\frac{-Q}{1073R} -\frac{-Q}{973R} ]\\18.182= \exp[\frac{-Q}{1073R} +\frac{Q}{973R} ]\\R = 8.314\\18.182= \exp[\frac{-Q}{1073*8.314} +\frac{Q}{973*8.314} ]\\18.182= \exp[0.0000115 Q]\\](https://tex.z-dn.net/?f=%5Cfrac%7B0.01%7D%7B5.5%20%2A%2010%5E%7B-4%7D%20%7D%20%3D%20%5Cexp%5B%5Cfrac%7B-Q%7D%7B1073R%7D%20-%5Cfrac%7B-Q%7D%7B973R%7D%20%5D%5C%5C18.182%3D%20%5Cexp%5B%5Cfrac%7B-Q%7D%7B1073R%7D%20%2B%5Cfrac%7BQ%7D%7B973R%7D%20%5D%5C%5CR%20%3D%208.314%5C%5C18.182%3D%20%5Cexp%5B%5Cfrac%7B-Q%7D%7B1073%2A8.314%7D%20%2B%5Cfrac%7BQ%7D%7B973%2A8.314%7D%20%5D%5C%5C18.182%3D%20%5Cexp%5B0.0000115%20Q%5D%5C%5C)
Take the natural log of both sides

Answer:
Absolute pressure=70.72 KPa
Explanation:
Given that Vacuum gauge pressure= 30 KPa
Barometer reading =755 mm Hg
We know that barometer always reads atmospheric pressure at given situation.So atmospheric pressure is equal to 755 mm Hg.
We know that P= ρ g h
Density of 
So P=13600 x 9.81 x 0.755
P=100.72 KPa
We know that
Absolute pressure=atmospheric pressure + gauge pressure
But here given that 30 KPa is a Vacuum pressure ,so we will take it as negative.
Absolute pressure=atmospheric pressure + gauge pressure
Absolute pressure=100.72 - 30 KPa
So
Absolute pressure=70.72 KPa
Answer:
a) the inductance of the coil is 6 mH
b) the emf generated in the coil is 18 mV
Explanation:
Given the data in the question;
N = 570 turns
diameter of tube d = 8.10 cm = 0.081 m
length of the wire-wrapped portion l = 35.0 cm = 0.35 m
a) the inductance of the coil (in mH)
inductance of solenoid
L = N²μA / l
A = πd²/4
so
L = N²μ(πd²/4) / l
L = N²μ(πd²) / 4l
we know that μ = 4π × 10⁻⁷ TmA⁻¹
we substitute
L = [(570)² × 4π × 10⁻⁷× ( π × (0.081)² )] / 4(0.35)
L = 0.00841549 / 1.4
L = 6 × 10⁻³ H
L = 6 × 10⁻³ × 1000 mH
L = 6 mH
Therefore, the inductance of the coil is 6 mH
b)
Emf ( ∈ ) = L di/dt
given that; di/dt = 3.00 A/sec
{∴ di = 3 - 0 = 3 and dt = 1 sec}
Emf ( ∈ ) = L di/dt
we substitute
⇒ 6 × 10⁻³ ( 3/1 )
= 18 × 10⁻³ V
= 18 × 10⁻³ × 1000
= 18 mV
Therefore, the emf generated in the coil is 18 mV
Answer:
(a) 0.12924
(b) Taking into consideration significance level of 0.05 yet the value of p is greater than 0.05, it suggests that the coin is fair hence the coin can be used at the beginning of any sport event.
Explanation:
(a)
n=200 for fair coin getting head, p= 0.5
Expectation = np =200*0.5=100
Variance = np(1 - p) = 100(1-0.5)=100*0.5=50
Standard deviation,
Z value for 108,
P( x ≥108) = P( z >1.13)= 0.12924
(b)
Taking into consideration significance level of 0.05 yet the value of p is greater than 0.05, it suggests that the coin is fair hence the coin can be used at the beginning of any sport event.