Answer:
The average velocity is 0.203 m/s
Explanation:
Given;
initial displacement, x₁ = 20 yards = 18.288 m
final displacement, x₂ = ¹/₃ x 18.288 = 6.096 m
change in time between 5:02 PM and 5:03 PM, Δt = 3 mins - 2 mins = 1 min = 60 s
The average velocity is given by;
V = change in displacement / change in time
V = (x₂ - x₁) / Δt
V = (18.288 - 6.096) / 60
V = 0.203 m/s
Therefore, the average velocity is 0.203 m/s
Answer:
Explanation:
Using the expression shown below as:
Where,
is the number of vacancies
N is the number of defective sites
k is Boltzmann's constant = 
is the activation energy
T is the temperature
Given that:
N = 10 moles
1 mole = 
So,
N = 
Temperature = 425°C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T = (425 + 273.15) K = 698.15 K
T = 698.15 K
Applying the values as:
![ln[\frac {2.3}{6.023}\times 10^{-11}]=-\frac {Q_v}{1.38\times 10^{-23}\times 698.15}](https://tex.z-dn.net/?f=ln%5B%5Cfrac%20%7B2.3%7D%7B6.023%7D%5Ctimes%2010%5E%7B-11%7D%5D%3D-%5Cfrac%20%7BQ_v%7D%7B1.38%5Ctimes%2010%5E%7B-23%7D%5Ctimes%20698.15%7D)
Answer:
Most hydraulic systems develops pressure surges that may surpass settings valve. by exposing the hose surge to pressure above the maximum operating pressure will shorten the hose life.
Explanation:
Solution
Almost all hydraulic systems creates pressure surges that may exceed relief valve settings. exposing the hose surge to pressure above the maximum operating pressure shortens the hose life.
In systems where pressure peaks are severe, select or pick a hose with higher maximum operating pressure or choose a spiral reinforced hose specifically designed for severe pulsing applications.
Generally, hoses are designed or created to accommodate pressure surges and have operating pressures that is equal to 25% of the hose minimum pressure burst.
Answer:
a bond issued by a bank or other financial institution, guaranteeing the fulfilment of a particular contract
