Answer:
be like good
Explanation:
maybe rubber like so you don't slip
Answer: A it is true.
Explanation: It is true under Subpart C—Information Security Responsibilities for Employees who Manage or Use Federal Information Systems
Number 3 and 4
(3) Program and functional managers must receive training in information security basics; management and implementation level training in security planning and system/application security management; and management and implementation level training in system/application life cycle management, risk management, and contingency planning.
(4) Chief Information Officers (CIOs), IT security program managers, auditors, and other security-oriented personnel (e.g., system and network administrators, and system/application security officers) must receive training in information security basics and broad training in security planning, system and application security management, system/application life cycle management, risk management, and contingency planning.
Answer:
G = 0.424
Explanation:
Ds = ( 0.278tr * V ) + (0.278 * V²)/ ( 19.6* ( f ± G))
Where Ds = stopping sight distance = 415miles = 126.5m
G = absolute grade road
V = velocity of vehicle = 52miles/hr
f = friction = 0 because the road is wet
tr = standard perception / reaction time = 2.5s
So therefore:
Substituting to get G
We have
2479.4G = 705.6G + 751.72
1773.8G = 751.72
G = 751.72/1773.8
G = 0.424
Answer:
See attachment for completed question
Explanation:
Given that; Brainly.com
What is your question?
mkasblog
College Engineering 5+3 pts
The dry unit weight of a soil sample is 14.8 kN/m3.
Given that G_s = 2.72 and w = 17%, determine:
(a) Void ratio
(b) Moist unit weight
(c) Degree of saturation
(d) Unit weight when the sample is fully saturated
See complete solving at attachment
Answer:
0.05 J/K
Explanation:
Given data in question
heat (Q) = 10 J
temperature (T) = 200 K
to find out
the change in entropy of the system
Solution
we will solve this by the entropy change equation
i.e ΔS = ΔQ/T ...................1
put the value of heat Q and Temperature T in equation 1
ΔS is the enthalpy change and T is the temperature
so ΔS = 10/200
ΔS = 0.05 J/K
