The acceleration and distance is related to the following expression:
y=v0*t + a*t^2/2 ; v0=0
y=44.1*100/2 = 2205m
hence, the speed will be
v=0 + a*t = 441m/s
from that height it will just be subjected to the gravitational acceleration
0=v_acc^2 -2g*y_free
y_free = v_acc^2/2g = 9922.5m
<span>y_max = y_acc+y_free = 441+9922.5 =10363.5m</span>
To model time-variant data, one must create a new entity in an m:n relationship with the original entity, is a False statement.
- Like the majority of software engineering initiatives, the ER process begins with gathering user requirements. What information must be retained, what questions must be answered, and what business rules must be implemented (For instance, if the manager column in the DEPARTMENT table is the only column, we have simply committed to having one manager for each department.)
- The end result of the E-R modeling procedure is an E-R diagram that can be roughly mechanically transformed into a set of tables. Tables will represent both entities and relationships; entity tables frequently have a single primary key, but the primary key for relationship tables nearly invariably involves numerous characteristics.
To know more about entity AND relationship visit : brainly.com/question/28232864
#SPJ4
Answer:
The contribution of the wavelets lying on the back of the wave front is zero because of something known as the Obliquity Factor. It is assumed that the amplitude of the secondary wavelets is not independent of the direction of propagation, Sources: byju's.com
Answer:
155.38424 K
2.2721 kg/m³
Explanation:
= Pressure at reservoir = 10 atm
= Temperature at reservoir = 300 K
= Pressure at exit = 1 atm
= Temperature at exit
= Mass-specific gas constant = 287 J/kgK
= Specific heat ratio = 1.4 for air
For isentropic flow
The temperature of the flow at the exit is 155.38424 K
From the ideal equation density is given by
The density of the flow at the exit is 2.2721 kg/m³
Answer:
the ans is D... good luck