You get a lot of detailed information. hope this helped :)
Answer: im not gonna give i to you just do 15+15=_+ 5.6+6.4 easy
Explanation: i took the test and got a 100%
Answer:
D = 527.31 Km
Explanation:
given,
angle of ship, θ = 23.5° N of W
distance travel in the direction = 575 Km
Distance of ship in west from harbor = ?
now,
Distance of the ship in the west direction
D = d cos θ
d = 575 Km
θ = 23.5°
inserting all the values
D = 575 x cos 23.5°
D = 575 x 0.91706
D = 527.31 Km
Hence, the distance travel by the ship in west from harbor is equal to D = 527.31 Km
Answer:
d = 1.954 Km
Explanation:
given,
total distance, D = 2.5 Km
in stretch A to B =
speed = 99 Km/h = 99 x 0.278 = 27.22 m/s time =t
in stretch B to C
time = 3.4 s
In stretch C to D
speed = 48 Km/h = 48 x 0.278 = 13.34 m/s time =t
we know,
distance = speed x time
distance of BC
using equation of motion
v = u + a t
27.22 = 13.34 - a x 3.4
a = 4.08 m/s²
uniform deceleration is equal to 4.08 m/s²
distance traveled in BC
s = 68.94 m
3000 = 27.5 t + 68.94 + 13.33 t
40.83 t = 2931.06
t = 71.79 s
distance travel in AB
distance = s x t
d = 27.22 x 71.79
d = 1954 m
d = 1.954 Km
distance between A and B is equal to 1.954 Km.
Answer:
This process of weighting attributes is used ______.remove the connection among the various attribute in the set of specification and to set priorities
Explanation:
The process of weighting attribute is used to remove the association among attributes in a sample set as some attributes are considered more important than others. The weighing serves to indicate the desirability for an attribute with respect o the other attributes
As such, the weighting is used to set the order of importance, and to indicate the priorities
