Answer : a. Community
Allows a system to be accessible by a group of organizations. It can be shared between several organizations. It may be managed by organizations or by the third party.
This should be chosen by Ryan, since this computing model is cost effective and best to share among companies and organizations.
Other options explained:
-Software model is accessible via a browser and multiple users can use it.
-Infrastructure model is based on providing services of computer architecture in a virtual environment
Answer:
The distance traveled by the ball is 8.5 m
Explanation:
Initial height of the ball, h₁ = 1.5 m above the ground
final height of the ball, h₂ = 5m
Upward distance = distance traveled by the ball from a height of 1.5m to 5m = 5m - 1.5m = 3.5 m
Downward distance = distance traveled by the ball from 5m height to the ground =5m - 0 = 5m
Total distance traveled = upward distance + downward distance
Total distance traveled = 3.5 m + 5m = 8.5 m
Therefore, the distance traveled by the ball is 8.5 m
Answer:
a) w = 7.27 * 10^-5 rad/s
b) v1 = 463.1 m/s
c) v1 = 440.433 m/s
Explanation:
Given:-
- The radius of the earth, R = 6.37 * 10 ^6 m
- The time period for 1 revolution T = 24 hrs
Find:
What is the earth's angular speed?
What is the speed of a point on the equator?
What is the speed of a point on the earth's surface located at 1/5 of the length of the arc between the equator and the pole, measured from equator?
Solution:
- The angular speed w of the earth can be related with the Time period T of the earth revolution by:
w = 2π / T
w = 2π / 24*3600
w = 7.27 * 10^-5 rad/s
- The speed of the point on the equator v1 can be determined from the linear and rotational motion kinematic relation.
v1 = R*w
v1 = (6.37 * 10 ^6)*(7.27 * 10^-5)
v1 = 463.1 m/s
- The angle θ subtended by a point on earth's surface 1/5 th between the equator and the pole wrt equator is.
π/2 ........... s
x ............ 1/5 s
x = π/2*5 = 18°
- The radius of the earth R' at point where θ = 18° from the equator is:
R' = R*cos(18)
R' = (6.37 * 10 ^6)*cos(18)
R' = 6058230.0088 m
- The speed of the point where θ = 18° from the equator v2 can be determined from the linear and rotational motion kinematic relation.
v2 = R'*w
v2 = (6058230.0088)*(7.27 * 10^-5)
v2 = 440.433 m/s
Answer:
This can be the FBD of the bag.
(my bag looks more like a box tho ^^")
Answer:
A & B
Explanation:
A & B Would be the right answer since Morse code cannot be represented through the height of the fire.