Answer:
1800 N
Solution:
Impulse = mΔv = m * (u - v) .
here m = 100 kg
u = 4 m/s
v = -5 m/s
impulse = 100 x ( 4 - ( -5 ) ) = 900 Kg m/s .
Average reaction Force ( Favg ) = impulse / Δt
Average reaction Force ( Favg ) = 900kg·m/s / 0.5s
Average reaction Force ( Favg ) = 1800 N
Well if the rock doesn't move, then there is no amount of work done. There is no work done on an object if a force is applied to the object but it DOES NOT change its position, in this case is the rock.
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
A)
It is a launch oblique, therefore the initial velocity in the vertical direction is zero. Space Hourly Equation in vertical, we have:
Through Definition of Velocity, comes:

B)
Using the Velocity Hourly Equation in vertical direction, we have:
The angle of impact is given by:

If you notice any mistake in my english, please let me know, because i am not native.
To solve this problem it is necessary to apply the concepts related to acceleration due to gravity, as well as Newton's second law that describes the weight based on its mass and the acceleration of the celestial body on which it depends.
In other words the acceleration can be described as

Where
G = Gravitational Universal Constant
M = Mass of Earth
r = Radius of Earth
This equation can be differentiated with respect to the radius of change, that is


At the same time since Newton's second law we know that:

Where,
m = mass
a =Acceleration
From the previous value given for acceleration we have to

Finally to find the change in weight it is necessary to differentiate the Force with respect to the acceleration, then:




But we know that the total weight (F_W) is equivalent to 600N, and that the change during each mile in kilometers is 1.6km or 1600m therefore:


Therefore there is a weight loss of 0.3N every kilometer.