What is the velocity (in m/s) of a 550 kg roller coaster cart at the bottom of the track if it started with 990,000 J of gravita
1 answer:
You might be interested in
The work done by the machine is equal to the product between the force applied and the distance over which the force is applieds, so in this case:
And the power of the machine is equal to the ratio between the work done by the machine and the time taken:
And the power of the machine is equal to the ratio between the work done by the machine and the time taken:
Answer:
Part A:
Part B:
Magnitude of displacement=44.721359 m
Explanation:
Given Data:
40 m in negative direction of the x axis
20 m perpendicular path to his left (considering +ve y direction)
25 m up a tower (Considering +ve z direction)
Required:
- In unit-vector notation, what is the displacement of the sign from start to end.
- The magnitude of the displacement of the sign from start to this new end=?
Solution:
Part A:
where:
i,j,k are unit vectors
Part B:
Sign falls to foot of tower so z=0
Magnitude of displacement: