Answer: 113.75
Explanation:
You know
acceleration = a = 3.5 m/s²
time = t = 5 seconds
initial velocity = u = 14 m/s
Unknown is distance = s = ?
Use equation: s = ut + at²
Substitute all the known values inside the equation:
s = (14*5) + 0.5 * 3.5 * 5²
s = 70 + 43.75 = 113.75 m
The car travels 113.75 metres.
Answer:
The smallest distance the student that the student could be possibly be from the starting point is 6.5 meters.
Explanation:
For 2 quantities A and B represented as
and
The sum is represented as
For the the values given to us the sum is calculated as
Now the since the uncertainity inthe sum is
The closest possible distance at which the student can be is obtained by taking the negative sign in the uncertainity
Thus closest distance equals meters
Answer:20.03 m/s
Explanation:
Given
velocity of Prototype
Taking Froude number same for both flow as it is a dimensionless number for different flow regimes in open Flow