F=ma (Newton's second law of motion).
Then,
a = F/m = Uk*mg/m = Uk*g = 0.0751*9.81 = 0.737 m/s^2 (should be negative since a deceleration is expected).
v = Sqrt (u^2-2ad), Where v = final velocity, u=initial velocity, a = acceleration, d = distance moved.
Therefore,
v^2-u^2 = -2ad => d = (v^2-u^2)/-2a = (2^2-3.18^2)/-2*0.737 = 4.15 m
Answer:
It is characterized by fast, quick passes down the court and using more players on the attack than the opposition has for their defense. Slower, more deliberate play characterizes the slow-break style. This technique calls for more thoughtful action; players maneuver carefully in order to shoot in this type of offense.
The Average acceleration is 3.25m/s².
Acceleration is the change in velocity of the body with respect to time.
Where,
a= acceleration
u= initial speed
v= final speed
t= time
Initially, u = 2m/s
v = 16m/s
t = 3.5s
a = v-u/t
a = 16-2/3.5
a = 4m/s²
Then, u = 26m/s
v = 18m/s
t = 4s
a = v-u/t
a = 26-18/4
a =2 m/s²
Now, u = 0m/s
v = 20m/s
t = 2s
a = v-u/t
a = 20/2
a = 10m/s²
Finally, u = 20m/s
v = 10m/s
t = 2s
a = v-u/t
a = 10-20/2
a = -5m/s²
The average acceleration is the acceleration during the entire journey = 4+2+10-5/4
Average acceleration = 13/4m/s²
Average acceleration = 3.25m/s²
The Average acceleration is 3.25m/s²
Learn more about Acceleration here, brainly.com/question/2303856
#SPJ4
Answer:
0.91437 m
0.22859 m
Explanation:
g = Acceleration due to gravity = 9.81 m/s² = a
When the time intervals are equal, if four drops are falling then we have 3 time intervals.
So, the time interval is
For second drop time is given by
Distance from second drop
Distance from second drop is 0.91437 m
Distance from third drop
Distance from third drop is 0.22859 m
