Answer:
The answer to your question is:
Explanation:
Data
Duane Albert
d = 5 m ; v = 3 m/s v = 4.2 m/s
a) b)
Duane's Albert's
d = 5 + (3)t d = 4.2t
d = 5 + 3t
c) 5 + 3t = 4.2t
4.2t - 3t = 5
1.2t = 5
t = 4.17 s
d)
Duane's
d= 5 + 3(4.17)
d = 17.51 m
Alberts
d = 4.2(4.17)
d = 17.51 m
Answer:
0.60 m/s
Explanation:
The average velocity from t = a to t = b is:
v_avg = (x(b) − x(a)) / (b − a)
Given that x(t) = 0.36t² − 1.20t, and the time is from 1.0 to 4.0:
v_avg = (x(4.0) − x(1.0)) / (4.0 − 1.0)
v_avg = [(0.36(4.0)² − 1.20(4.0)) − (0.36(1.0)² − 1.20(1.0))] / 3.0
v_avg = [(5.76 − 4.8) − (0.36 − 1.20)] / 3.0
v_avg = [0.96 − (-0.84)] / 3.0
v_avg = 0.60
The average speed is 0.60 m/s.
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Hope this helps! :)
Answer:
t=37 mins -> 2220sec
We want "T" which is the pendulum time constant
Using this equation
.5A=Ae^(-t/T)
The .5A is half the amplitude
Take ln of both sides to get ride of Ae
=ln(.5)=-2220/T
Now rearrange to = T
T=-2220/ln(.5) = 3202.78sec / 60 secs = 53.38 mins -> first part of the answer.
The second part is really easy. It took 37 mins to decay half way. meaning to decay another half of 50% which equals 25% it will take an additional 37 mins!
Explanation:
(a) Displacement of an object is the shortest path covered by it.
In this problem, a student is biking to school. She travels 0.7 km north, then realizes something has fallen out of her bag. She travels 0.3 km south to retrieve her item. She then travels 0.4 mi north to arrive at school.
0.4 miles = 0.64 km
displacement = 0.7-0.3+0.64 = 1.04 km
(b) Average velocity = total displacement/total time
t = 15 min = 0.25 hour
Hence, this is the required solution.
