Answer: a. Frequency. 20, 4, 3, 1, 1, 0, 1
Explanation:
Interval(Height in feet) - - - - - - - - frequency
14000 - 14999 - - - - - - - - - - - - - - - 20
15000 - 15999 - - - - - - - - - - - - - - - 4
16000 - 16999 - - - - - - - - - - - - - - - 3
17000 - 17999 - - - - - - - - - - - - - - - 1
18000 - 18999 - - - - - - - - - - - - - - - 1
19000 - 19999 - - - - - - - - - - - - - - - 0
20000 - 20999 - - - - - - - - - - - - - - 1
Frequency : (20, 4, 3 1, 1, 0, 1)
Answer:
A. speed = 7.14 Km/s
B. distance = 1820.7 Km
Explanation:
Given that: a = 14.0 m/, t = 8.50 minutes.
But,
t = 8.50 = 8.50 x 60
= 510 seconds
A. By applying the first equation of motion, the speed of the shuttle at the end of 8.50 minutes can be determined by;
v = u + at
where: v is the final velocity, u is the initial velocity, a is the acceleration and t is the time.
u = 0
So that,
v = 14 x 510
= 7140 m/s
The speed of the shuttle at the end of 8.50 minute is 7.14 Km/s.
B. the distance traveled can be determined by applying second equation of motion.
s = ut + a
where: s is the distance, u is the initial velocity, a is the acceleration and t is the time.
u = 0
s = a
= x 14 x
= 7 x 260100
= 1820700 m
The distance that the shuttle has traveled during the given time is 1820.7 Km.