launch a rocket ship weighs 4.5 million pounds. When it is launched from rest, it takes 8.00 s to reach 164 km/h ; at the end of the first 1.00 min , its speed is 1640 km/h . Part APart complete What is the average acceleration (in m/s2) of the rocket during the first 8.00 s? aav = 5.69 m/s2 Previous Answers Correct Part BPart complete What is the average acceleration (in m/s2) of the rocket between 8.00 s and the end of the first 1.00 min? aav = 7.88 m/s2 Previous Answers Correct Part CPart complete Assuming the acceleration is constant during each time interval (but not necessarily the same in both intervals), what distance does the rocket travel during the first 8.00 s? Express your answer with the appropriate units. x = 182 m Previous Answers Correct Part D Assuming the acceleration is constant during each time interval (but not necessarily the same in both intervals), what distance does the rocket travel during the interval from 8.00 s to 1.00 min? Express your answer with the appropriate units.

A) a_av = 5.695 m/s²

B) a_av = 7.88 m/s

C) x_f = 182.24 m

D) x_f = 13029.12 m

Explanation:

A) Initial velocity at 0 seconds; v_i = 0 m/s

Final velocity at 8 seconds;v_f = 164 km/h = 45.56 m/s

So, formula for average acceleration is;

a_av = change in velocity/change in time.

Thus, for the first 8 seconds;

a_av = (45.56 - 0)/(8 - 0)

a_av = 45.56/8

a_av = 5.695 m/s²

B) From part A, we now want to find the average acceleration from 8 seconds to 1 minute (60 seconds)

After the 60 seconds, the speed is 1640 km/h = 455.56 m/s

Thus;

Average acceleration is;

a_av = (455.56 - 45.56)/(60 - 8)

a_av = 7.88 m/s

C) Now we are told the acceleration is constant and we want to find the distance the rocket travels during the first 8 seconds.

So, distance formula we will use is;

x_f = x_i + ½(v_i + v_f)t

So at this stage, v_i = 0 m/s and v_f = 45.56 m/s and x_i = 0 m

Thus;

x_f = 0 + ½(0 + 45.56)8

x_f = 182.24 m

D) Now we want to find the distance the rocket travels during the 8 seconds to 1 minute (60 seconds) interval.

From part B, v_i = 45.56 m/s and v_f = 455.56 m/s

Also, x_i = 0 while time interval(t) = 60 - 8 = 52 s

Thus;

x_f = 0 + ½(45.56 + 455.56)52

x_f = 13029.12 m

a high pressure system is a dense air mass that is usually cooler and drier than the surrounding air. a low pressure system is a less dense air mass that is usually wetter and warmer than the surrounding air.

in general, areas that experience high atmospheric pressure also experience fair weather. low pressure systems can cause the formation of clouds and storms. air usually flows from areas of high pressure to areas of low pressure..