A glideron an air trackmoves in the +x direction with a constant acceleration. It has two flags, each exactly 0.02 mlong, with the midpoints of the flags separated by 0.150m. The first flag interrupts the photogate timer for a time 0.058s, and the second flag interrupts the photogate timer for a time 0.047s.1.What was the average velocity of the glider during the interval when the first flag was interrupted?2. What was the average velocity of the glider during the interval when the second flag was interrupted?3. Approximately how much time elapsed between the passage of the first flag and the passage of the second flag?4. Calculatethe approximate accelerationof the glider.

1)  Average Velocity V1 = 0.3448 m/s

2) Average Velocity V2 = 0.4255 m/s

3) time elapsed t = 0.39 seconds

4) Approximate acceleration of the glider = 0.207 m/s²

Explanation:  

Given the data in the question;

1) What was the average velocity of the glider during the interval when the first flag was interrupted

V1 ( first interruption) = distance travelled / elapsed time

= 0.02 m / 0.058 s = 0.3448 m/s

2) What was the average velocity of the glider during the interval when the second flag was interrupted?

V2( second interruption) = distance travelled / elapsed time

= 0.02 m / 0.047 s = 0.4255 m/s

3) Approximately how much time elapsed between the passage of the first flag and the passage of the second flag?

Average  Speed = ( v1 + v2 ) / 2 = ( 0.3448 m/s + 0.4255 m/s) ) / 2

= 0.7703 / 2 = 0.3851 m/s

so time elapsed will be

t = 0.150 m / 0.3851 m/s

t = 0.3895 ≈ 0.39 seconds

4) Calculate the approximate acceleration of the glider

acceleration = ( v2 - v1 ) / t

a =  ( 0.4255 m/s - 0.3448 m/s ) / 0.3895

a = 0.0807 m/s / 0.3895 s

a = 0.207 m/s²

answer: 2×10^-2kg

explanation: density= mass per volume

vol=mass/density

converting to s.i units we have;

3÷1000kg = 0.003

1.5÷100m = 0.15

0.003/0.15 = 0.02kg

uio

explanation: