Answer:
4.44s
Explanation:
A 34-kg child on an 18-kg swing set swings back and forth through small angles. If the length of the very light supporting cables for the swing is 4.9 m, how long does it take for each complete back-and-forth swing? Assume that the child and swing set are very small compared to the length of the cables
since the mass of the child and that of the swing is negligible, the masses wont be involved in the calculation
T=2π√L/g
g=acceleration due to gravity which is 9.81m/s2
the length of the supporting cable is 4.9m
T the period
period is the time required to make a complete oscillation
T=2*π√4.9/9.81
T=2*π*0.706
T=4.44s
4.44s
Answer:
Explanation:
The speed of light in these mediums shall be lower than that in vacuum thus the total time light needs to cross both the media are calculated as under
Total time = Time taken through ice + Time taken through quartz
Time taken through ice = Thickness of ice / (speed of light in ice)
Thus in the same time the it would had covered a distance of
![Distance_{vaccum}=Totaltime\times V_{vaccum}\\\\Distance_{vaccum}=10^{-2}[2.20\mu _{ice+1.50\mu _{quartz}}]](https://tex.z-dn.net/?f=Distance_%7Bvaccum%7D%3DTotaltime%5Ctimes%20V_%7Bvaccum%7D%5C%5C%5C%5CDistance_%7Bvaccum%7D%3D10%5E%7B-2%7D%5B2.20%5Cmu%20_%7Bice%2B1.50%5Cmu%20_%7Bquartz%7D%7D%5D)
we have
Applying values we have
![Distance_{vaccum}=10^{-2}[2.20\times 1.309+1.50\times 1.542]](https://tex.z-dn.net/?f=Distance_%7Bvaccum%7D%3D10%5E%7B-2%7D%5B2.20%5Ctimes%201.309%2B1.50%5Ctimes%201.542%5D)
Answer: Globalization is the process of interaction and integration among people, companies, and governments worldwide.
Explanation:
6N I think I’m pretty sure
<span>internet tension = mass * acceleration internet tension = 23 – Friction tension = 14 * acceleration Friction tension = µ * 14 * 9.8 = µ * 137.2 23 – µ * 137.2 = 14 * acceleration Distance = undemanding speed * time undemanding speed = ½ * (preliminary speed + very final speed) Distance = ½ * (preliminary speed + very final speed) * time Distance = 8.a million m, preliminary speed = 0 m/s, very final speed = a million.8 m/s 8.a million = ½ * (0 + a million.8) * t Time = 8.a million ÷ 0.9 = 9 seconds Acceleration = (very final speed – preliminary speed) ÷ time Acceleration = (a million.8 – 0) ÷ 9 = 0.2 m/s^2 23 – µ * 137.2 = 14 * 0.2 resolve for µ</span>
