An air-track glider attached to a spring oscillates between the 14.0 cm mark and the 71.0 cm mark on the track. The glider compl etes 12.0 oscillations in 34.0 s . You may want to review (Pages 391 - 393) . Part A What is the period of the oscillations
1 answer:
Answer:
A = 2,8333 s
Explanation:
El periodo es definido como el tiene que toma de dar una oscilación.
En este caso realiza varias osicilacion por lo cual debemos encontrar el promedio del perdono.
T = t/n
calculemos
A = 34,0/ 12,0
A = 2,8333 s
You might be interested in
I believe you square both numbers and take the square root. √34^2+14^2= so it would be 36.77
B) A ladybug crawling forward at constant rate of 2.5 m/s
Answer:
If gravity were stronger, houses and buildings would have to be more sturdy so they do not get crushed to the ground. If gravity were weaker, buildings and houses would have to be built stronger so they do not float away.
Hope this helps you!
50 N/cm^2. It is 75+525/4x3. Pls mark me brainliest
<u>Answer</u>
81.94 m
<u>Explanation</u>
The centripetal force of an object moving in a circular path is given by:
F = mv²/r Where m is the mass of the object, v is the constant velocity and r is the radius of the curve.
F = mv²/r
3,300 = (1600×13²)/r
3,300 = 270,400/r
r = 270,400/3,300
= 81.94 m
