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trasher [3.6K]
3 years ago
6

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
Physics
1 answer:
horsena [70]3 years ago
6 0

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

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Long Jump: inital center of mass height of 1.08 m, final center of mass height of 0.42 m, projection velocity of 8.7 m/s, projec
sammy [17]

Answer:

1) The maximum jump height is reached at A. 0.337s

2) The maximum center of mass height off of the ground is B. 1.64m

3) The time of flight is C. 0.834s

4) The distance of jump is B. 7.49m

Explanation:

First of all we need to decompose velocity in its rectangular components, so

v_{xi}=8.7m/s(cos 22.3\°)=8.05m/s= constant\\v_{yi}=8.7m/s(sin 22.3\°)=3.3m/s

1) We use, v_{fy}=v_{iy}-gt, as we clear it for t and using the fact that v_{fy}=0 at max height, we obtain t=\frac{v_{iy}}{g} =\frac{3.3m/s}{9,8m/s^{2}} =0.337s

2) We can use the formula y_{max}=y_{i}+v_{iy}t-\frac{gt^{2}}{2} for t=0.337s, so

y_{max}=1.08m+(3.3m/s)(0.337s)-\frac{(9.8m/s^{2})(0.337)^{2}}{2}=1.64m

3) We can use the formula y_{f}=y_{i}+v_{iy}t-\frac{gt^{2}}{2}, to find total time of fligth, so 0.42=1.08+3.3t-\frac{(9.8)t^{2}}{2}\\0=-4.9t^{2}+3.3t+0.66, as it is a second-grade polynomial, we find that its positive root is t=0.834s

4) Finally, we use x=v_{x}t=8.05m/s(0.834s)=6.71m, as it has an additional displacement of 0.77m due the leg extension we obtain,

x=6.71m+0.77m=7.48m, aprox x=7.49m

5 0
3 years ago
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