Answer:
You can make a horizontale transverse by moving a slinky vertically up and down!
Explanation:
For rectilinear motions, derived formulas all based on Newton's laws of motion are formulated. The equation for acceleration is
a = (v2-v1)/t, where v2 and v1 is the final and initial velocity of the rocket. We know that at the end of 1.41 s, the rocket comes to a stop. So, v2=0. Then, we can determine v1.
-52.7 = (0-v1)/1.41
v1 = 74.31 m/s
We can use v1 for the formula of the maximum height attained by an object thrown upwards:
Hmax = v1^2/2g = (74.31^2)/(2*9.81) = 281.42 m
The maximum height attained by the model rocket is 281.42 m.
For the amount of time for the whole flight of the model rocket, there are 3 sections to this: time at constant acceleration, time when it lost fuel and reached its maximum height and the time for the free fall.
Time at constant acceleration is given to be 1.41 s. Time when it lost fuel covers the difference of the maximum height and the distance travelled at constant acceleration.
2ax=v2^2-v1^2
2(-52.7)(x) = 0^2-74.31^2
x =52.4 m (distance it covered at constant acceleration)
Then. when it travels upwards only by a force of gravity,
d = v1(t) + 1/2*a*t^2
281.42-52.386 = (0)^2+1/2*(9.81)(t^2)
t = 6.83 s (time when it lost fuel and reached its maximum height)
Lastly, for free falling objects, the equation is
t = √2y/g = √2(281.42)/9.81 = 7.57 s
Therefore, the total time= 1.41+6.83+7.57 = 15.81 s
"increments of 8" means the major divisions are 0,8,16,24 ?
<span>x axis, calculate the moment arms from 0 </span>
<span>3x4, 2x12, 1x20 </span>
<span>from an arbitrary C </span>
<span>3(c-4) + 2(c-12) + (c-20) = 0 </span>
<span>3c - 12 + 2c -24 + c - 20 = 0 </span>
<span>6c = 56 </span>
<span>c = 9.33 </span>
<span>y axis </span>
<span>3x3, 1x12, 2x20 </span>
<span>3(c-4) + 1(c-12) +2 (c-20) = 0 </span>
<span>3c - 12 + c - 12 + 2c - 40 = 0 </span>
<span>6c = 64 </span>
<span>c = 10.67 </span>
<span>so center is x = 9.33, y = 10.67 </span>
The temperature will be the pressure of 130 ka
