Really could use help with this lab Need show the work for question 10 and the A

Question

Really could use help with this lab Need show the work for question 10 and the Analysis questions answered 1- 5Thank you.

PART B VERTICAL MOTION 6. Measure the height of the balcony Record the distance as lhetolal distance from the ground + 1 m 1 K 25 7. Hold the ball a the top of the vertical meter stick. You ll drop the batl from this height lo the floor Drop the bal so hat v0 8. Mcasure and rccord the ti it takes the ball to fall to the ground. Repeat for at least 4 trials 9. Determine the average time for the trials Distance (m) t (sec) t2 (sec) t; (sec) t4 (sec) uve (sec) .83 | .91 | .92 | .SiaS Using displacement =-vit-½ gt , calculate the acceleration of gravity of the falling ball. Include appropriate UNITS 10. (o)(8925)-(4(3425) 69045.MotionExperinenrevisudTRMOci2012dex

Explanation / Answer

Let's analyze the question 10, step by step

the expression for the distance is

s = Vi t - ½ g t2

The data given in the exercise are:

Vi = 0

s = 5 m

t = 0.8925 s

I write the equation Vi=0

s = - ½ g t2

calculate g

g = -2 s / t2

g = - 2 5 /0.89252 [m / s2]

g = - 12.55 m / s2

The negative sign means that the acceleration of gravity is in the negative direction of the Y axis

acceleration value a little high there are two possible sources of error problems in measuring time, trouble releasing the body (Vi = 0)

Discussion Questions

1) When a person walks the process involves climbing a foot and dropped under the action of gravity and just before hitting the place the other foot to stop the fall floor, we can assume from time to move the leg lift to keep it in the air and the floor is constant, therefore this distance between time would give the average speed of the entire interval step, but for each of the aforementioned steps have different speeds.

2) When a ball exits zero or very small initial velocity and is subjected to the acceleration of gravity going speed increased with increasing distance dropped, according to the equation

VF2 = 2 g (Y-Y0)

3) apply the percentage error formula

% Error = (1255 - 9797)! / 9,797 100

% Error = 2,753 / 9,797 100

% Error = 28%

4)

   The largest source of error is the time for two reasons

- This squared in the expression so any inaccuracy is squared

- The measurement of time depends on the reaction of the operator of the chronometer in the best case is reaction time is about 0.1 s and observe in the data that this is the variation of the measure, the second decimal meaningless because this reaction time

- A third observation is that the time error occurring at two points to start the movement and finalize

  There is another source but it is difficult to assess what happens with problems when releasing the body or is it starting with a different initial velocity of zero, this error generally occurs when one person does all the measuring, if the experiment is performed for two people it is unlikely to occur

5)

  To improve the experiment could improve timekeeping placing an activation system and stopping the chronometer in electronic form, they can be used as switches logic gates

A ball drop system to ensure that the initial velocity is zero and the height is the same for all measurements

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