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Levart [38]
2 years ago
9

Joe is measuring the time it takes for a ball to roll down a ramp. In this experiment Joe takes the measurement 5 times and gets

the following results: 24.8, 23.9, 26.1, 25.1, 24.5 seconds. Joe uses the standard deviation of these numbers as the "margin of error" on his measurement.
Does Joe's average time agree with the accepted value of 25.9 seconds, within his margin of error? Can anyone help me with this?
Physics
1 answer:
PolarNik [594]2 years ago
6 0

 Step by step solution :

standard deviation is given by :

\sigma = \sqrt\dfrac{{\sum (x-\bar{x})^2}}{n}

where, \sigma is standard deviation

\bar{x} is mean of given data

n is number of observations

From the above data, \bar{x}=24.88

Now, if x=24.8, then (x-\bar{x})^2=0.0064

If  x=23.9, then (x-\bar{x})^2=0.9604

if x=26.1, then (x-\bar{x})^2=1.4884

If x=25.1, then (x-\bar{x})^2=0.0484

If x=24.5, then (x-\bar{x})^2=0.1444

so, \sum (x-\bar{x})^2 =\frac{0.0064+0.9604+1.4884+0.0484+0.1444}{5}

\sum (x-\bar{x})^2 =2.648

\sqrt{\sum \frac{(x-\bar{x})^2}{n}}

\sigma =0.7277

No, Joe's value does not agree with the accepted value of 25.9 seconds. This shows a lots of errors.

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An airplane traveling 919 m above the ocean at 505 m/s is to drop a case of twinkies to the victims below. How much time before
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a. 13.7 s b. 6913.5 m

Explanation:

a. How much time before being directly overhead should the box be dropped?

Since the box falls under gravity we use the equation

y = ut - 1/2gt² where y = height of plane above ocean = 919 m, u = initial vertical velocity of airplane = 0 m/s, g = acceleration due to gravity = -9.8 m/s² and t = time it takes the airplane to be directly overhead.

So,

y = ut - 1/2gt²

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t = 13.69 s

t ≅ 13.7 s

So, the box should be dropped 13.69 s before being directly overhead.

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