Question

In Cynthia's physics class, she must turn in four lab reports. For each report, she earns extra credit if her percent error is less than 5%.5%. For her first lab report, she calculated the value of the acceleration due to gravity three times. She got the following three values. 9.56 ms2 9.56 ms2 9.72ms2 9.72ms2 9.88ms2 9.88ms2 The accepted value of the acceleration due to gravity is 9.8ms2.9.8ms2. Which option is closest to Cynthia's percent error? Remember: Take the average of the three measurements before calculating the percent error.
a. about 8%8%.
b. about 1%1%.
c. about 0.8%0.8%.
d. about 5%.

Answer 1

Answer:

$error = 0.82$%

Explanation:

The three readings of her gravity calculation is given as

$g_1 = 9.56 m/s^2$

$g_2 = 9.72 m/s^2$

$g_3 = 9.88 m/s^2$

now the mean value of gravity calculation is given as

$g_{mean} = \frac{9.56 + 9.72 + 9.88}{3}$

$g_{mean} = 9.72 m/s^2$

now the percentage error in her calculation is given as

$error = \frac{9.8 - 9.72}{9.8} \times 100$

$error = 0.82$%

Answer 2

An error is defined as an activity that is erroneous or wrong. In certain contexts, an error is equivalent to a blunder. The % error will be 0.8 %

What is an error?

An error is defined as an activity that is erroneous or wrong. In certain contexts, an error is equivalent to a blunder.

In statistics, "error" refers to the difference between the computed value and the correct value. An error can cause failure or deviate from the planned performance or behavior.

The three readings are given below;

g₁=9.56  m/sec²

g₂=9.72  m/sec²

g₃=9.88  m/sec²

The mean gravitational acceleration will be;

$\rm g_{mean}=\frac{g_1+g_2+g_3}{3} m/sec^2\\\\\rm g_{mean}=9.72 m/sec^2$

The percentage of error in the calculation will be

$\rm error =\frac{9.8-9.72}{9.8} \times100$

% error= 0.82 %≈0.8 %

Hence the % error will be 0.8 %

To learn more about the error refer to the link;

https://brainly.com/question/13286220