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3 years ago

A 50 kg skydiver is falling downwards and accelerating 6 m/s2 down. What is the net force on the skydiver?

Physics

1 answer:

Montano1993 [528]3 years ago

6 0

Net Force = (mass) x (acceleration) (Newton #2)

Net Force = (50 kg) x (6 m/s² down)

Net Force = (50 * 6) (kg-m/s² down)

<em>Net Force = 300 Newtons down</em>

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Complete Question

The complete question is shown on the first uploaded image

Compare the percent error between your value and the accepted value of 1.26 times 10-6 T · m/A. (Use the accepted value given to three significant figures in your calculation.). 100% % error = %

Answer:

The permeability of free space is $\mu_0 = 1.32*10^{-6} \ Tm/A$

The percentage error is % error = 5.25%

Explanation:

From the question we are told that

The slope is $s= 3.9\ G/A$ , and

Generally $1 \ gauss = 10^{-4 } tesla$

So $s = 3.9 *10^ {-4} T /A$

From the relation given in the question

$\mu_0 = \frac{2R}{N} [\frac{B}{I} ]$

Where R is the radius of the coil $=\frac{Diameter \ of \ coil }{2} = 0.017m$

N is the number of loops of the coil = 10

Now from the question we are told that

$s = \frac{B}{I}$

substituting into the equation above

$\mu_0 = \frac{2R }{N} s$

Substituting values

$\mu_0= \frac{2 * 0.017}{10 } * 3.9*10^{-4}$

$= 1.326 *10^ {-6} \ Tm/A$

Generally the % error is mathematically represented as

% $error = \frac{Measured \ value - accepted \ value}{accepted \ value } *100$

Given that the accepted value is $\mu_o_ {acc} = 1.26 *10 ^{-6} \ T \cdot m /A$

Hence substituting values

% $error = \frac{(1.326-1.26)*10^{-6}}{1.26 *10 ^ {-6}} *100$

$= 5.24$

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