All categories

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>

You might be interested in

Since all stars begin their lives with the same basic composition, what characteristic most determines how they will differ?

Dafna1 [17]

4. Mass they are formed with

7 0

3 years ago

When a piece of copper is taken to the moon , a change will be observed in it's ?

SpyIntel [72]

Weight. Because there is less gravity on the moon.

3 0

3 years ago

Which of the following is true about acids?

Stells [14]

i would say, are sour and corrode metals

8 0

3 years ago

Read 2 more answers

Rearrange the equation to solve for the permeability of free space (μ0). Remember that the slope is the ratio of magnetic field

IRINA_888 [86]

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

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

$= 5.24$

6 0

4 years ago

In an experiment conducted using the Young's double-slit apparatus, the separation between the slits is 20 µm. A first-order con

jeka57 [31]

To solve this problem, we use the formula
λ = s sin θ
where s is the separation and θ is the angle interference
So,
λ = 20 x 10^-6 sin 2.5
λ = 8.72 x 10^-7 m

The required angle for the fourth order bright fringe is
θb = sin⁻¹ (4λ / s) = sin⁻¹ (4 (8.72 x 10^-7 m)/ 20 x 10^-6 ) = 10.04°

The required angle for the fourth order dark fringe is
θd = sin⁻¹ (4.5 λ / s) = sin⁻¹ (4.5 (8.72 x 10^-7 m)/ 20 x 10^-6 ) = 11.31°

6 0

3 years ago