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3 years ago

The unit of length most suitable for measuring the thickness of a cell phone is a

Physics

1 answer:

Elis [28]3 years ago

8 0

centimeter or millimeter.

Explanation:

The unit of length most suitable for measuring the thickness of a cell phone is centimeter or millimeter.

Length refers to how long a material is.

There are different units of length.
Smaller units that are microscopic are used to measure length dimensions that are very small
For example, nanometre is used to express microscopic length values.
Centimeters and millimeters are used for small dimensions that we can easily see.
Longer lengths can take the order of a yard, kilometer, miles e.t.c

Learn more:

Length brainly.com/question/13035995

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3 years ago

What force must the deltoid muscle provide to keep the arm in this position?

ruslelena [56]

Answer:

Deltoid Force, $F_{d} = \frac {r_{a}mgsin\alpha_{a}}{r_{d}sin\alpha_{d}}$

Additional Information:

Some numerical information are missing from the question. However, I will derive the formula to calculate the force of the deltoid muscle. All you need to do is insert the necessary information and calculate.

Explanation:

The deltoid muscle is the one keeping the hand arm in position. We have two torques that apply to the rotating of the arm.

1. The torque about the point in the shoulder for the deltoid muscle, $T_{Deltoid}$

2. The torque of the arm, $T_{arm}$

Assuming the arm is just being stretched and there is no rotation going on,

$T_{Deltoid}$ = 0

$T_{arm}$ = 0

⇒ $T_{Deltoid} = T_{arm}$

$r_{d}F_{d}sin\alpha_{d} = r_{a}F_{a}sin\alpha_{a}$

Where,

$r_{d}$ is radius of the deltoid

$F_{d}$ is the force of the deltiod

$\alpha_{d}$ is the angle of the deltiod

$r_{a}$ is the radius of the arm

$F_{a}$ is the force of the arm , $F_{a} = mg$ which is the mass of the arm and acceleration due to gravity

$\alpha_{a}$ is the angle of the arm

The force of the deltoid muscle is,

$F_{d} = \frac {r_{a}F_{a}sin\alpha_{a}}{r_{d}sin\alpha_{d}}$

but $F_{a} = mg$ ,

∴ $F_{d} = \frac {r_{a}mgsin\alpha_{a}}{r_{d}sin\alpha_{d}}$

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3 years ago

Read 2 more answers

PART ONE

Lina20 [59]

Explanation:

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