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2 years ago

7

Estimate the number of dollar bills (15.5 cm wide), placed end to end, that it would take to circle the Earth (radius = 6.40 × 1

03 km).

1 answer:

marishachu [46]2 years ago

3 0

Answer:

#_bills = 2.59 10⁸ Bills

Explanation:

For this exercise we must calculate the length of the Earth's circle

L = 2π r

Let's reduce to the SI system

r = 6.40 10³ km (1000m / 1km) = 6.40 10⁶ m

d = 15.5 cm (1m / 100cm) = 0.155 m

L = 2π 6.40 10⁶

L = 40.21 10⁶ m

now we can use a direct rule of proportions (rule of three). If 1 bill has d = 0.155, how much bill will have L

#_bills = 1 L / d

#_bills = 1 40.21 10⁶6 / 0.155

#_bills = 2.59 10⁸ Bills

the total number of bills is 2.6 10⁸ bills

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Answer:

Part a)

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Part a)

Average acceleration is given as

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$a = \frac{6.26 - (-8.86)}{12.0 \times 10^{-3}}$

$a = 1260.35 m/s^2$

Part B)

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

ruslelena [56]

Answer:

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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

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⇒ $T_{Deltoid} = T_{arm}$

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

Where,

$r_{d}$ is radius of the deltoid

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

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

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$F_{a}$ 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 $F_{a} = mg$ ,

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2 years ago