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DerKrebs [107]
3 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).
Physics
1 answer:
marishachu [46]3 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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Explanation:

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What circumstance would allow an officer to search a home even if they didn’t have a warrant?
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An aluminum wire having a cross-sectional area equal to 2.20 10-6 m2 carries a current of 4.50 A. The density of aluminum is 2.7
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Answer:

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Explanation:

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v = \frac{I}{nqA}

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n: is the number of electrons

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n = \frac{6.022\cdot 10^{23} atoms}{1 mol}*\frac{1 mol}{26.982 g}*\frac{2.70 g}{1 cm^{3}}*\frac{(100 cm)^{3}}{1 m^{3}} = 6.03 \cdot 10^{28} atom/m^{3}                  

Now, we can find the drift speed:

v = \frac{I}{nqA} = \frac{4.50 A}{6.03 \cdot 10^{28} atom/m^{3}*1.6 \cdot 10^{-19} C*2.20 \cdot 10^{-6} m^{2}} = 2.12 \cdot 10^{-4} m/s              

Therefore, the drift speed of the electrons in the wire is 2.12x10⁻⁴ m/s.

I hope it helps you!      

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3 0
3 years ago
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