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Tems11 [23]
3 years ago
14

Determine how dollar bills, placed end to end, are required to go around Earth at the equator. You will need to measure the leng

th of a dollar bill.
** Must use unit analysis for all problems **
Physics
1 answer:
Mrac [35]3 years ago
4 0

I didn't have a dollar bill handy.  So I conducted an extensive, exhaustive, in-depth, 2-minute internet search to determine the length of one.  (I lived in Boston for 6 years and it had a profound effect on me. All during the 2 minutes of my research, I kept mumbling "dorla bill" to myself.)  Anyway, the length of a US dorla bill is 6.14 inches. We want to find out HOW MANY dorla bills it would take.  We don't know that number yet, but we need to use it in our math, so we have to use something to put in its place until we find out the actual number.  Most people would use 'x'.  I'm going to use ' M '. Length of 1 bill . . . . . 6.14 inches Length of ' M ' bills . . . . . (6.14 M) inches Circumference of the Earth at the equator . . . . .  40,075 km inches in 1 foot . . . . . 12 feet in 1 meter . . . . . 3.28084 meters in 1 km . . . . . 1,000 This is all the outside information we need.  The rest is all arithmetic. ========================================== length of 1 bill, in inches . . . . . 6.14 inches length of ' M ' bills, in inches . . . . . (6.14 M inches) length of ' M ' bills, in feet . . . . .(6.14 M inches) x (1 foot/12 inches) length of ' M ' bills, in meters . . . . . (6.14 M inches) x (1 foot/12 inches) x (1 meter/3.28084 foots) length of ' M ' bills, in km . . . . .  (6.14 M inches) x(1 foot/12 inches) x(1 meter/3.28084 foots) x(1 km/1,000 m) . . . . . . . . . . ^ these numbers are equal \/ . . . . . . . . . . Circumference of the equator . . . . . 40,075 km ============================================== Gather all the numbers together, and all the units together: (6.14M x 1 x 1 x 1)/(12 x 3.28084 x 1,000) x (inch-foot-meter-km/inch-foot-meter) = 40,075 km Do the arithmetic, and cancel out units that appear on both top and bottom: (0.000155956 M) x (km) = 40,075 km Divide each side by 0.000155956 km, and you'll have: M = (40,075 / 0.000155956) M = 256.96 million bills ============================================ Quick check, just to see if it makes sense: Round each dorla bill to 6 inches even: (256.96 million x 6 inches) x (1 ft/12 inch) x (1 mile/5280 ft) = 24,333 miles Well,  the same internet says the equator's circumference is  24,901 miles.  YAY ! ! !  As an engineer, I'd say the two numbers match perfectly.  So the answer to the question is confirmed to be 256.96 million bills (2.57 x 10⁸ bills)  

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A solution is prepared by dissolving 49.3 g of KBr in enough water to form 473 mL of solution. Calculate the mass % of KBr in th
son4ous [18]

Answer:

9.31%

Explanation:

We are given that

Mass of KBr=49.3 g

Volume of solution=473 mL

Density of solution =1.12g/mL

We have to find the mass% of KBr.

Mass =volume\times density

Using the formula

Mass of solution=1.12\times 473=529.76 g

Mass % of KBr=\frac{mass\;of\;KBr}{Total\;mass\;of\;solution}\times 100

Mass % of KBr=\frac{49.3}{529.76}\times 100

Mass % of KBr=9.31%

Hence, the mass% of KBr=9.31%

7 0
3 years ago
A long string is wrapped around a 6.6-cm-diameter cylinder, initially at rest, that is free to rotate on an axle. The string is
lys-0071 [83]

Answer:

\omega_f=571.42\ rpm

Explanation:

It is given that,

Diameter of cylinder, d = 6.6 cm

Radius of cylinder, r = 3.3 cm = 0.033 m

Acceleration of the string, a=1.5\ m/s^2

Displacement, d = 1.3 m

The angular acceleration is given by :

\alpha =\dfrac{a}{r}

\alpha =\dfrac{1.5}{0.033}

\alpha =45.46\ rad/s^2

The angular displacement is given by :

\theta=\dfrac{d}{r}

\theta=\dfrac{1.3}{0.033}

\theta=39.39\ rad

Using the third equation of rotational kinematics as :

\omega_f^2-\omega_i^2=2\alpha \theta

Here, \omega_i=0

\omega_f=\sqrt{2\alpha \theta}

\omega_f=\sqrt{2\times 45.46\times 39.39}

\omega_f=59.84\ rad/s

Since, 1 rad/s = 9.54 rpm

So,

\omega_f=571.42\ rpm

So, the angular speed of the cylinder is 571.42 rpm. Hence, this is the required solution.

5 0
3 years ago
At the end of a race a runner decelerates from a velocity of 8.90 m/s at a rate of 1.70 m/s2. (a) How far in meters does she tra
Ad libitum [116K]

Answer:

x=22.33m

Explanation:

Kinematics equation for constant deceleration:

x =v_{o}*t - 1/2*at^{2}=8.9*6.3-1/2*1.70*6.3^{2}=22.33m

7 0
3 years ago
The tuning fork has a frequency of 426.7 is found to cause resonance in a closed column of air measuring 0.186 (the diameter of
lidiya [134]

Answer:

The velocity of the tuning fork sound is 317.4648 m/s

Explanation:

The given parameters are;

The frequency of the fork, f = 426.7 Hz

The length of the closed air column, L = 0.186 m

The diameter of the tube, d = 0.15 m

At the fundamental frequency in a closed tube, we have;

λ = 4·L

Where;

λ = The wavelength of the wave

L = The length of the tube

From the equation for the velocity of a wave, we have;

v = f·λ

Where;

v = The velocity of the (sound) wave

f = The frequency of the wave = 426.7 Hz

λ = 4·L = 4 × 0.186 m = 0.744 m

∴ v = 426.7 Hz × 0.744 m = 317.4648 m/s

Therefore, the velocity of the sound produced by the tuning fork, v = 317.4648 m/s

3 0
3 years ago
Why is the teammate bond often so strong?
Kazeer [188]

Answer:

c

Explanation:

While people are working, they need to have a strong bond with their teammates,to face all the hardships .

thus, teammates work together, win and lose together, and often eat and live together.

3 0
3 years ago
Read 2 more answers
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