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Tems11 [23]
4 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]4 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 car travels forward with constant velocity. It goes over a small stone, which gets stuck in the groove of a tire. The initial
kolbaska11 [484]

Answer:

a) vertically up

Explanation:

A car travels forward with constant velocity. It goes over a small stone, which gets stuck in the groove of a tire. The initial acceleration of the stone, as it leaves the surface of the road is,

a) vertically up

b) horizontally forward

c) horizontally backwards

d) zero

e) greater than zero but less than 45 degrees below the horizontal

A vertically upward since that is the direction of centripetal acceleration at that time. as the stone enters the circular path of the tire, it does so with a centripetal acceleration which is directed upward

8 0
3 years ago
A daring ranch hand sitting on a tree limb wishes to drop vertically onto a horse galloping under the tree. The constant speed o
Natali5045456 [20]
The time it will take him to fall can be found from:-3m = -(g*t^2)/2
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3 years ago
A single slit of width 0.50 mm is illuminated with light of wavelength 500 nm, and a screen is placed 120 cm in front of the sli
Vaselesa [24]

Answer:

a) 2.4 mm

b) 1.2 mm

c) 1.2 mm

Explanation:

To find the widths of the maxima you use the diffraction condition for destructive interference, given by the following formula:

m\lambda=asin\theta

a: width of the slit

λ: wavelength

m: order of the minimum

for little angles you have:

y=\frac{m\lambda D}{a}

y: height of the mth minimum

a) the width of the central maximum is 2*y for m=1:

w=2y_1=2\frac{1(500*10^{-9}m)(1.2m)}{0.50*10^{-3}m}=2.4*10^{-3}m=2.4mm

b) the width of first maximum is y2-y1:

w=y_2-y_1=\frac{(500*10^{-9}m)(1.2m)}{0.50*10^{-3}m}[2-1]=1.2mm

c) and for the second maximum:

w=y_3-y_2=\frac{(500*10^{-9}m)(1.2m)}{0.50*10^{-3}m}[3-2]=1.2mm

4 0
3 years ago
A ball is thrown straight upward and returns to the thrower’s hand after 1.8 s in the air. A second ball is thrown at an angle o
zysi [14]

Answer:

U = 9.1 m/s

Explanation:

from the question we are given the following

time (t) = 1.8 s

angle = 23 degrees

acceleration due to gravity (g) = 9.8 m/s^{2}

let us first calculate the initial velocity (u) which too the first ball to its maximum height from the equation below

v = u + 0.5at

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therefore

0 = u - (0.5 x 9.8 x 0.9)

u = 7.9 m/s

for the second ball to get to the maximum height of the first ball, the vertical component of its initial velocity (U) must be the same as the initial velocity of the first ball. therefore

U sin 60 = 7.9

U = 7.9 ÷ sin 60

U = 9.1 m/s

5 0
3 years ago
In a particular beam of radiation, which is traveling in a vacuum, the amounts of energy per second at an ultraviolet wavelength
Darina [25.2K]

Answer: d)

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If both beams have the same power, this means Energy/time so the number of photons per second must be different. As consequence a) is wrong as  b) since it is not posible since UV photon  have more energy that IR photons. c) It is no necessary know the frequency since the wavelength is related in the form:

c=λν  c is the speed of light, λ the wavelegth and ν the frequency.

d) Certainly will be more more IR photons than UV photons to get the same beam power.

8 0
3 years ago
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