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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 light wave passes through an aperture (that is, a narrow slit). When it does so, the degree to which the wave spreads out will
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Explanation:

Single slit diffraction

Diffraction is the phenomenon of spreading out of waves as they pass through an aperture or around objects. Diffraction occurs when the size of the aperture or obstacle is of the same order of magnitude as the wavelength of the incident wave. For very small aperture sizes, the vast majority of the wave is blocked. in case of  large apertures the wave passes by or through the obstacle without any significant diffraction.

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3 years ago
The highest element in the hierarchical breakdown of the wbs is
Makovka662 [10]
Work package. Hope this helps!
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. Consider the equation =0+0+02/2+03/6+04/24+5/120, where s is a length and t is a time. What are the dimensions and SI units of
Olegator [25]

Answer:

See Explanation

Explanation:

Given

s=s_0+v_0t+\frac{a_0t^2}{2}+ \frac{j_0t^3}{6}+\frac{S_0t^4}{24}+\frac{ct^5}{120}

Solving (a): Units and dimension of s_0

From the question, we understand that:

s \to L --- length

t \to T --- time

Remove the other terms of the equation, we have:

s=s_0

Rewrite as:

s_0=s

This implies that s_0 has the same unit and dimension as s

Hence:

s_0 \to L --- dimension

s_o \to Length (meters, kilometers, etc.)

Solving (b): Units and dimension of v_0

Remove the other terms of the equation, we have:

s=v_0t

Rewrite as:

v_0t = s

Make v_0 the subject

v_0 = \frac{s}{t}

Replace s and t with their units

v_0 = \frac{L}{T}

v_0 = LT^{-1}

Hence:

v_0 \to LT^{-1} --- dimension

v_0 \to m/s --- unit

Solving (c): Units and dimension of a_0

Remove the other terms of the equation, we have:

s=\frac{a_0t^2}{2}

Rewrite as:

\frac{a_0t^2}{2} = s_0

Make a_0 the subject

a_0 = \frac{2s_0}{t^2}

Replace s and t with their units [ignore all constants]

a_0 = \frac{L}{T^2}\\

a_0 = LT^{-2

Hence:

a_0 = LT^{-2 --- dimension

a_0 \to m/s^2 --- acceleration

Solving (d): Units and dimension of j_0

Remove the other terms of the equation, we have:

s=\frac{j_0t^3}{6}

Rewrite as:

\frac{j_0t^3}{6} = s

Make j_0 the subject

j_0 = \frac{6s}{t^3}

Replace s and t with their units [Ignore all constants]

j_0 = \frac{L}{T^3}

j_0 = LT^{-3}

Hence:

j_0 = LT^{-3} --- dimension

j_0 \to m/s^3 --- unit

Solving (e): Units and dimension of s_0

Remove the other terms of the equation, we have:

s=\frac{S_0t^4}{24}

Rewrite as:

\frac{S_0t^4}{24} = s

Make S_0 the subject

S_0 = \frac{24s}{t^4}

Replace s and t with their units [ignore all constants]

S_0 = \frac{L}{T^4}

S_0 = LT^{-4

Hence:

S_0 = LT^{-4 --- dimension

S_0 \to m/s^4 --- unit

Solving (e): Units and dimension of c

Ignore other terms of the equation, we have:

s=\frac{ct^5}{120}

Rewrite as:

\frac{ct^5}{120} = s

Make c the subject

c = \frac{120s}{t^5}

Replace s and t with their units [Ignore all constants]

c = \frac{L}{T^5}

c = LT^{-5}

Hence:

c \to LT^{-5} --- dimension

c \to m/s^5 --- units

4 0
3 years ago
If a short-wave radio station broadcasts on a frequency of 9.065 megahertz (MHz), what is the wavelength of
Yakvenalex [24]

Answer:

If the radio wave is on an FM station, these are in Megahertz. A megahertz is one ... Typical radio wave frequencies are about 88~108 MHz .

Explanation:

To calculate the wavelength of a radio wave, you will be using the equation: Speed of a wave = wavelength X frequency.

Since radio waves are electromagnetic waves and travel at 2.997 X

10

8

meters/second, then you will need to know the frequency of the radio wave.

If the radio wave is on an FM station, these are in Megahertz. A megahertz is one million hertz. If the radio wave is from an AM radio station, these are in kilohertz (there are one thousand hertz in a kilohertz). Hertz are waves/second. Hertz is usually the label for the frequency of electromagnetic waves.

To conclude, to determine the wavelength of a radio wave, you take the speed and divide it by the frequency.

Typical radio wave frequencies are about

88

~

108

MHz

. The wavelength is thus typically about

3.41

×

10

9

~

2.78

×

10

9

nm

.

7 0
2 years ago
PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!
Sonbull [250]

Question 1  

In order to do work, the force vector must be

in the same direction as the displacement vector and the motion.

Question 2  

In which of the following cases is work being done on an object?

Pulling a trailer up a hill

Question 3  

Which situation is an example of NOT doing work?

carrying a box

Question 4  

Work is measured in

Joules

Question 5  

To find the work done, the force exerted and distance moved are multiplied. A couch is moved twice before you are happy with its placement. The same force was used to move the couch both times. If more work is done the first time it is moved, what do you know about the distance it was moved?

When more work was done, the couch was moved further.

Question 6  

A weight lifter raises a 1600 N barbell to a height of 2.0 meters. How much work was done? W = Fd

3200 Joules

Question 7  

You and a friend (Alex) are at a a tree-top adventure park .. . . part of the course requires you to climb up a rope. You both climb the same rope in the same amount of time. However, the tension in the rope is greater

when Alex climbs. Who did the most work?

Alex did - more tension means more force - more force means more work was done  

Question 8  

Doing work at a faster rate creates power.

more

Question 9  

In one challenge on the Titan Games, competitors have to lift 200 pounds up a long ramp. Angel is able to move the weight in 42 seconds. Anthony gets it done in only 38 seconds. Which statement is true?

Anthony has more power than Angel.  

Question 10  

A mountain climber exerts 41,000 J of work to climb a cliff. How much power does the climber need if she wants to finish in only 500 seconds? Power = Work / time

82 Watts

Question 11  

Your family is moving to a new apartment. While lifting a box 83 Joules of work is done to put the box on a truck, you exert an upward force of 75 N for 3 s. How much power is required to do this? (Hint: You only need two of the 3 numbers given!) Power = Work / time

27.7 Watts

<em>*100% CORRECT ANSWERS</em>

4 0
3 years ago
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