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viktelen [127]
2 years ago
14

Instructions Write a program that inputs the length of two pieces of fabric in feet and inches (as whole numbers) and prints the

total. Sample Run Enter the Feet: 3 Enter the Inches: 11 Enter the Feet: 2 Enter the Inches: 5 Sample Output Feet: 6 Inches: 4
Physics
1 answer:
diamong [38]2 years ago
4 0

Answer:

The code is written in python in the explanation section below

Explanation:

a_feet = int(input("Input the feet of the first fabric: "))

a_inches = int(input("input the size in inches for the first piece of fabric: "))

b_feet = int(input("input the size in Feet for the second piece of fabric: "))

b_inches = int(input("Enter the size in Inches for the second piece of fabric: "))

sum_inches = a_inches + b_inches

inches_to_feet = sum_inches // 12

rem_from_div = sum_inches % 12

sum_feet = a_feet + b_feet + inches_to_feet

print("Feet: {} Inches: {}".format(sum_feet, rem_from_div))

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Answer:

Clouds form when below the dew point

4 0
3 years ago
Starting with the definition 1.00 in. = 2.54 cm, find the number of kilometers in 8.00 mi .
velikii [3]
Use Factor-Label Method:

8miles 63360 inches
---------- X --------------------- X
1 1 mile

2.54cm 1 meter
X ------------ X ---------------- X
1 inch 100 cm

1 km
----------------- = 12.87 km
1000meters


8 miles = 12.87 km
8 0
3 years ago
Velocity: You leave on a 549-mi trip in order to attend a meeting that will start 10.8 h after you begin your trip. Along the wa
rewona [7]

Answer:

1.55 h

Explanation:

Let the time to spend over dinner be t.

Since I need to spend 10.8 h for the whole trip, then I have 10.8 - t hours left to drive.

Speed = distance/time

v = \dfrac{d}{10.8-t}

d = v(10.8-t)

The distance is 549 and maximum speed is 65 mi/h.

549 = 65(10.8 - t)

10.8-t = 8.45

t = 10.8 - 8.45 = 1.55 \text{ h}

5 0
2 years ago
Suppose an electron is trapped within a small region and the uncertainty in its position is 24.0 x 10-15 m. What is the minimum
Alina [70]

Answer:

  • Uncertainty in position (∆x) = 24 × 10⁻¹⁵ m
  • Uncertainty in momentum (∆P) = ?
  • Planck's constant (h) = 6.26 × 10⁻³⁴ Js

\longrightarrow \:  \:  \sf\Delta x .\Delta p =  \dfrac{h}{4\pi}

\longrightarrow \:  \:  \sf24 \times  {10}^{ - 15}  .\Delta p =  \dfrac{6.26 \times  {10}^{ - 34}} {4 \times  \frac{22}{7} }

\longrightarrow \:  \:  \sf24 \times  {10}^{ - 15}  .\Delta p =  \dfrac{6.26 \times  {10}^{ - 34}} { \frac{88}{7} }

\longrightarrow \:  \:  \sf24 \times  {10}^{ - 15}  .\Delta p =  \dfrac{6.26 \times  {10}^{ - 34} \times 7} { 8 }

\longrightarrow \:  \:  \sf\Delta p =  \dfrac{43.82 \times  {10}^{ - 34} } { 8  \times 24 \times  {10}^{ - 15} }

\longrightarrow \:  \:  \sf\Delta p =  \dfrac{43.82 \times  {10}^{ - 34} } { 192 \times  {10}^{ - 15} }

\longrightarrow \:  \:  \sf\Delta p =  \dfrac{43.82 \times  {10}^{ - 34}  \times  {10}^{15} } { 192}

\longrightarrow \:  \:  \sf\Delta p =  \dfrac{43.82 \times  {10}^{ -19}   } { 192}

\longrightarrow \:  \:  \sf\Delta p =  \dfrac{4382 \times  {10}^{ - 2}  \times  {10}^{ -19}   } { 192}

\longrightarrow \:  \:  \sf\Delta p =  \dfrac{4382 \times  {10}^{ - 21}   } { 192}

\longrightarrow \:  \:  \sf\Delta p = 22.822\times  {10}^{ - 21}

\longrightarrow \:  \:  \sf\Delta p = 2.2822 \times  {10}^{1} \times  {10}^{ - 21}

\longrightarrow \:  \: \underline{ \boxed{ \red{  \bf\Delta p = 2.2822 \times  {10}^{ - 20}  \:  kg/ms}}}

4 0
2 years ago
A race car makes one lap around a track of radius 50 m in 9.0 s. What is the average velocity? *
Oksi-84 [34.3K]

Given that,

Radius of track, r = 50 m

time , t = 9 s

velocity, v = ?

Distance covered by car in one lap around a track is equal to the circumference of the track.

C = 2 π r = 2 * 3.14 * 50

C = 314.159 m

Distance covered by car, s = 314.159 m

Velocity = distance/ time

V = 314.159 / 9

V = 34.9 m/s

The average velocity of car is 34.9 m/s.

7 0
2 years ago
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