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Juli2301 [7.4K]
1 year ago
13

A pipe of length 10.0 m increases in length by 1.5 cm when its temperature is increased by 90°F. What is its coefficient of line

ar expansion?
Physics
1 answer:
azamat1 year ago
3 0

The coefficient of linear expansion, given that the length of the pipe increased by 1.5 cm is 1.67×10¯⁵ /°F

<h3>How to determine the coefficient of linear expansion</h3>

From the question given above, the following data were obtained

  • Original diameter (L₁) = 10 m
  • Change in length (∆L) = 1.5 cm = 1.5 / 100 = 0.015 m
  • Change in temperature (∆T) = 90 °F
  • Coefficient of linear expansion (α) =?

The coefficient of linear expansion can be obtained as illustrated below:

α = ∆L / L₁∆T

α = 0.015 / (10 × 90)

α = 0.015 / 900

α = 1.67×10¯⁵ /°F

Thus, we can conclude that the coefficient of linear expansion is 1.67×10¯⁵ /°F

Learn more about coefficient of linear expansion:

brainly.com/question/28293570

#SPJ1

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A thick steel sheet of area 100 in.2 is exposed to air near the ocean. After a one-year period it was found to experience a weig
vladimir1956 [14]

Answer:

Therefore the rate of corrosion 37.4 mpy and 0.952 mm/yr.

Explanation:

The corrosion rate is the rate of material remove.The formula for calculating CPR or corrosion penetration rate is

CPR=\frac{KW}{DAT}

K= constant depends on the system of units used.

W= weight =485 g

D= density =7.9 g/cm³

A = exposed specimen area =100 in² =6.452 cm²

K=534 to give CPR in mpy

K=87.6  to give CPR in mm/yr

mpy

CPR=\frac{KW}{DAT}

        =\frac{534\times( 485g)\times( 10^3mg/g)}{(7.9g/cm^3) \times (100in^2)\times (24h/day)\times (365day/yr)\times 1yr}

        =37.4mpy

mm/yr

CPR=\frac{KW}{DAT}

        =\frac{87.6\times (485g)\times (10^3 mg/g)}{(7.9g/cm^3)\times (100in^2)\times(2.54cm/in)^2\times (24h/day)\times (365day/yr)\times 1yr}

       =0.952 mm/yr

Therefore the rate of corrosion 37.4 mpy and 0.952 mm/yr.

3 0
4 years ago
Please help me, it's a simple equation
castortr0y [4]

Because the elevator moves at a constant speed, it's in equilibrium and the net force acting on it is zero. Then the tension in the cable exactly equals the magnitude of the elevator's weight, which is

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2 years ago
An airplane flies at 40 m/s at an altitude of 50 meters. The pilot drops a heavy package which falls to the ground. Where, appro
igor_vitrenko [27]

Answer:

128 m

Explanation:

From the question given above, the following data were obtained:

Horizontal velocity (u) = 40 m/s

Height (h) = 50 m

Acceleration due to gravity (g) = 9.8 m/s²

Horizontal distance (s) =?

Next, we shall determine the time taken for the package to get to the ground.

This can be obtained as follow:

Height (h) = 50 m

Acceleration due to gravity (g) = 9.8 m/s²

Time (t) =?

h = ½gt²

50 = ½ × 9.8 × t²

50 = 4.9 × t²

Divide both side by 4.9

t² = 50 / 4.9

t² = 10.2

Take the square root of both side

t = √10.2

t = 3.2 s

Finally, we shall determine where the package lands by calculating the horizontal distance travelled by the package after being dropped from the plane. This can be obtained as follow:

Horizontal velocity (u) = 40 m/s

Time (t) = 3.2 s

Horizontal distance (s) =?

s = ut

s = 40 × 3.2

s = 128 m

Therefore, the package will land at 128 m relative to the plane

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