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The washer in the image has an inner diameter of 1/4 inch. The outer diameter measures 3/4 inch, and the washer is 1/4 inch thick. The density of the metal that washe
made ofis 0.285 pounds per cubic inch. What would the weight of 5 washers be in pounds?

The washer in the image has an inner diameter of 14 inch The outer diameter measures 34 inch and the washer is 14 inch thick The density of the metal that washe class=

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sqdancefan

Answer:

  0.14 lb . . .  closest choice is 0.11 lb

Step-by-step explanation:

The inside and outside radii are 1/8 in and 3/8 in, respectively. Then the area of the top of the washer is ...

  A = π(R² -r²) = π((3/8)² -(1/8)²) = π(9/64 -1/64) = π/8 . . . square inches

The height of the stack of 5 washers will be 5×(1/4 in) = 5/4 in. So, the volume of material in the stack of washers is ...

  V = Bh = (π/8)(5/4) = 5π/32 . . . cubic inches

The weight of material is the product of volume and density, so is ...

  W = (5π/32 in³)(0.285 lb/in³) ≈ 0.1399 lb ≈ 0.14 lb

_____

Comment on the question

Since this answer does not correspond to any of the offered choices, we suggest you ask your teacher to work this out for you. We suspect they will have difficulty justifying any of the answer choices shown here. (0.11 lb corresponds to 4 washers, not 5.)

boffeemadrid

The weight of 5 washers is required.

The weight of the 5 washers is 0.56 pounds

Density

R = Outer diameter = 3/4 inch

r = Inner diameter = 1/4 inch

h = Height of washer = 1/4 inch

[tex]\rho[/tex] = Density of metal = 0.285 pounds per cubic inch

The volume of the washer is

[tex]V=\pi R^2h-\pi r^2h\\\Rightarrow V=\pi h(R^2-r^2)[/tex]

Mass is given by

[tex]m=V\rho\\\Rightarrow m=\pi h(R^2-r^2)\rho\\\Rightarrow m=\pi \dfrac{1}{4}((\dfrac{3}{4})^2-(\dfrac{1}{4})^2)\times 0.285\\\Rightarrow m=0.1119\ \text{pounds}[/tex]

For 5 washers

[tex]5\times 0.11=0.5595=0.56\ \text{pounds}[/tex]

Learn more about density:

https://brainly.com/question/6838128

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