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(b) The density of aluminum is 2.70 g/cm3. The thickness of a rectangular sheet of aluminum foil varies
but is much less than 1 mm.
A student wishes to find the average thickness. She obtains the following measurements.
mass of sheet = 60.7 g
length of sheet = 50.0 cm
width of sheet = 30.0 cm
Calculate the student's values for
(i) the volume of the sheet,
volume =
[2]
(ii) the average thickness of the sheet.
thickness =
[2]

b The density of aluminum is 270 gcm3 The thickness of a rectangular sheet of aluminum foil varies but is much less than 1 mm A student wishes to find the aver class=

Respuesta :

Answer:

(i) [tex]22.48 cm^3[/tex]

(ii) [tex]1.5 mm[/tex]

Explanation:

Let t be the average thickness of the sheet.

Given that:

Density of the aluminum sheet is [tex]2.70 g/cm^3[/tex]

Mass of sheet = 60.7 g

Length of sheet  = 50.0 cm

Width of sheet  = 30.0 cm

(i) Using, Density=Mass/Volume

[tex]\Rightarrow \text{Volume}=\frac{\text{Mass}}{\text{Density}}[/tex]

[tex]\Rightarrow \text{Volume}=\frac{60.7}{2.7}=22.48 cm^3[/tex]

Hence, the volume of the sheet is [tex]22.48 cm^3[/tex].

(ii) Now, as this aluminum sheet is in the shape of a cuboid, so the volume of the sheet is

[tex]\text{Volume}=\text{Length}\times\text{Width}\times\text{Thickness}[/tex]

[tex]\Rightarrow 22.48=50\times 30 \times w[/tex]

[tex]\Rightarrow w=\frac{22.48}{50\times 30}=0.015 cm[/tex]

Hence, the average thickness of the sheet is [tex]1.5 mm[/tex].

snehashish65

(i) The required volume of sheet is [tex]22.48 \;\rm cm^{3}[/tex].

(ii)  The required thickness of the sheet is 0.015 cm.

Given data:

The density of Aluminum is, [tex]\rho = 2.70 \;\rm g/cm^{3}[/tex].

The mass of sheet is, m = 60.7 g.

The length of sheet is, L = 50.0 cm.

The width of sheet is, w = 30.0 cm.

(i)

The given problem is based on the concept of density of substance, which is equal to the mass per unit volume of substance. The mathematical expression for the density of substance is given as,

[tex]\rho =\dfrac{m}{V}[/tex]

here, V is the volume of sheet.

Solving as,

[tex]V =\dfrac{m}{\rho}\\\\V =\dfrac{60.7}{2.70}\\\\V = 22.48 \;\rm cm^{3}[/tex]

Thus, we can conclude that the required volume of sheet is [tex]22.48 \;\rm cm^{3}[/tex].

(ii)

Now, use the volume of substance to fond the thickness (t) of the sheet as,

[tex]V = L \times w \times t[/tex]

Solving as,

[tex]22.48 = 50 \times 30 \times t\\\\t = \dfrac{22.48}{50 \times 30}\\\\t = 0.015 \;\rm cm[/tex]

Thus, the required thickness of the sheet is 0.015 cm.

Learn more about the density of substance here:

https://brainly.com/question/6034174

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