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kykrilka [37]
2 years ago
8

A manufacturer selected a metal to use in producing a lightweight button for clothing. A metal that has a density of 2.71 g/cm3

was selected.
metal, data, metal, mass, grams, volume, centimenters, cubed, 1, 22.1, 3.00, 2, 42.0, 4.00, 3, 9.32, 5.00, 4, 8.13, 3.00,
Which of the metals was selected?
Metal 1
Metal 2
Metal 3
Metal 4
Physics
1 answer:
Natali5045456 [20]2 years ago
4 0

Just find the density of every metal and select the one with a density of 2.71 g/cm³ . This is:

Metal 1

ρ = m/V

ρ = 22.1 g / 3 cm³

ρ = 7.367 g / cm³

Metal 2

ρ = m/V

ρ = 42 g / 4 cm³

ρ = 10.5 g / cm³

Metal 3

ρ = m/V

ρ = 9.32 g / 5 cm³

ρ = 1.864 g / cm³

Metal 4

ρ = m/V

ρ = 8.13 g / 3 cm³

ρ = 2.71 g / cm³

<h2>R / Metal 4 was selected.</h2>
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The shape is connected in parallel so;

5.1) Ans;

\frac{1}{R}  =  \frac{1}{R1} +  \frac{1}{R2}   \\  \frac{1}{R}  =  \frac{1}{2}  +  \frac{1}{3}  \\  \frac{1}{R}  =  \frac{3 + 2}{6}  =  \frac{5}{6}  \\ R =  \frac{6}{5}  = 1.2 \:  \: ohm

5.2) Ans;

\frac{1}{R}  =  \frac{1}{R1} +  \frac{1}{R2}   \\  \frac{1}{R}  =  \frac{1}{8}  +  \frac{1}{10}  \\  \frac{1}{R}  =  \frac{5 + 4}{40}  =  \frac{9}{40}  \\ R =  \frac{40}{9}  = 4.4 \:  \: ohm

I hope I helped you^_^

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A tetrahedron has an equilateral triangle base with 25.0-cm-long edges and three equilateral triangle sides. The base is paralle
djverab [1.8K]

Answer:

a. 7.046 Nm²/C

b. 2.348 Nm²/C

Explanation:

Data given:

Base of equilateral triangle = 25.0 cm = 0.25 m

Strength of electric field = 260 N/C

In order to find the electric flux we first have to find out the area of triangle.

Area of triangle = \frac{\sqrt{3} }{4} a^{2}

                         = \frac{\sqrt{3} }{4} (0.25)^{2}

                         = 0.0271 m³

Lets find electric flux,

      Electric Flux = E. A

                          = 260×0.0271

                          = 7.046 Nm²/C

Now we can find the electric flux through each of the three sides.

Electric flux through three sides = \frac{7.046}{3}

                                                = 2.348 N m²/C

       

3 0
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