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Fofino [41]
1 year ago
15

In a sample of 18-karat gold, 75 percent of the total mass is pure gold, while the rest is typically 16 percent silver and 9 per

cent copper. If the density of pure gold is rhogold=19.3g/cm3, while the densitites of silver and copper are respectively rhosilver=10.5g/cm3 and rhocopper=8.90g/cm3, what is the overall density rho18kt of this alloy of 18-karat gold?
Physics
1 answer:
ivann1987 [24]1 year ago
8 0

Answer:

15.57 g/cm³

Explanation:

\rho_g = Density of gold = 19.3 g/cm³

\rho_s = Density of silver = 10.5 g/cm³

\rho_c = Density of copper = 8.9 g/cm³

Assuming total mass as 1000 g

Volume of gold

V_g=\frac{0.75\times 1000}{19.3}\\\Rightarrow V_g=38.86\ cm^3

Volume of silver

V_g=\frac{0.16\times 1000}{10.5}\\\Rightarrow V_g=15.238\ cm^3

Volume of copper

V_c=\frac{0.09\times 1000}{8.9}\\\Rightarrow V_c=10.11\ cm^3

Density of the alloy would be

\rho=\frac{M}{V_g+V_s+V_c}\\\Rightarrow \rho=\frac{1000}{38.86+15.238+10.11}\\\Rightarrow \rho=15.57\ g/cm^3

The overall density of this alloy is 15.57 g/cm³

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Answer:

d = 2021.6 km

Explanation:

We can solve this distance exercise with vectors, the easiest method s to find the components of the position of each plane and then use the Pythagorean theorem to find distance between them

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Height   y₁ = 800m

Angle θ = 25°

           cos 25 = x / r

           sin 25 = z / r

           x₁ = r cos 20

           z₁ = r sin 25

          x₁ = 18 103 cos 25 = 16,314 103 m = 16314 m

          z₁ = 18 103 sin 25 = 7,607 103 m= 7607 m

2 plane

Height   y₂ = 1100 m

Angle θ = 20°

          x₂ = 20 103 cos 25 = 18.126 103 m = 18126 m

          z₂ = 20 103 without 25 = 8.452 103 m = 8452 m

The distance between the planes using the Pythagorean Theorem is

         d² = (x₂-x₁)² + (y₂-y₁)² + (z₂-z₁)²2

Let's calculate

        d² = (18126-16314)²  + (1100-800)² + (8452-7607)²

        d² = 3,283 106 +9 104 + 7,140 105

        d² = (328.3 + 9 + 71.40) 10⁴

        d = √(408.7 10⁴)

        d = 20,216 10² m

        d = 2021.6 km

7 0
2 years ago
An organ pipe is tuned to exactly 384 Hz when the temperature in the room is 20°C. Later, when the air has warmed up to 25°C, th
maksim [4K]

Answer: A. Greater than 384 Hz

Explanation:

The velocity of sound is directly related to the temperature rather it is directly proportional meaning if the temperature decreases the velocity decreases and if temperature increases the velocity increases.

Now, we are given that temperature has risen from 20°C to 25°C meaning it has increases. So it implies that velocity must also increase.

Also, the velocity for organ pipe is directly proportional to its frequency. Now if velocity increases frequency must also increase. In this case, the original frequency is 384 Hz. Now increasing the temperature resulted in increase in velocity and thus increase in frequency.

So option a is correct. i.e. now frequency will be greater than 384 Hz.

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1 year ago
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2 years ago
En la etiqueta de un bote de fabada de 350 g, leemos que su aporte energético es de 1630 kj por cada 100 g de producto a) La can
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Answer:

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(b) 5705 kJ

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