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3 years ago

Why are metals sometimes added to gold?

Physics

2 answers:

MrRa [10]3 years ago

8 0

Answer: In order for the mixture of gold and added metals to become hard and suitable for processing.

Explanation: Gold is precious metal, it has a great value, and it used to serve as a means of payment, but also proof of wealth, status and power. Gold as such is soft, and in order to improve hardness it is mixed with another metal, such as nickel, copper, palladium, zinc, etc. For example, white gold is obtained with a mixture of pure gold and nickel, which gives the mixture a white color, and is sufficiently tough suitable for making rings. Based on the composition of the mixture that is the percentage of gold in the mixture, the number of karats is determined, which also means the value of gold itself. For example, the 999 gold has 99.9% of the gold in the mixture, which means 24 K (karats), and represents the so-called mixture with the most valuable percentage of gold, or the purest of which can be purchased.

Komok [63]3 years ago

7 0

Because solo gold isn't strong material. But with admixture it's stronger.

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3 years ago

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The conduction of heat from hot body to cold body is an example of what thermodynamics process?<br>

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2 years ago

A Neglecting air resistance, a ball projected straight upward so it remains in the air for 10 seconds needs an initial speed of

ololo11 [35]

Answer:

The initial velocity is 50 m/s.

(C) is correct option.

Explanation:

Given that,

Time = 10 sec

For first half,

We need to calculate the height

Using equation of motion

$v^2=u^2+2gh$

$h =\dfrac{v^2}{2g}$ ....(I)

For second half,

We need to calculate the time

Using equation of motion

$h =ut+\dfrac{1}{2}gt_{2}^2$

$h=0+\dfrac{1}{2}gt_{2}^2$

$t_{2}=\sqrt{\dfrac{2h}{g}}$

Put the value of h from equation (I)

$t_{2}=\sqrt{\dfrac{2\times v^2}{g^2}}$

$t_{2}=\dfrac{v}{g}$

According to question,

$t_{1}+t_{2}=10$

$t_{1}=t_{2}$

Put the value of t₁ and t₂

$\dfrac{v}{g}+\dfrac{v}{g}=10$

$\dfrac{2v}{g}=10$

$v=\dfrac{10\times g}{2}$

Here, g = 10

The initial velocity is

$v=\dfrac{10\times10}{2}$

$v=50\ m/s$

Hence, The initial velocity is 50 m/s.

3 0

3 years ago

A Foucault pendulum consists of a brass sphere with a diameter of 31.0 cm suspended from a steel cable 11.0 m long (both measure

kozerog [31]

Answer:

43.7 °C

Explanation:

$\alpha_b$ = Coefficient of linear expansion of brass = $18\times 10^{-6}\ ^{\circ}C$

$\alpha_s$ = Coefficient of linear expansion of steel = $11\times 10^{-6}\ ^{\circ}C$

$L_{0b}$ = Initial length of brass = 31 cm

$L_{0s}$ = Initial length of steel = 11 m

$\Delta L$ = Total change in length = 3 mm

Total change in length would be

$\Delta L=\Delta L_b+\Delta L_s\\\Rightarrow \Delta L=L_{0b}\alpha_b\Delta T+L_{0s}\alpha_b\Delta T\\\Rightarrow \Delta T=\frac{\Delta L}{L_{0b}\alpha_b+L_{0s}\alpha_b}\\\Rightarrow \Delta T=\frac{0.003}{0.31\times 18\times 10^{-6}+11\times 10^{-6}\times 11}\\\Rightarrow \Delta T=23.7\ ^{\circ}C$

$\Delta T=23.7\\\Rightarrow T_f-T_i=23.7\\\Rightarrow T_f=23.7+T_i\\\Rightarrow T_f=23.7+20\\\Rightarrow T_f=43.7\ ^{\circ}C$

The final temperature is 43.7 °C

6 0

3 years ago