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maksim [4K]
3 years ago
13

List 10 uses of metals​

Physics
1 answer:
algol [13]3 years ago
8 0

Answer:

Shiny metals such as copper, silver, and gold are often used for decorative arts, jewelry, and coins.

Strong metals such as iron and metal alloys such as stainless steel are used to build structures, ships, and vehicles including cars, trains, and trucks.

Some metals have specific qualities that dictate their use. For example, copper is a good choice for wiring because it is particularly good at conducting electricity. Tungsten is used for the filaments of light bulbs because it glows white-hot without melting.

Nonmetals are plentiful and useful. These are among the most commonly used:

Oxygen, a gas, is absolutely essential to human life. Not only do we breathe it and use it for medical purposes, but we also use it as an important element in combustion.

Sulfur is valued for its medical properties and as an important ingredient in many chemical solutions. Sulfuric acid is an important tool for industry, used in batteries and manufacturing.

Chlorine is a powerful disinfectant. It is used to purify water for drinking and fill swimming pools.

Explanation:

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A small water pump is used in an irrigation system. The pump takes water in from a river at 10oC, 100 kPa at a rate of 5 kg/s. T
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Answer:

0.98kW

Explanation:

The conservation of energy is given by the following equation,

\Delta U = Q-W

\dot{m}(h_1+\frac{1}{2}V_1^2+gz_1)-\dot{W} = \dot{m}(h_2+\frac{1}{2}V_2^2+gz_)

Where

\dot{m} = Mass flow

h_1 =Specific Enthalpy (IN)

h_2 = Specific Enthalpy (OUT)

g = Gravity

z_{1,2} = Heigth state (In, OUT)

V_{1,2} =Velocity (In, Out)

Our values are given by,

T_i = 10\°C

P_1 = 100kPa

\dot{m} = 5kg/s

z_2 = 20m

For this problem we know that as pressure, temperature as velocity remains constant, then

h_1 = h_2

V_1 = V_2

Then we have that our equation now is,

\dot{m}(gz_1) = \dot{m}(gz_2)+\dot{W}

\dot{W} = \frac{(5)(9.81)(0-20)}{1000}

\dot{W} = -0.98kW

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Douglas has a segment with endpoints I(5, 2) and J(9, 10) that is divided by a point K such that IK and KJ form a 2:3 ratio. He
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For the answer to the question above, 
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(2 parts from i to k and 3 parts from k to j) 

The y distance from i to j is 
10 - 2 = 8 

Each part is 8/5 = 1.6 
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From the y coordinate of I which is 2 plus the 3.2 to point k 
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Answer y =5.2 

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2 years ago
Suppose you could fit 100 dimes, end to end, between your card with the pinhole and your dime-sized sunball. how many suns could
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Answer: 100 suns

Explanation:

We can solve this with the following relation:

\frac{d}{x_{sunball-pinhole}}=\frac{D}{x_{sun-pinhole}}

Where:

d=17.91 mm =17.91(10)^{-3}  m is the diameter of a dime

D is the diameter of the Sun

x_{sun-pinhole}=150,000,000 km=1.5(10)^{11}  m is the distance between the Sun and the pinhole

x_{sunball-pinhole}=100 d=1.791 m is the amount of dimes that fit in a distance between the sunball and the pinhole

Finding D:

D=\frac{d}{x_{sunball-pinhole}}x_{sun-pinhole}

D=\frac{17.91(10)^{-3}  m}{1.791 m} 1.5(10)^{11}  m

D=1.5(10)^{9}  m This is roughly the diameter of the Sun

Now, the distance between the Earth and the Sun is one astronomical unit (1 AU), which is equal to:

1 AU=149,597,870,700 m

So, we have to divide this distance between D in order to find how many suns could it fit in this distance:

\frac{149,597,870,700 m}{1.5(10)^{9}  m}=99.73 suns \approx 100 suns

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