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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 circular loop of wire has radius of 9.50 cmcm. A sinusoidal electromagnetic plane wave traveling in air passes through the loo
m_a_m_a [10]

Answer:

The induced emf  is  \epsilon  = 0.1041 \  V  

Explanation:

From the question we are told that

   The  radius of the circular loop is  r =  9.50 \ cm  =  0.095 \ m

     The  intensity of the wave is  I  =  0.0215 \ W/m^2

      The wavelength is  \lambda =  6.90\ m

Generally the intensity is mathematically represented as

         I  =  \frac{ c *  B^2  }{ 2 * \mu_o  }

Here  \mu_o is the permeability of free space with value  

         \mu_o  =  4 \pi *10^{-7} N/A^2

B is the magnetic field which can be mathematically represented from the equation as

          B  =  \sqrt{ \frac{ 2 *  \mu_o  *  I  }{ c} }

substituting values

          B  =  \sqrt{ \frac{ 2 *  4\pi *10^{-7} *   0.0215  }{ 3.0*10^{8}} }

          B  =  1.342 *10^{-8} \  T

The  area is mathematically represented as

       A =  \pi r^2

substituting values

       A =  3.142 *   (0.095)^2

       A = 0.0284

The angular velocity is mathematically represented as

        w =  2 *  \pi  *  \frac{c}{\lambda }

substituting values          

       w =  2 *  3.142   *  \frac{3.0*10^{8}}{ 6.90 }  

        w =  2.732 *10^{8} rad  \ s^{-1}  

Generally the induced emf is mathematically represented as

        \epsilon  =  N *  B  *  A  *  w * sin (wt )

At maximum induced emf  sin (wt)  =  1

    So

         \epsilon  =  N *  B  *  A  *  w

substituting values

         \epsilon  = 1  *    1.342 *10^{-8}   *  0.0284  *2.732 *10^{8}  

         \epsilon  = 0.1041 \  V  

         

7 0
3 years ago
Which type of energy is released when a bond between atoms is broken
Murrr4er [49]

Answer: gases

Explanation: because gases move around freely and they would be the only one to make sense because solid are compacted together and liquid are not so fast at moving but gases are wild                                          

dont use this this is a bad explanation

4 0
3 years ago
Two ships of equal mass are 110 m apart. What is the acceleration of either ship due to the gravitational attraction of the othe
dusya [7]

Answer:

Acceleration of the ship, a=2.14\times 10^{-7}\ m/s^2

Explanation:

It is given that,

Mass of both ships, m=39000\ metric\ tons=39\times 10^6\ kg

Distance between two ships, d = 110 m

The gravitational force between two ships is given by :

F=G\dfrac{m^2}{d^2}

F=6.67\times 10^{-11}\ Nm^2/kg^2\times \dfrac{(39\times 10^6\ kg)^2}{(110\ m)^2}

F = 8.38 N

Let a is the acceleration. Now, using second law of motion as :

a=\dfrac{F}{m}

a=\dfrac{8.38\ N}{39\times 10^6\ kg}

a=2.14\times 10^{-7}\ m/s^2

So, the acceleration of either ship due to the gravitational attraction of the other is 2.14\times 10^{-7}\ m/s^2. Hence, this is the required solution.

7 0
3 years ago
If the maximum tension in the simulation is 10.0 N, what is the linear mass density (m/L) of the string
vovikov84 [41]

Complete Question

The speed of a transverse wave on a string of length L and mass m under T is given by the formula

     v=\sqrt{\frac{T}{(m/l)}}

If the maximum tension in the simulation is 10.0 N, what is the linear mass density (m/L) of the string

Answer:

(m/l)=\frac{10}{V^2}

Explanation:

From the question we are told that

Speed of a transverse wave given by

v=\sqrt{\frac{T}{(m/l)}}

Maximum Tension is T=10.0N

Generally making (m/l) subject from the equation mathematically we have

v=\sqrt{\frac{T}{(m/l)}}

v^2=\frac{T}{(m/l)}

(m/l)=\frac{T}{V^2}

(m/l)=\frac{10}{V^2}

Therefore the Linear mass in terms of Velocity is given by

(m/l)=\frac{10}{V^2}

8 0
3 years ago
Which of the following is the equation for calculating daily caloric needs? A. BMR calories + Digestion calories B. BMR calories
labwork [276]

Explanation:

To determine your total daily calorie needs, multiply your BMR by the appropriate activity factor, as follows: If you are sedentary (little or no exercise) : Calorie-Calculation = BMR x 1.2. If you are lightly active (light exercise/sports 1-3 days/week) : Calorie-Calculation = BMR x 1.375.

<em><u>HAPPY TO HELP</u></em>

8 0
3 years ago
Read 2 more answers
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