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Mrrafil [7]
2 years ago
12

A sample of gold has a volume of 2 cm3 and a mass of 38.6 grams. What would be the density, and three other properties of the sa

mple that would be considered physical properties.
Physics
1 answer:
lukranit [14]2 years ago
3 0

The density of the gold is calculated to be "19,300 kg/m³".

<u>Explanation:</u>

Given:

Volume = 2cm³

Mass = 38.6 grams.

To Find:

Density of the gold = ?

Solution:

Density is obtained by dividing mass of the sample by its volume and it is given in the units of kg/m³.

Mass in grams is converted into kg as,

1 g = 0.001 kg

38. 6 g = \frac{38.6}{1000} = 0.0386 kg

Now we have to convert cm³ to m³ as,

1 cm³ = 10⁻⁶ m³

2 cm³ = 2 × 10⁻⁶ m³

So Density = \frac{0.0386 k g}{2 \times 10^{-6} m^{3}}=19,300 \mathrm{kg} / \mathrm{m}^{3}

Physical properties of Gold:

  • Gold is a heavy metal.
  • It is Malleable and ductile.
  • It is a corrosion resistant element.
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mariarad [96]

Answer:

S_i=\frac{9}{10} =0.9

Explanation:

Given:

  • volume of oil in the cylinder, V_o=60\ cm^2
  • volume of the oil level when the ice is immersed, V=90\ cm^3
  • the volume level of oil when the ice melted, V'=87\ cm^3

<u>Now, therefore the volume of ice:</u>

V_i=V-V_o

V_i=90-60

V_i=30\ cm^3

<u>Now the volume of water:</u>

V_w=V'-V_o

V_w=87-60

V_w=27\ cm^3

As we know that the relative density is the ratio of density of the substance to the density of water.

<u>So, the relative density of ice:</u>

S_i=\frac{\rho_i}{\rho_w} .....................(1)

as we know that density is given as:

\rm \rho=\frac{mass}{volume}

now eq. (1)

S_i=\frac{m}{V_{i}}\div  \frac{m}{V_w}

where, m = mass of the water or the ice which remains constant in any phase

S_i=\frac{V_w}{V_i}

S_i=\frac{27}{30}

S_i=\frac{9}{10} =0.9

7 0
2 years ago
Which law of motion accounts for the following statement? "When a marble and a billiard ball are impacted by the same force, the
Komok [63]
The second law explains this.
4 0
2 years ago
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A circular coil has a 10.0 cm radius and consists of 30.0 closely wound turns of wire. an externally produced magnetic field of
IrinaVladis [17]
Magnetic flux can be calculated by the product of the magnetic field and the area that is perpendicular to the field that it penetrates. It has units of Weber or Tesla-m^2. For the first question, when there is no current in the coil, the flux would be:

ΦB = BA
          A = πr^2
          A = π(.1 m)^2
          A = π/100 m^2   

 ΦB = 2.60x10^-3 T (π/100 m^2 ) ΦB = 8.17x10^-5 T-m^2 or Wb (This is only for one loop of the coil)

The inductance on the coil given the current flows in a certain direction can be calculated by the product of the total number of turns in the coil and the flux of one loop over the current passing through. We do as follows:

L = N (ΦB ) / I
L = 30 (8.17x10^-5 T-m^2) / 3.80 = 6.44x10^-4 mH

6 0
2 years ago
Two long, parallel, current-carrying wires lie in an xy-plane. The first wire lies on the line y = 0.340 m and carries a current
sladkih [1.3K]

Answer:

The y-value of the line in the xy-plane where the total magnetic field is zero  U = 0.1355 \ m

Explanation:

From the question we are told that

    The distance of wire one from two along the y-axis is    y = 0.340 m

   The current on the first wire is  I_1 =  (27.5i) A

    The force per unit length on each wire is  Z =  295 \mu N/m = 295*10^{-6}  N/m

Generally the force per unit length is mathematically represented as

         Z = \frac{F}{l}  =  \frac{\mu_o I_1I_2}{2\pi y}

=>      \frac{\mu_o I_1I_2}{2\pi y}  =  295

Where  \mu_o is the permeability of free space with a constant value of  \mu_o  =  4\pi *10^{-7} \ N/A2

substituting values

       \frac{ 4\pi *10^{-7} 27.5 * I_2}{2\pi * 0.340}  =  295 *10^{-6}

=>    I_2 =  18.23 \ A

Let U  denote the  line in the xy-plane where the total magnetic field is zero

So  

      So the force per unit length of  wire 2  from  line  U is equal to the force per unit length of wire 1  from  line  (y - U)      

   So  

         \frac{\mu_o  I_2  }{2 \pi U} =  \frac{\mu_o  I_1  }{2 \pi(y -  U) }

substituting values

          \frac{  18.23  }{ U} =  \frac{ 27.5 }{(0.34 -  U) }

         6.198 -18.23U = 27.5U

          6.198=45.73U

          U = 0.1355 \ m              

5 0
2 years ago
You have a device that needs a voltage reference of 3.0 V, but you have only a 9.0 V battery. Fortunately, you also have several
Nimfa-mama [501]

Answer:

Explanation:

Solution:

- We are to develop a circuit that has an input of available battery 9.0 V and has an output potential difference of 3.0 V

- We are given:

             Battery ..... 9.0 V

             Resistors ... 10 Kohms

- We will develop a potential divider circuit by placing a few resistors in series and then connecting in between resistors to get our desired voltage.

- How many resistors should we use ?

  We know that if we add series resistance in a circuit the current decreases proportionally. However, the potential difference across resistors also changes.

- Our desired voltage is a ratio of input battery voltage.

                                   Input / Output = 9 / 3 = 3

We can use this ratio as the number of "Identical resistors" that can be placed in series to give us the desired voltage. Note: This would not be true if we did not had any identical resistors.

- We will place 3, 10 Kohms resistors in series.

- To verify we will calculate the potential difference across each resistor. The current of the total circuit is:

                                   I = V / R_eq

                                   R_eq = 3*R = 30 kohms

                                   I = 9 / 30,000 = 0.0003 Amps

- Now the potential difference for each resistor:

                                   V = I*R_each

                                   V = 0.0003*(10,000)

                                   V = 3.0 V

- We can take two leads across any 10 kohms resistor and the potential difference across the leads would be the desired voltage 3.0 V.

8 0
2 years ago
Read 2 more answers
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