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zavuch27 [327]
3 years ago
9

the density of dead sea surface water is about 1.166 g/ml. how much mass, in grams, would 2 liters of this salty water have?

Physics
2 answers:
kogti [31]3 years ago
6 0
<span>Density is a value for mass, such as kg, divided by a value for volume, such as m^3. Density is a physical property of a substance that represents the mass of that substance per unit volume. The mass of the salty water is calculated as follows:

Mass = 1.166 g/mL ( 2000 mL ) = 2332 g of salty water</span>
Dennis_Churaev [7]3 years ago
4 0
<span>Density is a value for mass, such as kg, divided by a value for volume, such as m^3. Density is a physical property of a substance that represents the mass of that substance per unit volume. The mass of the salty water is calculated as follows:

Mass = 1.166 g/mL ( 2000 mL ) = 2332 g of salty water</span>
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A spherical capacitor is formed from two concentric sphericalconducting shells separated by vacuum. The inner sphere has radius1
zubka84 [21]

Explanation:

(1).  Formula to calculate the potential difference is as follows.

       \Delta V = -\int E dr

                  = -\int \frac{kq}{r^{2}} dr

                 = \frac{kq}{r_{f}} - \frac{kq}{r_{i}}

                 = \frac{kq(r_{f} - r_{i})}{r_{f}r_{i}}

                 = \frac{9 \times 10^{9} \times 3.30 \times 10^{-9}(0.1 - 0.015)}{0.1 \times 0.015}

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Therefore, magnitude of the potential difference between the two spheres is 38.7 volts.

(2).  Now, formula to calculate the energy stored in the capacitor is as follows.

           E = \frac{1}{2}QV

              = \frac{1}{2} \times 3.30 \times 10^{-9} \times 3.87 V

              = 6.39 \times 10^{-8} J

Thus, the electric-field energy stored in the capacitor is 6.39 \times 10^{-8} J.

7 0
3 years ago
Read 2 more answers
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