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Eduardwww [97]
3 years ago
9

If the density of a diamond is 3.5 /cm^ 3 what would be the mass of a diamond whose volume is 0.5 cm ^ 3 ?

Physics
1 answer:
Elan Coil [88]3 years ago
7 0

Answer:

1.75g

Explanation:

Given parameters:

Density  = 3.5g/cm³

Volume  = 0.5cm³  

Unknown:

Mass of the diamond  = ?

Solution:

Density is the mass per unit volume of a substance

  Density  = \frac{mass}{volume}  

So;

  Mass  = 3.5 x 0.5  = 1.75g

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Sound level at distance of 15 m is given as 20 dB

so intensity at this distance is given as

L = 10 Log\frac{I}{I_0}

20 = 10 Log{I}{10^{-12}}

I_1 = 10^{-10}W/m^2

now if we move closer to some some distance the sound level is now 50 dB

now the intensity is given as

L = 10 Log\frac{I}{I_0}

50 = 10Log\frac{I}{10^{-12}}

I_2 = 10^{-7}W/m^2

now we know that

\frac{I_1}{I_2} = \frac{r_2^2}{r_1^2}

\frac{10^{-10}}{10^{-7}} = \frac{r_2^2}{15^2}

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A large power plant heats 1917 kg of water per second to high-temperature steam to run its electrical generators.
erastova [34]

Complete Question

A large power plant heats 1917 kg of water per second to high-temperature steam to run its electrical generators.

(a) How much heat transfer is needed each second to raise the water temperature from 35.0°C to 100°C, boil it, and then raise the resulting steam from 100°C to 450°C? Specific heat of water is 4184 J/(kg · °C), the latent heat of vaporization of water is 2256 kJ/kg, and the specific heat of steam is 1520 J/(kg · °C).

J

(b) How much power is needed in megawatts? (Note: In real power plants, this process occurs under high pressure, which alters the boiling point. The results of this problem are only approximate.)

MW

Answer:

The heat transferred is  Q = 5.866 * 10^9 J

The power is  P = 5866\  MW

Explanation:

From the question we are told that

      Mass of the water per second is m = 1917 \ kg

      The initial temperature of the water is T_i  = 35^oC

      The boiling point of water is  T_b = 100^oC

      The final temperature T_f = 450^oC

      The latent heat of vapourization of water is  c__{L}} = 2256*10^3 J/kg

      The specific heat of water c_w = 4184 J/kg^oC

      The specific heat of stem is C_s =1520 \ J/kg ^oC

Generally the heat needed each second is mathematically represented as

         Q = m[c_w (T_i - T_b) + m* c__{L}}  + m* c__{S}} (T_f - T_b)]

Then substituting the value

        Q = m[c_w [T_i - T_b] + c__{L}}  + C__{S}} [T_f - T_b]]

         Q = 1917 [(4184) [100 - 35] + [2256 * 10^3]  +[1520]  [450 - 100]]

         Q = 1917 * [3.05996 * 10^6]

         Q = 5.866 * 10^9 J

The power required is mathematically represented as

         P = \frac{Q}{t}

From the question t = 1\ s

So  

        P = \frac{5.866 *10^9}{1}

        P = 5866*10^6 \ W

        P = 5866\  MW

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Answer:

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Explanation:

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KengaRu [80]

Answer:

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Explanation:

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When Irina rides to work with her bike, she travels at a speed of 16 mph. So we can find the time she takes by dividing the total distance (10.8 miles) by her speed:

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