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jok3333 [9.3K]
3 years ago
11

A ball has a diameter of 3.79 cm and average density of 0.0838 g/cm3.

Physics
1 answer:
suter [353]3 years ago
3 0

Answer: 0.258 N

Explanation:

As the density of the object is much less than the density of water, it’s clear that the buoyant force, is greater than the weight of the object, which means that in normal conditions, it would float in water.

So, in order to get the ball submerged in water, we need to add a downward force, that add to the weight, in order to compensate the buoyant force, as follows:

F = Fb – Fg

Fb= δH20* 4/3*π*(d/2)³  * g

Fg = δb* 4/3*π*(d/2)³ *g

F= (δH20- δb) * 4/3*π*(d/2)³*g

Replacing by the values of the densities, and the ball diameter, we finally get:

F= 0.258 N

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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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masya89 [10]
The equation for work (W) done by an electric field is:

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Answer: equation for the reaction is given below

PCL2+PCL3=PCL5

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∆G=-8.314*298in(0.0029/0.40*.27)

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Explanation:

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