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Mademuasel [1]
3 years ago
14

When the plutonium bomb was tested in New Mexico in 1945, approximately 1 gram of matter was converted into energy. Suppose anot

her bomb is tested, and 3.6 grams of matter are converted into energy. How many joules of energy are released by the explosion?
Physics
1 answer:
Softa [21]3 years ago
3 0

Answer:

1.08 * 10^{14} J

Explanation:

Energy and mass are related by the famous equation developed by Albert Einstein:

E = mc^2

where m = mass and c = speed of light

This equation explains that an object with very small mass can produce a large amount of energy in reactions such as a nuclear reaction.

Hence, the energy produced by the explosion of a Plutonium bomb containing 3.6 grams of matter is:

E = 3.6 * 10^{-3} * (3 * 10^8)^2

E = 1.08 * 10^{14} J

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Vlada [557]
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2 years ago
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18 kilogram Mass Blokus addressed a level surface if the coefficient of static friction between the Block in the surface is 0.6
Tasya [4]

Question: 18 kilogram Mass Block rest on level surface if the coefficient of static friction between the Block and the surface is 0.6 what  horizontal force is required to just move the blcok ( take gravity as 10m/s2 )

Answer:

108 N

Explanation:

From the question,

Applying

F' = mgμ................ Equation 1

Where F' = Frictional force = horizontal  force required to just move the block,  m = mass of the block, g = acceleration due to gravity, μ = coefficient of static friction.

From the question,

Given: m = 18 kg, μ = 0.6, g = 10 m/s²

Substitute these values into equation 1

F' = 18×0.6×10

F' = 108 N

4 0
2 years ago
How is velocity different from speed?
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2 years ago
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A 3-m-high, 7-m-wide rectangular gate is hinged at the top edge and is restrained by a fixed ridge. Determine the hydrostatic fo
Shalnov [3]

Answer:

The Hydrostatic force is   F  =  137.2 kN

The location of pressure center is  Z  = 1.333 \ m  

Explanation:

From the question we are told that

   The height of the gate is  h =  3 \ m

     The weight of the gate is  w =  7 \  m

      The height of the water is  h_w  =  2 \ m

       The density of water is \rho_w  =  1000 \ kg/m^3

Note used h_w for height of water and height of gate immersed by water since both have the same value

The area of the gate immersed in water  is mathematically represented as

         A =  h_w  * w

substituting values

         A =  2*  7

         A =  14  \ m^2

The hydrostatic force is mathematically represented as

          F  =  \rho_w * g * h_f * A

Where

            h_f =h-  h_w

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So  

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            F  =  137.2 kN

The center of pressure is mathematically represented as

        Z  =  h_f + \frac{I_g}{h_f * A}

Where I_g is the moment of inertia of the gate which mathematically represented as

            I_g =  \frac{w * h_w^2}{12}

The h_w is the height of gate immersed in water

            I_g =  \frac{7  * 2^2 }{12}

             I_g = 4.667\ kg  m^2

Thus  

        Z  = 1  + \frac{4.66}{1 * 14}

        Z  = 1.333 \ m

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