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lisov135 [29]
2 years ago
8

Find the energy released when there is a decrease of 0.3kg of material in a nuclear reaction

Physics
1 answer:
MrRissso [65]2 years ago
4 0

Answer:

E = 2.7 x 10¹⁶ J

Explanation:

The release of energy associated with the mass can be calculated by Einstein's mass-energy relation, as follows:

E = mc^2

where,

E = Energy Released = ?

m = mass of material reduced = 0.3 kg

c = speed of light = 3 x 10⁸ m/s

Therefore,

E = (0.3\ kg)(3\ x\ 10^8\ m/s)^2

<u>E = 2.7 x 10¹⁶ J</u>

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zhenek [66]
W=2pi/T=>w=2
Vmax=AxW=2x1=2
5 0
2 years ago
the average american receives 2.28 mSv dose equivalent from radon each year. Assuming you receive this dose, and it all comes fr
Dmitry [639]

Answer:

The approximate number of decays  this represent  is  N= 23*10^{10}  

Explanation:

 From the question we are told that

    The amount of Radiation received by an average american is I_a = 2.28 \ mSv

     The source of the radiation is S = 5.49 MeV \ alpha \ particle

 Generally

            1 \  J/kg = 1000 mSv

   Therefore  2.28 \ mSv = \frac{2.28}{1000} = 2.28 *10^{-3} J/kg

Also  1eV = 1.602 *10^{-19}J

  Therefore  2.28*10^{-3} \frac{J}{kg} = 2.28*10^{-3} \frac{J}{kg}  * \frac{1ev}{1.602*10^{-19} J} = 1.43*10^{16} ev/kg

           An Average american weighs 88.7 kg

      The total energy received is mathematically evaluated as

        1 kg ------> 1.423*10^{16}ev \\88.7kg  --------> x

Cross-multiplying and making x the subject

           x = 88.7 * 1.423*10^{16} eV

              x = 126.2*10^{16}eV

Therefore the total  energy  deposited is x = 126.2*10^{16}eV

The approximate number of decays  this represent  is mathematically evaluated as

            N = \frac{x}{S}

Where n is the approximate number of decay

   Substituting values

                             N = \frac{126 .2*10^{16}}{5.49*10^6}  

                                  N= 23*10^{10}  

                     

             

7 0
3 years ago
A barbell consists of two small balls, each with mass m at the ends of a very low mass rod of length d. The barbell is mounted o
sveta [45]

The total angular momentum of the system about point B is L=m_1r_1\omega_1+m_2r_2\omega_2

Angular momentum, also known as moment of momentum or rotational momentum, is the rotating counterpart of linear momentum.

A rigid object's angular momentum is defined as the product of its moment of inertia and its angular velocity. If there is no external torque on the object, it is analogous to linear momentum and is subject to the fundamental constraints of the conservation of angular momentum principle. The vector quantity angular momentum It is derived from the expression for a particle's angular momentum.

Given,

mass of ball 1 = m1

m₂ mass of ball 2=m2

v₁ is the velocity of ball=r₁ω₁

v₂ is the velocity of ball 2=r₂ω₂

The total angular momentum is given as;

V_{total}=r_1\omega_1+r_2\omega_2\\\\L=m_1r_1\omega_1+m_2r_2\omega_2

Hence the total angular momentum  will be L=m_1r_1\omega_1+m_2r_2\omega_2

To learn more about angular momentum refer here

brainly.com/question/29512279

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6 0
1 year ago
Temperature is a measure of work.<br> True<br> False
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The answer is false.
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3 years ago
A gas is contained in a cylinder with a frictionless moveable piston at a pressure of 3.50 × 105 pascals and a volume of 0.02 cu
Softa [21]

W=PΔV

W=3.5 x 10⁵ x (0.15-0.02)

W=4.55 x 10⁴ J

3 0
2 years ago
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