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kirza4 [7]
2 years ago
14

The reaction energy of a reaction is the amount of energy released by the reaction. It is found by determining the difference in

mass between the reactants and products and then using E=mc2 to convert the lost mass into the released energy. Look again at the equation for the decay of radium (Ra), given in the introduction. Use the following table of masses to determine the reaction energy of this reaction.Nucleus or particle Mass in atomic mass unitsradium-226 226.025402radon-222 222.017571alpha particle 4.002602Express your answer in joules to three significant figures.
Physics
1 answer:
solmaris [256]2 years ago
5 0

Answer: the correct answer is 7.8026035971 x 10^(-13) joule

Explanation:

Use Energy Conservation. By ``alpha decay converts'', we mean that the parent particle turns into an alpha particle and daughter particles. Adding the mass of the alpha and daughter radon, we get

m = 4.00260 u + 222.01757 u = 226.02017 u .

The parent had a mass of 226.02540 u, so clearly some mass has gone somewhere. The amount of the missing mass is

Delta m = 226.02540 u - 226.02017 u = 0.00523 u ,

which is equivalent to an energy change of

Delta E = (0.00523 u)*(931.5MeV/1u)

Delta E = 4.87 MeV

Converting  4.87 MeV to Joules

1 joule [J] = 6241506363094 mega-electrón voltio [MeV]

4 mega-electrón voltio = 6.40870932 x 10^(-13) joule

4.87 mega-electrón voltio = 7.8026035971 x 10^(-13) joule

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Two long conducting cylindrical shells are coaxial and have radii of 20 mm and 80 mm. The electric potential of the inner conduc
xxMikexx [17]

Answer: 14.52*10^6 m/s

Explanation: In order to explain this problem we have to consider the energy conservation for the electron within the coaxial cylidrical wire.

the change in potential energy for the electron; e*ΔV is  equal to energy kinetic gained for the electron so:

e*ΔV=1/2*m*v^2  v^=(2*e*ΔV/m)^1/2= (2*1.6*10^-19*600/9.1*10^-31)^1/2=14.52 *10^6 m/s

3 0
2 years ago
A sports car accelerates from 0 to 30 mph in 1.5 s. How long would it take to accelerate from 0 to 60 mph, assuming the power of
Crank

Answer:

6 s

Explanation:

given,

Sports car accelerate from 0 to 30 mph in 1.5 s

time taken to accelerate  0 to 60 mph = ?

The power of the engine is independent of velocity and neglecting friction

power =

P = constant  

the kinetic energy for 60 mph larger than this of 30 mph

 = \dfrac{\dfrac{1}{2}mv_1^2}{\dfrac{1}{2}mv_2^2}

 = \dfrac{v_1^2}{v_2^2}

 = \dfrac{60^2}{30^2}

 = 4

gain in kinetic energy  = P x t

time = 4 x 1.5

       = 6 s

8 0
2 years ago
As a runner crosses the finish line of a race, she starts decelerating from a velocity of 9 m/s at a rate of 2 m/s^2. Take the r
Ksivusya [100]

Answer:

- 1 m/s, 20 m

Explanation:

u = 9 m/s, a = - 2 m/s^2, t = 5 sec

Let s be the displacement and v be the velocity after 5 seconds

Use first equation of motion.

v = u + a t

v = 9 - 2 x 5 = 9 - 10 = - 1 m/s

Use second equation of motion

s = u t + 1/2 a t^2

s = 9 x 5 - 1/2 x 2 x 5 x 5

s = 45 - 25 = 20 m

4 0
2 years ago
A cylindrical rod of steel (E = 207 GPa, 30 × 10 6 psi) having a yield strength of 310 MPa (45,000 psi) is to be subjected to a
Yanka [14]

Answer:

Diameter of the cylinder will be d=2.998\times 10^4m

Explanation:

We have given young's modulus of steel E=207GPa=207\times 10^9Pa  

Change in length \Delta l=0.38mm

Length of rod l=500mm

Load F = 11100 KN

Strain is given by strain=\frac{\Delta l}{l}=\frac{0.38}{500}=7.6\times 10^{-4}

We know that young's modulus E=\frac{stress}{strain}

So 207\times 10^9=\frac{stress}{7.6\times 10^{-4}}

stress=1573.2\times 10^{-5}N/m^2

We know that stress =\frac{force}{artea }

So 1573.2\times 10^{-5}=\frac{11100\times 1000}{area}

area=7.055\times 10^{8}m^2

So \frac{\pi }{4}d^2=7.055\times 10^{8}

d=2.998\times 10^4m          

6 0
2 years ago
A flywheel of diameter 1.2 m has a constant angular acceleration of 5.0 rad/s2. the tangential acceleration of a point on its ri
vodka [1.7K]
We know that tangential acceleration is related with radius and angular acceleration according the following equation:  
at = r * aa  
where at is tangential acceleration (in m/s2), r is radius (in m) aa is angular acceleration (in rad/s2)  
So the radius is r = d/2 = 1.2/2 = 0.6 m  
Then at = 0.6 * 5 = 3 m/s2  
Tangential acceleration of a point on the flywheel rim is 3 m/s2
5 0
2 years ago
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