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IceJOKER [234]
2 years ago
7

5. the combustion of a single molecule of methane produces about 10 ev of energy. a methane molecule has a mass of 16 amu. the f

ission of a uranium-235 atom
produces about 189 mev of energy. based on these numbers, about what mass of methane will produce as much energy as a single gram of uranium-235? (one mel
is equal to one million ev.)
oa 19 grams of methane
ob 304 grams of methane.
o c. 71.4 kilograms of methane
od 1,287 kilograms of methane
Physics
1 answer:
Alchen [17]2 years ago
7 0

The mass of methane will produce as much energy as a single gram of uranium-235 1,287 kilograms of methane. Option D is correct.

<h3>What is uranium?</h3>

Uranium, with the atomic number 92 and the symbol U in the periodic table, is a weakly radioactive element. One of the heavy metals that may be used as a concentrated energy source is uranium.

Given data;

⇒One mole of U-235 = 235 grams

=> 1 gram of U-235 = 1/235 moles

⇒1 mole of U-235 = 6.023 x 10²³ atoms

The no of atoms is;

=> 1/235 moles of U-235

⇒N = 6.023 x 10²³/235 atoms

⇒N=25.6 x 10²⁰ atoms

If one atom of U-235 gives is 189 x 10⁶ eV energy,25.6 x 10²⁰ atoms of U-235 gives;

=> 25.6 x 10²⁰ atoms of U-235 gives;

E = 25.6 x 10²⁰ x 189 x 10⁶

E= 4.8 x 10²⁹ eV energy

One methane molecule produces;

E = 10 eV of energy

=> To produce 4.8 x 10²⁹ eV energy, no. of molecules required;

= 4.8 x 10²⁸

6.032 x 10²³ molecules of methane = 16 gms

=> 4.8 x 10²⁸ molecules of methane = 16 x 4.8 x 10²⁸/ 6.032 x 10²³ gms

m= 12.75 x 10⁵ gms

m= 1275 kilograms of methane

m≅ 1287 kilograms of methane

1 gram of U-235 has the same amount of energy as 1287 kilograms of methane.

Hence option D is correct,

To learn more about uranium refer to the link;

brainly.com/question/9099776

#SPJ1

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Answer:

a) h_m=74625\ m

b) t=265.55\ s

Explanation:

Given:

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<u>During the phase of fuel exhaustion:</u>

<u>velocity attained by the rocket just as the fuel ends:</u>

v_f=u+a.t_f

where:

u= initial velocity of the rocket = 0

v_f=0+30\times 35

v_f=1050\ m.s^{-1} this will be the initial velocity for the phase of ascending of the rocket's height under the influence of gravity.

<u>height at which the fuel finishes:</u>

v_f^2=u^2+2a.h_f

1050^2=0^2+2\times 30\times h_f

h_f=18375\ m

<u>During the phase of ascend in height of rocket after the fuel is over:</u>

<u>Time taken to reach the top height after the fuel is over:</u>

v=v_f+g.t'

at top v = (final velocity during this course of motion )= 0 m.s^{-1}

0=1050-9.8\times t'

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<u>Height ascended by the rocket after the fuel is over:</u>

v^2=v_f^2+2g.h'

at the top height the velocity is zero

0^2=1050^2-2\times 9.8\times h' (-ve sign denotes that the direction of motion is opposite to that of acceleration)

h'=56250\ m

<u>Therefore the maximum altitude attained by the rocket:</u>

h_m=h_f+h'

h_m=18375+56250

h_m=74625\ m

b)

time taken by the rocket to fall back to the earth:

h_m=v.t_m+\frac{1}{2} g.t_m^2

where:

v= initial velocity of the rocket during the course of free fall from the top height.

74625=0+4.9\times t_m^2

t_m=123.41\ s

Now the total time for which the rocket is in the air:

t=t_f+t'+t_m

t=35+107.1429+123.41

t=265.55\ s

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Answer:

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