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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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2 Pont
Reptile [31]

Answer:

A. It makes astronauts weightless.

Explanation:

Gravity does not make astronauts feel weightless. Astronauts are weightless because they are orbiting at the same rate as their shuttle.

Although the force of gravity weakens as one moves away from the earth surface, it does not mean that this force is absent in orbit

  • Gravitational force has a constant acceleration value near the earth surface which is commonly known to be 9.8m/s².
  • It is a force of attraction tending to hold and bind bodies together so far they have mass.
  • This force keeps every thing from escaping space-ward from the earth surface.
6 0
3 years ago
g A large tank, at 400 kPa and 450 K, supplies air to a converging-diverging nozzle of throat area 4 cm2 and exit area 5 cm2. Fo
mariarad [96]

Answer:

A) ≥ 325Kpa

B) ( 265 < Pe < 325 ) Kpa

C) (94 < Pe < 265 )Kpa

D)  Pe < 94 Kpa

Explanation:

Given data :

A large Tank : Pressures are at 400kPa and 450 K

Throat area = 4cm^2 ,  exit area = 5cm^2

<u>a) Determine the range of back pressures that the flow will be entirely subsonic</u>

The range of flow of back pressures that will make the flow entirely subsonic

will be ≥ 325Kpa

attached below is the detailed solution

<u>B) Have a shock wave</u>

The range of back pressures for there to be shock wave inside the nozzle

= ( 265 < Pe < 325 ) Kpa

attached below is a detailed solution

C) Have oblique shocks outside the exit

= (94 < Pe < 265 )Kpa

D) Have supersonic expansion waves outside the exit

= Pe < 94 Kpa

7 0
3 years ago
What is happening in the picture?
Veseljchak [2.6K]
The gravity is pushing rhe boat down
3 0
3 years ago
Read 2 more answers
A cannonball is fired vertically upwards at 100.0 m/s a) How long will it take to return to the cannon? b) what is it's maximum
V125BC [204]
Answer:
a) 20s
b) 500m

Explanation:
Given the initial velocity = 100 m/s, acceleration = -10m/s^2 (since it is moving up, acceleration is negative), and at the maximum height, the ball is not moving so final velocity = 0 m/s.

To find time, we apply the UARM formula:

v final = (a x t) + v initial

Replacing the values gives us:

0 = (-10 x t) + 100

-100 = -10t

t = 10s

It takes 10s for the the ball to reach its max height, but it must also go down so it takes 2 trips, once going up and then another one going down, both of which take the same time to occur

So 10s going up and another 10s going down:

10x2 = 20s

b) Now that we have v final = 0, v initial = 100, a = -10, t = 10s (10s because maximum displacement means the displacement from the ground to the max height) we can easily find the displacement by applying the second formula of UARM:

Δy = (1/2)(a)(t^2) + (v initial)(t)

Replacing the values gives us:

Δy = (1/2)(-10)(10^2) + (100)(10)

= (-5)(100) + 1000

= -500 + 1000

= 500 m

Hope this helps, brainliest would be appreciated :)
7 0
2 years ago
uppose a wheel is initially rotating at 10.0 rad/s while undergoing constant angular acceleration reaching a speed of 30.0 rad/s
Paul [167]

Answer:

t = 22.2 s

Explanation:

angular distance covered in the 36.0 s is

θ = ω(avg)t = ½(10.0 + 30.0)36 = 720 radians

720/2 = 360 radians

α = Δω/t = (30 - 10)/36 = 5/9 rad/s²

θ = ω₀t + ½αt²

360 = 10.0t + ½(5/9)t²

   0 = (5/18)t² + 10.0t - 360

   0 = t² + 36t - 1296

t = (-36 ±√(36² - 4(1)(-1296))) / 2

t = (-36 ±√(6480)) / 2

t = -18 ±√1620

we ignore the negative time result as it occurs before we care.

t = -18 + √1620 = 22.249223... s

8 0
3 years ago
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