1) In the nuclear reaction shown, one deuterium (hydrogen- 2 ) and one tritium (hydrogen-3) fuse to form one helium-4 and one neutron.
Data: Hydrogen-2 mass: 3.34450 10⁻²⁷ kg
Hydrogen-3 mass: 5.00827 10⁻²⁷ kg
Helium-4 mass: 6.64648 10⁻²⁷ kg
Neutron mass: 1.67493 10⁻²⁷ kg
a) What is the total mass of the pieces going into the reaction?
b) What is the total mass of the pieces coming out of the reaction?
c) How much energy is converted from rest energy to other forms of energy if one mole of deuterium (about 2.0 g ) and one mole of tritium (about 3.0 g ) are converted?

1) In The Nuclear Reaction Shown, One Deuterium (hydrogen- 2 ) And One Tritium (hydrogen-3) Fuse To Form

Answers

Answer 1

When helium-3 and deuterium combine in this nuclear process, regular helium and a proton are produced, wasting less energy and making the reaction easier to contain.

What energy forms of energy if one mole of deuterium?

The fusion of deuterium and tritium is the most promising of the hydrogen fusion processes that make up the deuterium cycle.

The reaction produces 17.6 MeV of energy, but in order to induce fusion, one must tunnel through the coulomb barrier, which requires extremely high temperatures.

Therefore, 8.4 1020 joules of energy would be released if one ton of deuterium were to undergo the fusion reaction with tritium. This is comparable to the 2.9 1010 joules of energy found in a ton of coal.

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Related Questions

The map below shows the path of a river. The arrow
shows the direction the river is flowing. LettersA
and B identify the banks of the river.The water depth is greater near bank A than bank B
because the water velocity near bank A isA) faster, causing deposition to occur
B) faster, causing erosion to occur
C) slower, causing deposition to occur
D) slower, causing erosion to occurAnswer: B

Answers

Different landforms are created during the process of water erosion, which flows water downhill. Rills, gullies, streams, rivers, tributaries, waterfalls, floodplains, meanders, lakes, and other natural features

What are the names of a river's beginning and end?

The source and mouth of a river are its beginning and closing, respectively. Before they get to the river's mouth, numerous rivers and streams will converge. The smaller streams and rivers are referred to as tributaries.

A river's beginning is at its mouth, correct?

The mouth of a river is where it flows into an ocean, a lake, or another river. River mouths are extremely dynamic places. The flow of a river collects material from the river, debris through bank erosion, and

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Rank these automobiles based on the magnitude of the impulse needed to stop them, from largest to smallest. Rank from largest to smallest. To rank items as equivalent, overlap them. If the ranking cannot be determined, check the box below. View Available Hint(s) 2000 kg 1000 kg 500 kg 4000 kg 500 kg 1000 kg 5 m/s 10 m/s 20 m/s 5 m/s 10 m/s 20 m/s largest smallest The correct ranking cannot be determinar Submit

Answers

Since impulse is defined as a change in momentum, we may conclude that since all autos are based on magnitude, their initial momentum and ultimate momentum are the same. Automobile 6 = Automobile 4>(Automobile 2 = Automobile 3>Automobile 1> Automobile 5

What is magnitude?

Magnitude is a term used in physics to describe an object's maximal size and direction. Magnitude is a factor that is shared by both scalar and vector quantities. We are aware that by definition, scalar quantities are those with only magnitude. It displays an object's size, direction, or motion in absolute or relative terms. It is used to describe something's size or scope. Magnitude in physics typically refers to a size or quantity.

How much is a magnitude?

One is roughly 10 times greater than the other if two integers are separated by one order of magnitude. They differ by a factor of around 100 if they are separated by two orders of magnitude. The greater value is less than 10 times the smaller value when two numbers are of the same order of magnitude.

Briefing:

Automobile-1 Mass= 2000kg = 5m/s

Automobile-2 Mass= 1000kg = 10m/s

Automobile-3 Mass= 500kg = 20m/s

Automobile-4 Mass= 4000kg = 5m/s

Automobile-5 Mass= 500kg = 10m/s

Automobile-6 Mass= 1000kg = 20m/s

Since we now understand that each car's momentum is the result of its mass and velocity, we will have

Automobile-1

P₁ = m × v

P₁ = (2000)(5)

P₁ = 2 × 10⁴kgm/s

Automobile-2

P₂ = m × v

P₂ = (1000)(10)

P₂ = 10⁴kgm/s

Automobile-3

P₃ = m × v

P₃ = (500)(20)

P₃ = 10⁴kgm/s

Automobile-4

P₄ = m × v

P₄ = (4000)(5)

P₄ = 2 × 10⁵kgm/s

Automobile-5

P₅ = m × v

P₅ = (500)(10)

P₅ = 10³kgm/s

Automobile-6

P₆ = m × v

P₆ = (1000)(20)

P₆ = 2 × 10⁵kgm/s

Then the momentum is:

Automobile 6 = Automobile 4>(Automobile 2 = Automobile 3>Automobile 1> Automobile 5

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establish the relation among energy, density, strain and stress

Answers

Strain energy is defined as the energy stored in a body due to deformation. The strain energy per unit volume is known as strain energy density and the area under the stress-strain curve towards the point of deformation.

Two physical science classes are playing tug of war. Mrs. Hankinson’s class pulls left with a force of 100 N, while Mrs. Wade’s class pulls with a force of 105 N in the opposite direction.


What is the net force, and which class won?


A. 205 N, Mrs. Wade's class

B. 5 N, Mrs. Wade's class

C. 205 N, it was a tie

D. 5 N, Mrs. Hankinson's class

Answers

The net force is 5 N and the class that won is Mrs. Wade's class (Option B)

How do I determine the net force?

We can obtain the net force by using the following formula:

Net force = Opposite force - force of pull

Now, we shall determine the net force as illustrated below:

Force by Mrs. Hankinson’s class = 100 NForce by Mrs. Wade’s class = 105 NNet force =?

Net force = Opposite force - force of pull

Net force = Mrs. Wade’s class - Mrs. Hankinson’s class

Net force = 105 - 100

Net force = 5 N

From the above calculation, we can say that the net force is 5 N and the class that won given the above data is Mrs. Wade’s class

Therefore, the correct answer to the question is:

5 N, Mrs. Wade's class (Option B)

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If it requires 9 J of work to stretch a spring by 2 cm from its equilibrium length, how much more work will be required to stretch it an additional 4 cm

Answers

A spring will therefore need 72 J of work to stretch an additional 4 cm from its equilibrium length if it takes 9 J to stretch it that far.

What is equilibrium length?

The length of a multi-mode optical fiber required to achieve a static mode distribution from a particular excitation condition is known as the equilibrium length, and it is sometimes used to characterize stationary mode distributions. There is a location where the weight and spring force are equal in magnitude but in the opposite direction. The equilibrium position is where you are at this moment. There is a net force known as the restoring force that is directed toward the equilibrium position if the mass is in any other position.

How do you find the equilibrium length of a spring?

F = -kx. The proportional constant k is referred to as the "spring constant". It gauges the stiffness of the spring. When a spring is stretched or compressed to a length that differs by an amount equal to or greater than x, it produces a force F = -kx in the direction of its equilibrium position. From its equilibrium length.

Briefing:

Knowing that the results of the work done to lengthen the springs will be reported as

U=1/2kx

so here we have

9= 1/2 k〖(0.02)〗^2

so we have

K=45000N/m

Now, we must locate the work completed to extend the spring from x = 2 cm to x = 4 cm

here we have

U=1/2 k(x 2/1-x 2/2

now we have

U=1/2(45000)(〖0.06)〗^2-〖0.02〗^2)

U= 72J

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In a physics lab, Ray Zuvlite arranges two mirrors with a right angle orientation as shown. Ray then directs a laser line at one of the mirrors. The light reflects off both mirrors as shown. If angle A is 38°, determine the what is the angle measure of angles B, C, and D?

Answers

The angle measure of angle B = 25° and angle C = 65° and angle D = 25°.

What is meant by measure?Through the use of a protractor, angles are measured in degrees (°). In order to calculate or draw angles in terms of degrees, a protractor is a measuring tool that is utilized. Using a protractor, for instance, we can observe that the black arrow in the image below crosses the 90° line at 100°. Thus, the angle has a 100° measurement. A degree is a unit of measurement for a plane angle in which one complete rotation equals 360 degrees, generally indicated by the symbol °. It is included in the SI brochure as an acceptable unit even though it is not a SI unit—the radian is the SI unit of angular measure.

The given parameters are;

The orientation of the two mirrors = At right angle to each other

The laser light is directed at one of the mirror

The measure of angle, A = 25°

The measures of angle B, C, and D are found as follows;

We have;

∠A = ∠B = 25°, by angle of incidence equals angle of reflection

∠B = 25°

∠B + ∠C = 90° by sum of the acute angles of a right triangle

25° + ∠C = 90°

∴ ∠C = 90° - 25° = 65°

∠C = 65°

∠E = ∠C = 65° by angle of incidence equals angle of reflection

∴ ∠E = 65°

Line 'L' is perpendicular to the second mirror, therefore, the angle between line 'L' and the second mirror = 90° = ∠E + ∠D

∠E + ∠D = 90°,  by angle sum property

Therefore;

65° + ∠D = 90°

∴ ∠D = 90° - 65° = 25°

∠D = 25°

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A new pump can drain your pool 4 times faster than the old pump.

Running the two together, it takes 4 hours to drain the pool.

How long will it take the old pump on its own just to drain the pool water?​

Answers

it is 20 4×5 trust me i did this for like 30 min

Based on current evidence, decide whether each of the following statements apply to the concept of dark matter, dark energy, both, or neither. Drag each statement into the appropriate bin.
Dark Matter Only:
as mass.
most common form of mass in the halo of the Milky Way Galaxy.
detected through its gravitational attraction.
likely to consist of tiny, weakly interacting particles.
Dark Energy Only:
inferred to exist from the measured acceleration of the universe's expansion.
the proposed source of a universal repulsive force.
Both Dark Matter and Dark Energy:
the universe contains more of this than it contains matter made from atoms.
we do not know what it is made of.
Neither Dark Matter nor Dark Energy
blocks light from stars behind it.
"dark" because it emits only infrared light.

Answers

The concept that the information applies to is illustrated below:

Dark matter:

Detected through its gravitational attraction

Has mass

Most common forms of mass in the halo of the Milky way galaxy, likely to consist of subatomic particles.

Dark energy:

Inferred to exist from the measured acceleration of the universe's expansion.

The proposed source of a universal repulsive force.

Dark matter and dark energy:

The universe contains more of this than it contains matter made from atoms.

We do not know what it is made of.

Neither:

Blocks light from stars behind it, 'dark' because it emits only infrared light.

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A 3.0 cm diameter water line splits into two 1.0 cm diameter pipes. All pipes are circular and at the same elevation.In the larger diameter section, the water speed is 2.0 and the gauge pressure is 50. What is the gauge pressure at a point in one of the narrower branches?

Answers

If all pipelines are round and of the same elevation as the question suggests, then point B's gauge pressure is 11.5 KPa.

The definition of elevation

Distance above water level is known as elevation. Elevations are often expressed in feet or meters. On charts, they can be represented by contour lines that link points of the same elevation, by colours, or by numbers that indicate the precise elevations of specific places on the Planet's surface.

Briefing:

From continuity equation:

Velocity of water at B = (2*pi*3²)/[2*pi*1²] =9 m/s

then from Bernoulli equation;

v² /2 + φ + p/ρ =constant

ρ=1000 kg/m³

3²/2 +φ +50000/1000 = 9²/2 +φ +P/1000

Therefore,

Pressure at B = 11.5 KPa

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The complete question is-

The 3.0 cm-diameter water line in the figure splits into two 1.0 cm-diameter pipes. All pipes are circular and at the same elevation. At point A, the water speed is 2.0 m/s and the gauge pressure is 50 kPa.

What is the gauge pressure at point B?

Marvin is moving furniture in his house , including the 52.8 kg dresser. he is applying a force of 90 N to the left to overcome 24 N of friction to accelerate the dresser.

1. What is the horizontal acceleration of the dresser?(value only, round to two decimal places, no units, or direction. example: 97.50)

Answers

Answer: 3.81 m/s^2.

Explanation: use the equation for acceleration, which is a = (F - f)/m. In this case, we are given the force applied by Marvin (F), the force of friction (f), and the mass of the dresser (m). Plugging in the given values, we get a = (90 N - 24 N)/(52.8 kg) = 3.81 m/s^2.

A straight, nonconducting plastic wire 9.50cm {\rm cm} long carries a charge density of 125nC/m {\rm nC/m} distributed uniformly along its length. It is lying on a horizontal tabletop.
A) Find the magnitude and direction of the electric field this wire produces at a point 5.50 {\rm cm} directly above its midpoint.
B) If the wire is now bent into a circle lying flat on the table, find the magnitude and direction of the electric field it produces at a point 5.50cm {\rm cm} directly above its center.

Answers

Part A. The midway magnitude of the electric field is 17.2 * 10³ N/C.

Part B. The electric field intensity is 47.17 * 10³ N/C N/C when the wire shape is a circle.

The intensity of an Electric Field

Given this, the wire length is 9.50 cm and the charge density is 125 nC/m.

Part A

The electric field at the wires halfway is

E = q/4πε * 1/ (z(z²/a²+1)¹/²)

Where E is the intensity of the electric field, 8.85*10⁻¹² is the permittivity, z is 5.5 cm, a is the midpoint, 9.5/2 cm, and q is the charge density.

Substituting the value in the above equation

E = 17.2 * 10³N/C

The midway magnitude of the electric field is 17.2 * 10³ N/C

Part B

When the wire is a circle then, the charge density at the wire is

Q = q*l

Q= 1.235 * 10⁻⁸ * 0.095

Q = 1.235*10⁻⁸ C

And to calculate the radius

r = l/2π

r = 9.50/ 2*3.14

r = 1.512cm

r = 0.015m

The electric field density at the midpoint of the circle is,

E = Q/4π * z(z²+r²)¹/² * E

Substituting the values in the above equation, we get

E = 47.17 * 10³ N/C

The electric field intensity is 47.17 * 10³ N/C N/C when the wire shape is a circle.

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Given that Shapley calculated the size of the Milky Way based on the stars' apparent brightness, how did not including the effects of gas and dust affect Shapley's calculation?
A. The distance calculated to each globular cluster was too large, and thus his size for the Milky Way was too small.
B. The distance calculated to each globular cluster was too large, and thus his size for the Milky Way was too large.
C. The distance calculated to each globular cluster was too small, and thus his size for the Milky Way was too large.
D. The distance calculated to each globular cluster was too small, and thus his size for the Milky Way was too small.Answer:B. The distance calculated to each globular cluster was too large, and thus his size for the Milky Way was too large.

Answers

The correct option is, The distance calculated to each globular cluster was too large, and thus his size for the Milky Way was too large.

What main conclusion did Shapley draw from his measurements of the distances to the globular clusters?
Shapley
concluded (and other astronomers have since verified) that the center of the distribution of globular clusters is the center of the Milky Way as well, so our galaxy looks like a flat disk of stars embedded in a spherical cloud, or 'halo,' of globular clusters.

How did Shapley estimate the location of the Sun in the Milky Way?
Harlow Shapley
determined the position of the Sun in the galaxy by measuring the distances to 93 globular clusters of stars.

What is a globular star cluster?
Globular
clusters are stable, tightly bound clusters of tens of thousands to millions of stars. They are associated with all types of galaxies. Globular clusters are typically much larger than open clusters and are tightly gravitationally bound.

What are the characteristics of the stars in globular clusters?
Globular clusters
are densely packed collections of ancient stars. Roughly spherical in shape, they contain hundreds of thousands, and sometimes millions, of stars. Studying them helps astronomers estimate the age of the universe or figure out where the center of a galaxy lies.

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