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scoundrel [369]
3 years ago
11

A proton (1H) and a deuteron (2H, "heavy hydrogen) start out far apart. An experimental apparatus shoots them toward each other

(with equal and opposite momenta). If they get close enough to make actual contact with each other, they can react to form a Helium-3 nucleus and a gamma ray (a high-energy photon, which has kinetic energy but zero rest energy):1H+2H \rightarrow 3He + \gammaThis is one of the thermonuclear or fusion reactions that takes place inside a star such as our sun.The mass of the proton is 1.0073 u (unified atomic mass unit, 1.7x10-27kg), the mass of the deunteron is 2.0136 u, the mass of the helium-3 nucleus is 3.0155 u, and the gamma ray is massless. Although in most problems you solve in this course it is adequate to use values of constants rounded to two or three significatn figures, in this problem you must keep at least six significant figures throughout your calculation. Problems involving mass changes require many significant figures becasue the changes in mass are small compared to the total mass. (a) The strong interaction has a very short range and is essentially a contact interaction. For this fusion reaction to take place, the proton and deuteron have to come close enough together to touch. The approximate radius of a proton or neutro is about 1x10-15m. What is the approximate initial total kinetic energy of the proton and deuteron required for the fusion reaction to proceed, in juoles and electron volts (1 e V = 1.6x10-19J)? (b) Given the initial conditions found in part (a), what is the kinetic energy of the 3He plus the energy of the gamma ray, in juoles and in electron volts? (c) The net energy released is the kinetic energy of the 3He plus the energy of the gamma ray found in part (b), minus the energy input that you calculated in part (a). What is the net energy release, in juoles and electron volts? Note that you do get back the energy investment made in part (a). (d) Kinetic energy can be used to drive motors and do other useful things. If a mole of hydrogen and a mole of deuterium underwent this fusion reaction, how much kinetic energy would be generated? (For comparison, around 1x106 J are obtained from burning a mole of gasoline). (e) Which of the following potential energy curves (1-4) in FIgure 6.87 is a reasonable representation of the interaction in this fusion reaction? Why?

Physics
1 answer:
TEA [102]3 years ago
4 0

Answer:

Explanation:

solution is attached below

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An electron moves through a uniform electric field vector E = (2.80î + 5.20ĵ) V/m and a uniform magnetic field vector B = 0.400k
alina1380 [7]

Answer:

1.758820×10^11(-2.5i-0.8j) m/s^2

Explanation:

From the question, the parameters given are; E=(2.80i+ 5.20j) v/m, a uniform magnetic field,B= 0.400K T, acceleration, a= ??? and velocity vector, v= 11.0i metre per seconds (m/s)...

We can solve this problem using the formula below;

Ma= q[E+V × B] ---------------(1).

Note: q is negative, m= mass of electron.

Making acceleration,a the subject of the formula and substituting the parameters into equation (1);

a= -e/m × (2.5i + 5.2j +11.0i × 0.400K)

a= -e/m × (2.5i+5.2j-4.4j)

a= e/m × (-2.5i - 0.8j)

e/m= 1.758820×10^11 c/kg

Therefore, slotting in the value of charge to mass(e/m) ratio;

a= 1.7588×10^11×(-2.5i-0.8j) m/s^2

7 0
3 years ago
Consider a balloon of mass 0.030kg being inflated with a gas of density 0.54kg/m. What will be the volume of the balloon when it
Nonamiya [84]

the weight of the balloon is .030 * 10 = 0.3 N

the weight of the gas of volume v is 0.54*10 N

The lifting force of a volume of v m³ of displaced air is 1.29v N

so, we need

1.29*10*v = 0.3 + 0.54*10*v

or

1.29v = 0.03+0.54v

7 0
1 year ago
On a highway, a car is driven 80. kilometers
nydimaria [60]

The average speed of the car for the entire trip can be calculate by using:

v=\frac{S}{t}

where S is the total distance covered by the car, and t is the total time taken.


The total distance travelled by the car is:

S=80 km+50 km+40 km=170 km

while the total time taken is:

t=1.00 h+0.50 h+0.50 h=2.00 h


so, the average speed of the car is:

v=\frac{S}{t}=\frac{170 km}{2.00 h}=85 km/h


so, the correct answer is (3) 85 km/h.

7 0
3 years ago
Read 2 more answers
For these pictures is more or less friction needed?
antiseptic1488 [7]

Answer:

8: More

9: More

10: More

11: Less

12: Less

12: More

4 0
2 years ago
In a simple electric circuit, ohm's law states that v=irv=ir, where vv is the voltage in volts, ii is the current in amperes, an
Tju [1.3M]
We take the derivative of Ohm's law with respect to time: V = IR
Using the product rule:
dV/dt = I(dR/dt) + R(dI/dt)
We are given that voltage is decreasing at 0.03 V/s, resistance is increasing at 0.04 ohm/s, resistance itself is 200 ohms, and current is 0.04 A. Substituting:
-0.03 V/s = (0.04 A)(0.04 ohm/s) + (200 ohms)(dI/dt)
dI/dt = -0.000158 = -1.58 x 10^-4 A/s
8 0
3 years ago
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