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mr Goodwill [35]
3 years ago
6

In a nuclear fusion reaction two 2H atoms are combined to produce one 4He.

Physics
1 answer:
Mrrafil [7]3 years ago
5 0

Answer:

a)= 0.025602u

b) = 23.848MeV

c) N = 1.546 × 10¹³

Explanation:

The reaction is

²₁H   +   ²₁H   ⇄   ⁴₂H + Q

a) The mass difference is

Δm = 2m(²₁H) - m (⁴₂H)

       = 2(2.014102u) - 4.002602u

        = 0.025602u

b) Use the Einstein mass energy relation ship

The enegy  release is the mass difference times 931.5MeV/U

E = (0.025602) (931.5)

   = 23.848MeV

c)

the number of reaction need per seconds is

N = Q/E

     = 59W/ 23.848MeV

  = \frac{59}{(23.848 \times 10^6 )(1.6 \times 10^1^9) } \\\\= 1.546 \times 10^1^3

N = 1.546 × 10¹³

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An electron in the n = 5 level of an h atom emits a photon of wavelength 1282.17 nm. to what energy level does the electron move
lions [1.4K]
This is an interesting (read tricky!) variation of Rydberg Eqn calculation.
Rydberg Eqn: 1/λ = R [1/n1^2 - 1/n2^2]
Where λ is the wavelength of the light; 1282.17 nm = 1282.17×10^-9 m
R is the Rydberg constant: R = 1.09737×10^7 m-1
n2 = 5 (emission)
Hence 1/(1282.17 ×10^-9) = 1.09737× 10^7 [1/n1^2 – 1/25^2]
Some rearranging and collecting up terms:
1 = (1282.17 ×10^-9) (1.09737× 10^7)[1/n2 -1/25]
1= 14.07[1/n^2 – 1/25]
1 =14.07/n^2 – (14.07/25)
14.07n^2 = 1 + 0.5628
n = √(14.07/1.5628) = 3
8 0
3 years ago
Read 2 more answers
A planet exerts a gravitational force of magnitude 9e22 N on a star. If the planet were 2 times closer to the star (that is, if
Dmitrij [34]

To solve this problem we will use the related concepts in Newtonian laws that describe the force of gravitational attraction. We will use the given value and then we will obtain the proportion of the new force depending on the Radius. From there we will observe how much the force of attraction increases in the new distance.

Planet gravitational force

F_p = 6*10^{22}N

F_p = \frac{GMm}{R^2}

F_p = 9*10^{22}N

Distance between planet and star

r = \frac{R}{2}

Gravitational force is

F = \frac{GMm}{r^2}

Applying the new distance,

F = \frac{GMm}{(\frac{R}{2})^2}

F =  4\frac{GMm}{R^2}

Replacing with the previous force,

F = 4F_p

Replacing our values

F= 4(9*10^{22}N)

F = 36*10^{22}N

Therefore the magnitude of the force on the star due to the planet is  36*10^{22}N

5 0
3 years ago
Unpolarized light with an average intensity of 845 W/m2 enters a polarizer with a vertical transmission axis. The transmitted li
RideAnS [48]

The concept to develop this problem is the Law of Malus. Which describes what happens with the light intensity once it passes through a polarized material.

Mathematically this can be expressed as

I = I_0 cos^2\theta

Where

I = New intensity after pass through the Polarizer

I_0= Original intensity

\theta = Indicates the angle between the axis of the analyzer and the polarization axis of the incident light.

When the light passes perpendicularly through the first polarizer, the light intensity is reduced by half which will cause the intensity to be 225W / m ^ 2 at the output of the new polarizer, mathematically:

I= \frac{I_0}{2} cos^2\theta

225 = \frac{845}{2}cos^2\theta

Solving to find the angle we have

\theta = 43.11\°

The orientation angle of the second polarizer relative to the first one is 43.11°

5 0
3 years ago
Determine the amount of work done when a crane lifts a 100-N block form 2m above the ground to 6m above the ground
Dima020 [189]

Data given:

F=100N

Δx=6m-2m=4m

A=?

Formula needed:

A=F×s

Solution:

A=100N×4m

A=400J

4 0
3 years ago
A 1200 kg car passes traffic light at a velocity of 10.2 m/s to the north and accelerates at a rate of 2.45 m/s^2. Calculate the
kumpel [21]

The car’s momentum after 4.21s is 24617.4 kgm/s

<h3>Newton's Second Law of Motion.</h3>

Newton's second law state that, the rate of change of momentum, is directly proportional to the applied force.

Given that a 1200 kg car passes traffic light at a velocity of 10.2 m/s to the north and accelerates at a rate of 2.45 m/s^2. To calculate the car’s momentum after 4.21 s, Let us first list all the parameters involved.

  • Velocity u = 10.2 m/s
  • Acceleration a =  2.45 m/s²
  • Mass m = 1200Kg
  • Time t = 4.21 s

From Newton's second law,

F = (mv - mu) / t

ma = (mv - mu) / t

Substitute all the parameters into the formula above.

1200 × 2.45 = ( mv - 1200 × 10.2 ) / 4.21

2940 = ( mv - 12240 ) / 4.21

Cross multiply

12377.4 = mv - 12240

Make mv the subject of the formula

mv = 12377.4 + 12240

mv = 24617.4 kgm/s

Therefore, the car’s momentum after 4.21s is 24617.4 kgm/s

Learn more about Momentum here: brainly.com/question/25121535

#SPJ1

3 0
1 year ago
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