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nirvana33 [79]
3 years ago
15

I need help with 1-4

Physics
1 answer:
mylen [45]3 years ago
3 0
Do the adding up on the chart
You might be interested in
Find the energy released in the fission reaction¹₀n + ²³⁵₉₂U → ⁹⁸₄₀Zr + ¹³⁵₅₂Te +3(¹₀n) The atomic masses of the fission product
ikadub [295]

Fission reaction is given

¹₀n + ²³⁵₉₂U → ⁹⁸₄₀Zr + ¹³⁵₅₂Te +3(¹₀n)

The atomic masses of the fission products are 97.912735 u for ⁹⁸₄₀Zr and 134.916450 u for ¹³⁵₅₂Te. Hence, the energy released in fission reaction is 191.715 MeV.

How to find the energy released in the given fission reaction?

We know, that the atomic mass of the elements are as follows:

  • ⁹⁸₄₀Zr - 97.9120u
  • ¹³⁵₅₂Te - 134.9087u
  • ²³⁵₉₂U  - 235.0483923u
  • n = 1.008665u

In order to find the mass difference, we will calculate the initial mass and the mass of products.

The equation for the reaction:

n + ²³⁵₉₂U → ⁹⁸₄₀Zr + ¹³⁵₅₂Te +3n

Mass of initial reagents is 1.008665u + 235.0483923u = 233.052588u

Product of the reagents is 97.9120u + 134.9087u + 3(1.008665u) = 235.846773u

Now, using the formula of

E=(\triangledown m)c^2

E=(0.205815u)\frac{931.494MeV/c^2}{u} c^2=191.715MeV

Hence, the energy released in fission reaction is 191.715 MeV.

To learn more about fission reactor, refer to:

brainly.com/question/23276812

#SPJ4

4 0
2 years ago
A resistor resists the flow of electricity and usually converts electrical energy to heat energy. True or False?
Sidana [21]
Your answer is: True!
7 0
3 years ago
Read 2 more answers
Imagine tying a string to ball and twirling it around you.how is this similar to the moon orbiting earth?in this example what is
guapka [62]
The changing force is you swinging the STRING
————-

this is similar because it’s on a set path and you are the earth and the gravitational pull is from the earth which the string represents
5 0
4 years ago
Why does air pressure decrease with increasing altitude? (Select all that apply.)
Harrizon [31]

Can't help but add me friend i will give you 25 points!

6 0
3 years ago
Please help 50 points
PilotLPTM [1.2K]

Answer:

3a 0.7m/s

3b partially inelastic

4a 8.33 m/s

4b completely inelastic

5. puck: 10.59 m/s octopus: 10.59m/s

6. car: 117.44 m/s truck: 17.44m/s

7a - 1 m/s , the red cart travels to the left

7b) elastic

Explanation:

For all these questions, you have find momentum (P=mv) always remember initial P is always equal to final P

3.

Initial P:

mass of first ball x velocity of first ball + mass of second ball x velocity of second ball

0.5*3.5 + 0.5*0 = 1.75 kg.m/s

final P: also 1.75Kg.m/s

let x be the velocity for first ball

0.5*2.8+0.5*x=1.75

0.5x=1.75-1.4

     x=0.35/0.5

      x=0.7m/s

b) collision is elastic when KEi = KEf, so calculate and see if that's true

using 1/2mv² we know KEi does not equal to KEf

so the collision is partially inelastic, partially because the ball did not stick together

4.

Initial P:

2575*11 + 825*0 = 28,325 kg.m/s

final P: also 28,325 Kg.m/s

let x be the velocity for both vehicles

(2575+825)x=28325

                   x=28325/3400

                    x=8.33 m/s

b) collision is elastic when KEi = KEf, so calculate and see if that's true

using 1/2mv² we know KEi does not equal to KEf

so the collision is completely inelastic, completely because the vechicles slides off together

5

Initial P:

0.115*35 + 0.265*0 = 4.025 kg.m/s

final P: also 4.025Kg.m/s

since they slides off together, the velocity will be the same, so let x be the velocity for the puck and octopus,

(0.115+0.265)x=4.025

                     x=4.025/0.38

                     x=10.59

6. Initial P:

565*25+785*12 = 23,545 kg.m/s

final P: also 23,545 Kg.m/s

since they slides off together, the velocity will be the same, so let x be the velocity for the car and truck,

(565+785)x=23545

                 x=23545/1350

                 x=17.44

7.

Initial P:

0.25*2 + 0.75*0 = 0.5 kg.m/s

final P: also 0.5 Kg.m/s

let x be the velocity for red cart

0.75*1+0.25*x=0.5

0.25x=0.5-0.75

     x=-0.25/0.25

      x=-1m/s

b) collision is elastic when KEi = KEf, so calculate and see if that's true

using 1/2mv² we know KEi equals to KEf

so the collision is elastic

7 0
3 years ago
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