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maw [93]
3 years ago
7

After the fission of U-235 takes place, how many neutrons does the missing nucleus have?

Physics
2 answers:
motikmotik3 years ago
4 0

Answer:

B) 85

Explanation:

In the nuclear fission of uranium-235, a nucleus of uranium-235 absorbs a slow-moving neutron and then decays into a nucleus of Kripton-92, a nucleus of Barium-141 (which is the missing nucleus in the figure) and three more neutrons.

Knowing that:

- Uranium has atomic number 92, so it has 92 protons

- Kripton has atomic number 36, so it has 36 protons

- Barium has atomic number 56, so it has 56 protons

We can calculate the number of neutrons in the nucleus of barium, since we know that the total number of neutrons must be conserved. We have:

- On the left side of the reaction: (235-92) neutrons of the uranium + 1 neutron absorbed = 144 neutrons

- On the right side of the reaction: (92-36) neutrons of the krypton + X neutrons of the barium + 3 neutrons = 144

So we get:

56 + X + 3 = 144

X = 85

So, the nucleus of barium has 85 neutrons.


Ivahew [28]3 years ago
3 0
I just finished this question on my test and got it right. 85 is your answer.
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A power supply has an open-circuit voltage of 40.0 V and an internal resistance of 2.00 V. It is used to charge two storage batt
Natali [406]

Complete Question

A power supply has an open-circuit voltage of 40.0 V and an internal resistance of 2.00 \Omega. It is used to charge two storage batteries connected in series, each having an emf of 6.00 V and internal resistance of 0.300\Omega . If the charging current is to be 4.00 A, (a) what additional resistance should be added in series? At what rate does the internal energy increase in (b) the supply, (c) in the batteries, and (d) in the added series resistance? (e) At what rate does the chemical energy increase in the batteries?

Answer:

a

The additional resistance is R_z =  4.4 \Omega

b

The rate at which internal energy increase at the supply is Z_1 = 32 W

c

The rate at which internal energy increase in the battery  is  Z_1 = 32 W

d

The rate at which internal energy increase in the added series resistance is  Z_3 = 70.4 W

e

the increase rate of the chemically energy in the battery is C =  48 W

Explanation:

From the question we are told that

    The  open circuit voltage is  V =  40.0V

     The internal resistance is R = 2 \Omega

     The emf of each battery is e =  6.00 V

      The internal resistance of the battery is  r = 0.300V

      The  charging current is  I = 4.00 \ A

Let assume the the additional resistance to to added to the circuit is  R_z

 So this implies that

        The total resistance in the circuit is

                              R_T =  R + 2r +R_z

Substituting values

                             R_T = 2.6 +R_z

And  the difference in potential in the circuit is  

                         E = V -2e

                 =>   E =  40 - (2 * 6)

                        E =  28 V

Now according to ohm's law

            I = \frac{E}{R_T}

Substituting values

           4 = \frac{28}{R_z + 2.6}        

Making R_z the subject of the formula

So    R_z =  \frac{28 - 10.4}{4}

           R_z =  4.4 \Omega

The  increase rate of   internal energy at the supply is mathematically represented as

        Z_1  = I^2 R

Substituting values

     Z_1  = 4^2 * 2

     Z_1 = 32 W

The  increase rate of   internal energy at the batteries  is mathematically represented as

         Z_2 = I^2 r

Substituting values

         Z_2 = 4^2 * 2 * 0.3

         Z_2 = 9.6 \ W

The  increase rate of  internal energy at the added  series resistance  is mathematically represented as

        Z_3 = I^2 R_z

Substituting values

       Z_3 = 4^2 * 4.4

      Z_3 = 70.4 W

Generally the increase rate of the chemically energy in the battery is  mathematically represented as

         C = 2 * e * I

Substituting values

       C =  2 * 6  * 4

      C =  48 W

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Answer:

18 m

Explanation:

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Solving for the average velocity. avg v

avg v = (vo + v) / 2

avg v = (0 m / s + 12 m/s) / 2

avg v = 6 m / s

Solving for the distance traveled after 3 s

d = avg v * t

d = 6 m / s * 3 s

d = 18 meters

In the first 3s the car travels 18 meters.

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A 23.3-kg mass is attached to one end of a horizontal spring, with the other end of the spring fixed to a wall. The mass is pull
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In spring mass system we know that angular frequency is given as

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f = 8.38 Hz

\omega = 2\pi(8.38)

\omega = 52.65  rad/s

now we know that speed of SHM at its extreme position is given by

v = A\omega

here we know that

A = 17.5 cm

v = 0.175 (52.65)

v = 9.21 m/s

so maximum speed is 9.21 m/s

7 0
3 years ago
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