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-Dominant- [34]

3 years ago

10

The nuclear equation is incomplete. Superscript 239 Subscript 94 Baseline P u + Superscript 1 Subscript 0 Baseline n yields Supe

rscript 100 Subscript 40 Baseline Z r + blank + 2 Superscript 1 Subscript 0 Baseline n What particle completes the equation?

1 answer:

Crank3 years ago

3 0

Answer:

The particle which completes the given equation is : $_{54}^{138}\textrm{Xe}$

Explanation:

The given reaction is of a fission reaction:

$_{94}^{239}\textrm{Pu}+_0^1\textrm{n}\rightarrow _{40}^{100}\textrm{Zr}+_Z^A\textrm{X}+2_0^1\textrm{n}$

Total mass on the reactant side is equal to the total mass on the product side:

239 + 1 = 100 +A+ 2

A = 138

Sum of atomic numbers on the reactant side is equal to the sum of atomic number on the product side:

94 + 1(0) = 40 + Z + 2(0)

Z = 54

So atomic number 54 id of Xenon.

The particle which completes the given equation is :

$_{94}^{239}\textrm{Pu}+_0^1\textrm{n}\rightarrow _{40}^{100}\textrm{Zr}+_{54}^{138}\textrm{Xe}+2_0^1\textrm{n}$

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Careful measurements have been made of Olympic sprinters in the 100-meter dash. A quite realistic model is that the sprinter's v

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Answer:

a.

$\displaystyle a(0 )=8.133\ m/s^2$

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Explanation:

Modeling With Functions

Careful measurements have produced a model of one sprinter's velocity at a given t, and it's is given by

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For Carl Lewis's run at the 1987 World Championships, the values of a and b are

$\displaystyle a=11.81\ ,\ b=-0.6887$

Please note we changed the value of b to negative to make the model have sense. Thus, the equation for the velocity is

$\displaystyle V(t)=11.81(1-e^{-0.6887t})$

a. What was Lewis's acceleration at t = 0 s, 2.00 s, and 4.00 s?

To compute the accelerations, we must find the function for a as the derivative of v

$\displaystyle a(t)=\frac{dv}{dt}=11.81(0.6887\ e^{0.6887t})$

$\displaystyle a(t)=8.133547\ e^{-0.6887t}$

For t=0

$\displaystyle a(0)=8.133547\ e^o$

$\displaystyle a(0 )=8.133\ m/s^2$

For t=2

$\displaystyle a(2)=8.133547\ e^{-0.6887\times 2}$

$\displaystyle a(2)=2.05\ m/s^2$

$\displaystyle a(4)=8.133547\ e^{-0.6887\times 4}$

$\displaystyle a(4)=0.52\ m/s^2$

b. Find an expression for the distance traveled at time t.

The distance is the integral of the velocity, thus

$\displaystyle X(t)=\int v(t)dt \int 11.81(1-e^{-0.6887t})dt=11.81(t+\frac{e^{-0.6887t}}{0.6887})+C$

$\displaystyle X(t)=11.81(t+1.45201\ e^{-0.6887t})+C$

To find the value of C, we set X(0)=0, the sprinter starts from the origin of coordinates

$\displaystyle x(0)=0=>11.81\times1.45201+C=0$

Solving for C

$\displaystyle c=-17.1482\approx -17.15$

Now we complete the equation for the distance

$\displaystyle X(t)=11.81(t+1.45\ e^{-0.6887t})-17.15$

c. Find the time Lewis needed to sprint 100.0 m.

The equation for the distance cannot be solved by algebraic procedures, but we can use approximations until we find a close value.

We are required to find the time at which the distance is 100 m, thus

$\displaystyle X(t)=100=>11.81(t+1.45\ e^{-0.6887t})-17.15=100$

Rearranging

$\displaystyle t+1.45\ e^{-0.6887t}=9.92$

We define an auxiliary function f(t) to help us find the value of t.

$\displaystyle f(t)=t+1.45\ e^{-0.687t}-9.92$

Let's try for t=9 sec

$\displaystyle f(9)=9+1.45\ e^{-0.687\times 9}-9.92=-0.92$

Now with t=9.9 sec

$\displaystyle f(9.9)=9.9+1.45\ e^{-0.687\times 9.9}-9.92=-0.0184$

That was a real close guess. One more to be sure for t=10 sec

$\displaystyle f(10)=10+1.45\ e^{-0.687\times 10}-9.92=0.081$

The change of sign tells us we are close enough to the solution. We choose the time that produces a smaller magnitude for f(t).

$At t\approx 9.9\ sec, \text{ Lewis sprinted 100 m}$

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Answer:

v = 57.2 m/s

Explanation:

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v = s/t

where,

s = distance covered = 460 km = (460 km)(1000 m/1 km) = 4.6 x 10⁵ m

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Now, we convert it into minutes:

t = (2 h)(60 min/1 h) + 14 min

t = 120 min + 14 min = (134 min)(60 s/1 min)

t = 8040 s

Therefore, the value of velocity will be:

v = (4.6 x 10⁵ m)/8040 s

<u>v = 57.2 m/s</u>

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