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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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c. $t=9.9 \ sec$

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:

The average velocity of the train can be defined as the total distance covered by the train divided by the time taken by the train to cover that distance. Therefore, we will use the following formula to find the average velocity of the train:

v = s/t

where,

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

t = time taken to cover the distance = 2 h 14 min

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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