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3 years ago

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Radioactive plutonium−239 (t1/2 = 2.44 × 105 yr) is used in nuclear reactors and atomic bombs. If there are 3.50 × 102 g of the

isotope in a small atomic bomb, how long will it take for the substance to decay to 1.00 × 102 g, too small an amount for an effective bomb? (Hint: Radioactive decays follow first-order kinetics.) .0000441 yr

1 answer:

madam [21]3 years ago

7 0

Answer:

t = 4.41 10⁻⁴ years

Explanation:

For this exercise we must use the concept of average life time, which is the time in which the quantity and substance decays in half

$T_{1/2}$ = ln2 / λ

Let's calculate the decay constant of plutonium

λ = ln2 / $T_{1/2}$

λ = ln 2 / 2.44 10⁵

λ = 2.84 10⁻⁶ s⁻¹

Radioactive decay is a first order process

N = No e (-λ t)

Where N is the number of nuclei, the mass is this by molecular weight

m = N PM

m / PM = m₀ / PM e (- λ t)

m / m₀ = e (- λ t)

-λ t = ln (m / m₀)

t = -1 /λ ln (m/m₀)

t = - 1 / 2.84 10⁻⁶ ln (0.1 / 0.35)

t = 4.41 10⁻⁴ years

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A bicyclist starts at rest and speeds up to 30 m/s while accelerating at 4 m/s^2. Determine the distance traveled.

raketka [301]

Answer:

Distance, d = 112.5 meters

Explanation:

Initially, the bicyclist is at rest, u = 0

Final speed of the bicyclist, v = 30 m/s

Acceleration of the bicycle, $a=4\ m/s^2$

Let s is the distance travelled by the bicyclist. The third equation of motion is given as :

$v^2-u^2=2as$

$s=\dfrac{v^2-u^2}{2a}$

$s=\dfrac{(30)^2}{2\times 4}$

s = 112.5 meters

So, the distance travelled by the bicyclist is 112.5 meters. Hence, this is the required solution.

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3 years ago

Six artificial satellites circle a space station at constant speed. The mass m of each satellite, distance L from the space stat

nikklg [1K]

Answers:

a) $T_{2}>T_{5}>T_{1}>T_{3}=T_{6}>T_{4}$

b) $a_{4}>a_{6}>a_{1}>a_{3}>a_{5}>a_{2}$

Explanation:

a) Since we are told the satellites circle the space station at constant speed, we can assume they follow a uniform circular motion and their tangential speeds $V$ are given by:

$V=\omega L=\frac{2\pi}{T} L$ (1)

Where:

$\omega$ is the angular frequency

$L$ is the radius of the orbit of each satellite

$T$ is the period of the orbit of each satellite

Isolating $T$ :

$T=\frac{2 \pi L}{V}$ (2)

Applying this equation to each satellite:

$T_{1}=\frac{2 \pi L}{V_{1}}=261.79 s$ (3)

$T_{2}=\frac{2 \pi L}{V_{2}}=1570.79 s$ (4)

$T_{3}=\frac{2 \pi L}{V_{3}}=196.349 s$ (5)

$T_{4}=\frac{2 \pi L}{V_{4}}=98.174 s$ (6)

$T_{5}=\frac{2 \pi L}{V_{5}}=785.398 s$ (7)

$T_{6}=\frac{2 \pi L}{V_{6}}=196.349 s$ (8)

Ordering this periods from largest to smallest:

$T_{2}>T_{5}>T_{1}>T_{3}=T_{6}>T_{4}$

b) Acceleration $a$ is defined as the variation of velocity in time:

$a=\frac{V}{T}$ (9)

Applying this equation to each satellite:

$a_{1}=\frac{V_{1}}{T_{1}}=0.458 m/s^{2}$ (10)

$a_{2}=\frac{V_{2}}{T_{2}}=0.0254 m/s^{2}$ (11)

$a_{3}=\frac{V_{3}}{T_{3}}=0.4074 m/s^{2}$ (12)

$a_{4}=\frac{V_{4}}{T_{4}}=1.629 m/s^{2}$ (13)

$a_{5}=\frac{V_{5}}{T_{5}}=0.101 m/s^{2}$ (14)

$a_{6}=\frac{V_{6}}{T_{6}}=0.814 m/s^{2}$ (15)

Ordering this acceerations from largest to smallest:

$a_{4}>a_{6}>a_{1}>a_{3}>a_{5}>a_{2}$

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3 years ago

At what displacement of a sho is the energy half kinetic and half potential? what fraction of the total energy of a sho is kinet

expeople1 [14]

As we know that KE and PE is same at a given position

so we will have as a function of position given as

$KE = \frac{1}{2}m\omega^2(A^2 - x^2)$

also the PE is given as function of position as

$PE = \frac{1}{2}m\omega^2x^2$

now it is given that

KE = PE

now we will have

$\frac{1}{2}m\omega^2(A^2 - x^2) = \frac{1}{2}m\omega^2x^2$

$A^2 - x^2 = x^2$

$2x^2 = A^2$

$x = \frac{A}{\sqrt2}$

so the position is 0.707 times of amplitude when KE and PE will be same

Part b)

KE of SHO at x = A/3

we can use the formula

$KE = \frac{1}{2}m\omega^2(A^2 - x^2)$

now to find the fraction of kinetic energy

$f = \frac{KE}{TE} = \frac{A^2 - x^2}{A^2}$

$f = \frac{A^2 - (\frac{A}{3})^2}{A^2}$

$f_k = \frac{8}{9}$

now since total energy is sum of KE and PE

so fraction of PE at the same position will be

$f_{PE} = 1 - f_k$

$f_{PE} = 1 - (8/9) = 1/9$

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3 years ago

What is the acceleration of a car going 50 mph that slows down to rest over 10 seconds?

VladimirAG [237]

Answer:

If the velocity is constant, then there is no acceleration. That is, the value of the acceleration is 0.

Explanation:

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3 years ago

A pail in a water well is hoisted by means of a frictionless winch, which consists of a spool and a hand crank. When Jill turns

Katena32 [7]

Answer:

166 W

Explanation:

Power is the rate at which work is done.

$\text{Power} = \dfrac{\text{Work done}}{\text{time}}$

The work done by Jill is the product of the weight of the pail and the height it moves.

The weight is the product of the mass and acceleration of gravity, <em>g</em>. Taking <em>g</em> as 9.81 m/s², the weight is

<em>W</em> = (6.90 kg)(9.81 m/s²) = 67.689 N

Work done = (67.689 N)(27.0 m) = 1827.603 J

Power = (1827.603 J) ÷ (11.0 s) = 166 W

4 0

3 years ago