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grandymaker [24]

2 years ago

5

Isla’s change in velocity is 30 m/s, and Hazel has the same change in velocity. Which best explains why they would have differen

t accelerations?
A) Isla had negative acceleration, and Hazel had positive.
B) Isla had a different time than Hazel.
C) Isla had positive acceleration, and Hazel had negative.
D) Isla went a farther distance than Hazel.

1 answer:

Mashcka [7]2 years ago

5 0

Answer:

The answer is B) Isla had a different time than hazel

Explanation:

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A practical rule is that a radioactive nuclide is essentially gone after 10 half-lives. What percentage of the original radioact

ArbitrLikvidat [17]

Answer:

0.09 % of the original radioactive nucllde its left after 10 half-lives
It will take 241,100 years for 10 half-lives of plutonium-239 to pass.

Explanation:

The equation for radioactive decay its:

$N ( t) \ = \ N_0 \ e^{ \ - \frac{t}{\tau}}$ ,

where N(t) its quantity of material at time t, $N_0$ its the initial quantity of material and $\tau$ its the mean lifetime of the radioactive element.

The half-life $t_{\frac{1}{2}}$ its the time at which the quantity of material its the half of the initial value, so, we can find:

$N (t_{\frac{1}{2} }) \ = \ N_0 \ e^{ \ - \frac{t_{\frac{1}{2}}}{\tau}} \ = \frac{N_0}{2}$

so:

$\ N_0 \ e^{ \ - \frac{t_{\frac{1}{2}}}{\tau}} \ = \frac{N_0}{2}$

$e^{ \ - \frac{t_{\frac{1}{2}}}{\tau}} \ = \frac{1}{2}$

$- \frac{t_{\frac{1}{2}}}{\tau}} \ = - \ ln( 2 )$

$t_{\frac{1}{2}}\ = \tau ln( 2 )$

So, after 10 half-lives, we got:

$N ( 10 \ t_{\frac{1}{2}}) \ = \ N_0 \ e^{ \ - \frac{10 \ t_{\frac{1}{2}}}{\tau}}$

$N ( 10 \ t_{\frac{1}{2}}) \ = \ N_0 \ e^{ \ - \frac{10 \ \tau \ ln( 2 ) }{\tau}}$

$N ( 10 \ t_{\frac{1}{2}}) \ = \ N_0 \ e^{ \ - 10 \ \ ln( 2 ) }$

$N ( 10 \ t_{\frac{1}{2}}) \ = \ N_0 \ * \ 9.76 * 10^{-4}$

So, we got that a 0.09 % of the original radioactive nucllde its left.

Putonioum-239 has a half-life of 24,110 years. So, 10 half-life will take to pass

$10 \ * \ 24,110 \ years \ = \ 241,100 \ years$

It will take 241,100 years for 10 half-lives of plutonium-239 to pass.

7 0

2 years ago

Read 2 more answers

A 1.0 meter long rod is held horizontally in the east-west direction on a distant planet and is dropped from a height of 1.23 m.

enyata [817]

Answer:

The induced emf between two end is $34.02 \times 10^{-5}$ V

Explanation:

Given:

Length of rod $l = 1$ m

Height $h = 1.23$ m

Magnetic field $B = 6.93 \times 10^{-5}$ T

For finding induced emf,

$\epsilon = Blv$

Where $v =$ velocity of rod,

For finding the velocity of rod.

From kinematics equation,

$v^{2} = v_{o} ^{2} + 2gh$

Where $v_{o} =$ initial velocity, $g = 9.8 \frac{m}{s^{2} }$

$v = \sqrt{2gh}$

$v = \sqrt{2 \times 9.8 \times 1.23}$

$v = 4.91$ $\frac{m}{s}$

Put the velocity in above equation,

$\epsilon = 6.93 \times 10^{-5} \times 1 \times 4.91$

$\epsilon = 34.02 \times 10^{-5}$ V

Therefore, the induced emf between two end is $34.02 \times 10^{-5}$ V

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Explanation:

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