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3 years ago

Describe the difference between balanced forces and action/reaction forces

Physics

2 answers:

Alecsey [184]3 years ago

5 0

Balanced forces act on the same object and Action-Reaction forces act on different objects.

NeX [460]3 years ago

3 0

Balanced forces act on the same object and Action-Reaction forces act on different objects.

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Which statement best explains some consequences of early sexual activity? Consequences to early sexual activity are not importan

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3 years ago

Determine the value of n so that the vectors A and B are perpendicular: Ā = î + 5j + nk and B = 2î - j + k

Tanya [424]

If $\vec A$ and $\vec B$ are perpendicular, then their dot product is zero. This means

$\vec A \cdot \vec B = (\vec\imath + 5\,\vec\jmath + n\,\vec k) \cdot (2\,\vec\imath - \vec\jmath + \vec k) = 2 - 5 + n = 0$

Solving for n is trivial; it follows that n = 3.

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2 years ago

When radioactive tracers are used for medical scans, the tracers both decay in the body via radiation, as well as being excreted

irinina [24]

Answer:

Effective half-time of the tracer is 3.6 days

Explanation:

The formula for calculating the decay due to excretion for the first process is ;

$\frac{dN_1}{dt } = - \lambda _e N_o$

here ;

$N_o$ = initial number of tracers

Then to the second process ; we have :

$\frac{dN_2}{dt } = - \lambda _e N_o$

The total decay is as a result of the overall process occurring ; we have :

$\frac{dN_{total}}{dt } = \frac{dN_1}{dt} + \frac{dN_2}{dt}$ ------ (1)

here ;

$\frac{dN_{total}}{dt } = \lambda _{total} N_o$

Putting the values in (1);we have :

$- \lambda _{Total} N_o = - \lambda_e N_o + ( -\lambda r N_o})$

$\lambda _{Total} = \lambda_e + \lambda r$

As we also know that:

$\frac{1}{t_{1/2}} = \frac{[t_{1/2}]_{radiation}+[t_{1/2}]_{excretion}}{[t_{1/2}]_{radiation}*[t_{1/2}]_{excretion}}$

$\frac{1}{t_{1/2}}_{effective}} = \frac{[t_{1/2}]_{radiation}+[t_{1/2}]_{excretion}}{[t_{1/2}]_{radiation}*[t_{1/2}]_{excretion}}$

$\frac{1}{t_{1/2}}_{effective}} = \frac{9+6}{9*6}$

$\frac{1}{t_{1/2}_{effective}}}=\frac{15}{54}$

$t_{1/2}_{effective}} = \frac{54}{15}$

= 3.6 days

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