2. Only power tools with a 3-wire cord with

In 8,450 seconds, the number of radioactive nuclei decreases to 1/16 of the number present initially. What is the half-life (in

almond37 [142]

Answer:

<h2>2113 seconds</h2>

Explanation:

The general decay equation is given as $N = N_0e^{-\lambda t} \\\\$ , then;

$\dfrac{N}{N_0} = e^{-\lambda t} \\$ where;

$N/N_0$ is the fraction of the radioactive substance present = 1/16

$\lambda$ is the decay constant

t is the time taken for decay to occur = 8,450s

Before we can find the half life of the material, we need to get the decay constant first.

Substituting the given values into the formula above, we will have;

$\frac{1}{16} = e^{-\lambda(8450)} \\\\Taking\ ln\ of \both \ sides\\\\ln(\frac{1}{16} ) = ln(e^{-\lambda(8450)}) \\\\\\ln (\frac{1}{16} ) = -8450 \lambda\\\\\lambda = \frac{-2.7726}{-8450}\\ \\\lambda = 0.000328$

Half life f the material is expressed as $t_{1/2} = \frac{0.693}{\lambda}$

$t_{1/2} = \frac{0.693}{0.000328}$

$t_{1/2} = 2,112.8 secs$

Hence, the half life of the material is approximately 2113 seconds

7 0

3 years ago

A box weighing 52.4 N is sliding on a rough horizontal floor with a constant friction force of magnitude LaTeX: ff. The box's in

german

Answer:

The magnitude of the friction force exerted on the box is 2.614 newtons.

Explanation:

Since the box is sliding on a rough horizontal floor, then it is decelerated solely by friction force due to the contact of the box with floor. The free body diagram of the box is presented herein as attachment. The equation of equilbrium for the box is:

$\Sigma F = -f = m\cdot a$ (Eq. 1)

Where:

$f$ - Kinetic friction force, measured in newtons.

$m$ - Mass of the box, measured in kilograms.

$a$ - Acceleration experimented by the box, measured in meters per square second.

By applying definitions of weight ( $W = m\cdot g$ ) and uniform accelerated motion ( $v = v_{o}+a\cdot t$ ), we expand the previous expression:

$-f = \left(\frac{W}{g} \right)\cdot \left(\frac{v-v_{o}}{t}\right)$

And the magnitude of the friction force exerted on the box is calculated by this formula:

$f = -\left(\frac{W}{g} \right)\cdot \left(\frac{v-v_{o}}{t}\right)$ (Eq. 1b)

Where:

$W$ - Weight, measured in newtons.

$g$ - Gravitational acceleration, measured in meters per square second.

$v_{o}$ - Initial speed, measured in meters per second.

$v$ - Final speed, measured in meters per second.

$t$ - Time, measured in seconds.

If we know that $W = 52.4\,N$ , $g = 9.807\,\frac{m}{s^{2}}$ , $v_{o} = 1.37\,\frac{m}{s}$ , $v = 0\,\frac{m}{s}$ and $t = 2.8\,s$ , the magnitud of the kinetic friction force exerted on the box is:

$f = -\left(\frac{52.4\,N}{9.807\,\frac{m}{s^{2}} } \right)\cdot \left(\frac{0\,\frac{m}{s}-1.37\,\frac{m}{s} }{2.8\,s} \right)$

$f = 2.614\,N$

The magnitude of the friction force exerted on the box is 2.614 newtons.

5 0

3 years ago

The equation r (t )=(2t + 4)⋅i + (√ 7 )t⋅ j + 3t ²⋅k the position of a particle in space at time t. Find the angle between the v

velikii [3]

Answer:

$\theta = n\pi/2, {\rm where~n~is~an~integer.}$

Explanation:

We should first find the velocity and acceleration functions. The velocity function is the derivative of the position function with respect to time, and the acceleration function is the derivative of the velocity function with respect to time.

$\vec{v}(t) = \frac{d\vec{r}(t)}{dt} = (2)\^i + (\sqrt{7})\^j + (6t)\^k$