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2 years ago

Which of the following is NOT a level of home monitoring?

Physics

1 answer:

Bingel [31]2 years ago

5 0

<h2>Answer:</h2>

d. electronic monitoring

<h2>Explanation:</h2>

Home monitoring in its simplest term is the act of restricting the movement of an entity, object or an individual to a specified location, usually their homes.

In criminology, there are various levels of home monitoring. Some of them are;

i. <em>Curfew</em>: This type of home monitoring requires that the persons being monitored be indoors/at home at specified hours.

ii. <em>Home incarceration</em>: This level of home monitoring requires that the persons being monitored be indoors/at home at all times except in times of medical urgencies.

iii. <em>Home detention</em>: This seems to be one of the less strict forms of home monitoring. It requires that the persons being monitored be indoors/at home at all times except when necessary. For example, they could go to school, their workplace, to the market and so on.

Electronic monitoring is not a form of home monitoring as it allows the person being monitored to move freely. They might just contacted via telephone or any other means, at intervals to know their whereabouts.

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$V_{ft}= 317 cm/s$

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Where:

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$P_f = M_cV_{fc} + M_tV_{ft}$

Now:

$M_cV_{ic} + M_tV_{it} = M_cV_{fc} + M_tV_{ft}$

Where $M_c$ is the mass of the car, $V_{ic}$ is the initial velocity of the car, $M_t$ is the mass of train, $V_{fc}$ is the final velocity of the car and $V_{ft}$ is the final velocity of the train.

Replacing data:

$(1.1 kg)(4.95 m/s) + (3.55 kg)(2.2 m/s) = (1.1 kg)(1.8 m/s) + (3.55 kg)V_{ft}$

Solving for $V_{ft}$ :

$V_{ft}= 3.17 m/s$

Changed to cm/s, we get:

$V_{ft}= 3.17*100 = 317 cm/s$

b) The kinetic energy K is calculated as:

K = $\frac{1}{2}MV^2$

where M is the mass and V is the velocity.

So, the initial K is:

$K_i = \frac{1}{2}M_cV_{ic}^2+\frac{1}{2}M_tV_{it}^2$

$K_i = \frac{1}{2}(1.1)(4.95)^2+\frac{1}{2}(3.55)(2.2)^2$

$K_i = 22.06 J$

And the final K is:

$K_f = \frac{1}{2}M_cV_{fc}^2+\frac{1}{2}M_tV_{ft}^2$

$K_f = \frac{1}{2}(1.1)(1.8)^2+\frac{1}{2}(3.55)(3.17)^2$

$K_f = 19.61 J$

Finally, the change in the total kinetic energy is:

ΔK = Kf - Ki = 22.06 - 19.61 = 2.45 J

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