All categories

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.

You might be interested in

A suspicious-looking man runs as fast as he can along a moving sidewalk from one end to the other, taking 2.50 s. Then security

Zigmanuir [339]

13:9 because he starts back then runs slower

6 0

3 years ago

Read 2 more answers

A hockey puck initially travelling to the right at 34 m/s. It moves for 7 before

goldfiish [28.3K]

Answer:

$x=119m$

Explanation:

Hello,

In this case, since the hockey puck was moving at 34 m/s and suddenly stopped (final velocity is zero) in 7 seconds, we can first compute the acceleration via:

$a=\frac{v_f-v_o}{t}=\frac{0m/s-34m/s}{7s}\\ \\a=-4.86m/s^2$

In such a way, we can compute the displacement via:

$x=\frac{v_f^2-v_o^2}{2a}\\ \\x=\frac{0^2-(34m/s)^2}{2*-4.86m/s^2}\\ \\x=119m$

Best regards.

3 0

2 years ago

An object of mass 2kg raised to a height 10m possess potential energy of 200J. What is the kinetic energy and potential energy a

Shtirlitz [24]

Explanation:

${\bold{\sf{\underline{Understanding \: the \: concept}}}}$

✠ This question says that there is an object and its mass is 2 kg ; it's raised to a height 10 m and possess potential energy of 200 J. Now this question ask us to find the kinetic energy and the potential energy at a height 4 metre.

$\bold{↬{ }}$ ${\bold{\sf{\underline{Given \: that}}}}$

✰ Mass = 2 kilograms

✰ Raised height = 10 metres

✰ Posses potential energy = 200 Joules

$\bold{↬{ }}$ ${\bold{\sf{\underline{To \: find}}}}$

✰ Kinetic energy at a height 4 metre

✰ Potential energy at a height 4 metre

${\bold{\sf{\underline{Solution}}}}$

✰ Kinetic energy at a height 4 metre = 120 J

✰ Potential energy at a height 4 metre = 80 J

${\bold{\sf{\underline{Using \: concepts}}}}$

✰ Potential energy formula.

${\bold{\sf{\underline{Using \: formula}}}}$

✰ Potential energy = mgh

${\bold{\sf{\underline{We \: also \: write \: these \: as}}}}$

✰ Potential energy as P.E

✰ Mass as m

✰ Joules as J

✰ Height as h

✰ Raised height as g

${\bold{\sf{\underline{Full \: solution}}}}$

<h3>✠ Let us find the Potential energy.</h3>

↦ Potential energy = mgh

↦ Potential energy = 2 × 10 × 4

↦ Potential energy = 20 × 4

↦ Potential energy = 80 J

<h3>✠ Now according to the question let us find the kinetic energy</h3>

↦ Kinetic energy = Posses potential energy - Finded potential energy

↦ Kinetic energy = 200 J - 80 J

↦ Kinetic energy = 120 Joules

4 0

3 years ago

Read 2 more answers

What happens if you add additional,solid NaCl after the maximum has been reached?

azamat

<span>it would bond to the phosphate

</span>

3 0

3 years ago

Read 2 more answers

A rock dropped on the moon will increase it's speed from 0 m/s to 8.15 m/s in about 5 seconds what is the acceleration of the ro

Lunna [17]

Using the formula:

a = (Vf - Vi) / t

Our initial velocity is 0 m/s, and our final velocity is 8.15 m/s, with a time period of 5 seconds:

a = (8.15 - 0.0) / 5

a = 1.63 m/s^2

If you know the acceleration due to gravity on the Moon, you can confirm this answer. The recorded gravitational acceleration on the Moon is 1.62 m/s^2.

5 0

3 years ago