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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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The predictions of Rutherford's scattering formula failed to correspond with experimental data when the energy of the incoming a

Vera_Pavlovna [14]

Answer:

The radius of the gold nucleus is 7.1x10⁻¹⁵m

Explanation:

The nearest distance is:

$r=\frac{kze^{2} }{mpV^{2} }$ (eq. 1)

Where

z = atomic number of gold = 79

e = electron charge = 1.6x10⁻¹⁹C

k = electrostatic constant = 9x10⁹Nm²C²

energy of the particle = 32 MeV = 5.12x10⁻¹²J

At the potential energy is zero, all the energy will be kinetic energy:

$E_{k} =\frac{1}{2} m V^{2}$

Where

m = 4 mp = mass of proton

$5.12x10^{-12} =\frac{1}{2} *4*m_{p}* V^{2} \\m_{p}* V^{2} = 2.56x10^{-12}$

Replacing in equation 1

$r=\frac{9x10^{9}*79*(1.6x10^{-19})^{2} }{2.56x10^{-12} } =7.1x10^{-15} m$

8 0

2 years ago

there is a fish called an archer fish that shoots drops of water at insects resting on branches above the water. If the Archer f

Mariulka [41]

Answer:

=3.5 m/s

Explanation:

y = x tanθ - 1/2 g x² / (u²cos²θ )

y = 0.25 , x = 0.5, θ = 40°

.25 = .50 tan40 - .5 x 9.8x x²/ u²cos²40

.25 = .42 - 2.0875/u²

u = 3.5 m / s.

4 0

3 years ago

A solid, horizontal cylinder of mass 18.0 kg and radius 1.70.0 m rotates with an angular speed of 40 rad/s about a fixed vertica

Radda [10]

Answer:39.88 rad/s

Explanation:

Given

mass of cylinder m_1=18 kg

radius R=1.7 m

angular speed $\omega =40rad/s$

mass of $m_2=0.8 kg$ dropped at r=0.3 m from center

let $\omega _2$ be the final angular velocity of cylinder

Conserving Angular momentum

$L_1=L_2$

$\left ( \frac{m_1R^2}{2}\right )\omega =\left ( \frac{m_1R^2}{2}+m_2r^2\right )\omega _2$

$\left ( \frac{18\cdot 1.7^2}{2}\right )\cdot 40=\left ( \frac{18\cdot 1.7^2}{2}+0.8\cdot 0.3^2\right )\omega _2$

$26.01\times 40=26.082\times \omega _2$

$\omega _2=39.88 rad/s$

3 0

3 years ago

John says that the value of the function cos[ω(t + T) + ϕ], obtained one period T after time t, is greater than cos(ωt + ϕ) by 2

Svetllana [295]

Answer:

No one is right

Explanation:

John Case:

The function $cos(\omega t +\phi)$ is defined between -1 and 1, So it is not possible obtain a value $2\pi$ greater.

In addition, if you move the function cosine a T Value, and T is the Period, the function take the same value due to the cosine is a periodic function.

Larry case:

Is you have $f=1+cos(\omega t +\phi)$ , the domain of this is [0,2].

it is equivalent to adding 1 to the domain of the $f=1+cos(\omega t +\phi)$ , and its mean that the function $f=cos(\omega t +\phi)$ , in general, is not greater than $cos(\omega t +\phi)$ .