Which of these is the best description of the Frye Standard?

2 answers:

6 0

Number 1 is D) Deciding if the procedure, principles or techniques in question in a forensic case are generally accepted by the relevant scientific community.

trasher [3.6K]3 years ago

3 0

Answer:

1. Option D) Deciding if the procedure, principles or techniques in question in a forensic case are generally accepted by the relevant scientific community.

2. Option A) thin-layer chromatography

3. Option A) when the jury considers the verdict

Explanation:

1. The Frye Standard is a procedure to determine if the piece of evidence is admissible in a court of law. This is a rigorous test to test how well the evidence can be used against a criminal or prove a case in the court of law. This came after a ruling that an experts advice only was insufficient to prove a case.

2. Thin layer chromatography is used to separate the ink components. This is because the ink is made of different components that have different volatlitities and solublies in the alcohol solution. The inks can be compared by looking at the partitioning coefficient and comparing the two distances traveled by the components.

3. The guilty cannot plead a case when the verdict has been given. At that moment, the case is closed.

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Two runners start at the same point on a straight track. The first runs with constant acceleration so that he covers 98 yards in

charle [14.2K]

Answer:

94.13 ft/s

Explanation:

Given:

$t$ = time interval in which the rock hits the opponent = 10 s - 5 s = 5 s

$s$ = distance to be moved by the rock long the horizontal = 98 yards

$y$ = displacement to be moved by the rock during the time of flight along the vertical = 0 yard

Assume:

$u$ = magnitude of initial velocity of the rock

$\theta$ = angle of the initial velocity with the horizontal.

For the motion of the rock along the vertical during the time of flight, the rock has a constant acceleration in the vertically downward direction.

$\therefore y = u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow 0 = u\sin \theta 5 +\dfrac{1}{2}(-9.8)\times 5^2\\\Rightarrow u\sin \theta 5 =\dfrac{1}{2}(9.8)\times 5^2......(1)\\$

Now the rock has zero acceleration along the horizontal. This means it has a constant velocity along the horizontal during the time of flight.

$\therefore u\cos \theta t = s\\\Rightarrow u\cos \theta 5 = 98.....(2)\\$

On dividing equation (1) by (2), we have

$\tan \theta = \dfrac{25}{20}\\\Rightarrow \tan \theta = 1.25\\\Rightarrow \theta = \tan^{-1}1.25\\\Rightarrow \theta = 51.34^\circ$

Now, putting this value in equation (2), we have

$u\cos 51.34^\circ\times 5 = 98\\\Rightarrow u = \dfrac{98}{5\cos 51.34^\circ}\\\Rightarrow u =31.38\ yard/s\\\Rightarrow u =31.38\times 3\ ft/s\\\Rightarrow u =94.13\ ft/s$