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2 years ago

What are indicative acts? FORENSICS

Physics

1 answer:

puteri [66]2 years ago

6 0

Answer:

Actions or behaviors that may have taken place before death.

Do you wanna know how I know? No.... You do not wanna know how I know that you know what I known that you know I know

Explanation:

You Know?

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A thermometer is randomly selected and tested. Draw a sketch and find the temperature reading corresponding to Upper P 89, the

andrew11 [14]

Answer:

$X_{1}=1.23$

Explanation:

Given :

μ = 0

σ = 1

For 89th percentile

using invariant norm ( area, μ, σ )

= inv. norm ( 0.89, 0, 1 )

= 1.23

P ( x > $X_{1}$ ) = 0.89

$P\left ( \frac{X-N}{\sigma } >\frac{X_{1}-0}{1}\right )=0.89$

$P\left ( Z>Z_{1} )=0.89$

Now using Normal table, Z = 1.23

Therefore, $\frac{X_{1}-0}{1}=1.23$

$X_{1}=1.23$

8 0

3 years ago

Point charges q1=+2.00μC and q2=−2.00μC are placed at adjacent corners of a square for which the length of each side is 5.00 cm

mihalych1998 [28]

Point charges q1=+2.00μC and q2=−2.00μC are placed at adjacent corners of a square for which the length of each side is 5.00 cm.?

Point a is at the center of the square, and point b is at the empty corner closest to q2. Take the electric potential to be zero at a distance far from both charges.

(a) What is the electric potential at point a due to q1 and q2?

(b) What is the electric potential at point b?

(c) A point charge q3 = -6.00 μC moves from point a to point b. How much work is done on q3 by the electric forces exerted by q1 and q2?

Answer:

a) the potential is zero at the center .

Explanation:

a) since the two equal-magnitude and oppositely charged particles are equidistant

b)(b) Electric potential at point b, v = Σ kQ/r

r = 5cm = 0.05m

k = 8.99*10^9 N·m²/C²

Q = -2 microcoulomb

v= (8.99*10^9) * (2*10^-6) * (1/√2m - 1) / 0.0500m

v = -105 324 V

c)workdone = charge * potential

work = -6.00µC * -105324V

work = 0.632 J

6 0

2 years ago

HELP What is the goal of a theory?

Dafna1 [17]

Hi,

The goal of a theory is to explain natural and physical phenomena.

Hope it helps you...

Pls mark brainliest if it helps you...

(Answered by Benjemin)

4 0

2 years ago

What is the mass of metal if it has a density of 12.459 hg/cm^3 and displaces 28.7cm^3 of water

natima [27]

Answer:

357.6g

Explanation:

Given parameters:

Density = 12.459g/cm³

Volume of metal = 28.7cm³

Unknown:

Mass of metal = ?

Solution:

The density of a substance is its mass per unit volume.

To find the mass;

Mass of metal = density x volume

Now insert the parameters and solve;

Mass of metal = 12.459 x 28.7 = 357.6g

7 0

2 years ago

A farmer lifts his hay bales into the top loft of his barn by walking his horse forward with a constant velocity of 1 ft/s. Dete

Lesechka [4]

Answer:

The velocity of the hay bale is - 0.5 ft/s and the acceleration is $6.25\times 10^{- 3} ft/s^{2}$

Solution:

As per the question:

Constant velocity of the horse in the horizontal, $v_{x} = 1 ft/s$

Distance of the horse on the horizontal axis, x = 10 ft

Vertical distance, y = 20 ft

Now,

Apply Pythagoras theorem to find the length:

$20^{2} + 10^{2} = l^{2}$

$l^{2}= 500$

Now,

$x^{2} + y^{2} = 500$ (1)

Differentiating equation (1) w.r.t 't':

$2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$

$x\frac{dx}{dt} = - y\frac{dy}{dt}$

where

$\frac{dx}{dt}$ = Rate of change of displacement along the horizontal

$\frac{dy}{dt}$ = Rate of change of displacement along the vertical

$v_{x}$ = velocity along the x-axis.

$v_{y}$ = velocity along the y-axis

$xv_{x} = -yv_{y}$

$v_{y} = - 10\times \frac{1}{20} = - 0.5 ft/s$

$|v_{y}| = 0.5\ ft/s$

Acceleration of the hay bale is given by the kinematic equation:

$v_{y}^{2} = u_{y} + 2ay$

$(-0.5)^{2} =0 + 2ay$

$0.25 = 2ay$

$\frac{0.25}{2y} = a$

$a = \frac{0.25}{2\times 20} = 6.25\times 10^{- 3} ft/s^{2}$

7 0

3 years ago