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Marta_Voda [28]
3 years ago
13

What type of evidence should be collected first ?

Physics
1 answer:
grigory [225]3 years ago
6 0

Answer:

The evidence that should be collected first is the evidence such as fingerprints, footprints, tire tracks, blood and other body fluids, hairs, fibers and fire debris. NIJ funds projects to improve: Identification of blood and other body fluids at the scene.

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A 0.600 m long pendulum is used to determine the acceleration due to gravity on a distant plane. If 20 oscillations are complete
katrin2010 [14]

Answer:

7.50 m/s^2

Explanation:

The period of a pendulum is given by:

T=2\pi \sqrt{\frac{L}{g}} (1)

where

L = 0.600 m is the length of the pendulum

g = ? is the acceleration due to gravity


In this problem, we can find the period T. In fact, the frequency is equal to the number of oscillations per second, so:

f=\frac{N}{t}=\frac{20}{35.5 s}=0.563 Hz

And the period is the reciprocal of the frequency:

T=\frac{1}{f}=\frac{1}{0.563 Hz}=1.776 s

And by using this into eq.(1), we can find the value of g:

g=\frac{4 \pi^2 L}{T^2}=\frac{4 \pi^2 (0.600 m)}{(1.776 s)^2}=7.50 m/s^2

6 0
3 years ago
A closely wound search coil has an area of 3.13 cm2, 135 turns, and a resistance of 61.1 Ω. It is connected to a charge-measurin
erastovalidia [21]

Answer:

Explanation:

Let the magnitude of magnetic field be B .

flux passing through the coil's  = area of coil x field x no of turns

Φ = 3.13 x 10⁻⁴ x B x 135 = 422.55 x 10⁻⁴ B .

emf induced = dΦ / dt , Φ is magnetic flux.

current i = dΦ /dt x 1/R

charge through the coil = ∫ i dt

= ∫   dΦ /dt x 1/R dt

= 1 / R ∫ dΦ

= Φ / R

Total resistance R = 61.1 + 44.4 = 105.5 ohm .

3.44 x 10⁻⁵ = 422.55 x 10⁻⁴ B / 105.5

B = 3.44 x 10⁻⁵ x 105.5  / 422.55 x 10⁻⁴

= .86 x 10⁻¹

= .086 T .

8 0
3 years ago
Force F acts between a pair of charges, q1 and q2, separated by a distance d. For each of the statements, use the drop-down menu
lora16 [44]

The initial force between the two charges is given by:

F=k \frac{q_1 q_2}{d^2}

where k is the Coulomb's constant, q1 and q2 the two charges, d their separation. Let's analyze now the other situations:

1. F

In this case, q1 is halved, q2 is doubled, but the distance between the charges remains d.

So, we have:

q_1' = \frac{q_1}{2}\\q_2' = 2 q_2\\d' = d

So, the new force is:

F'=k \frac{q_1' q_2'}{d'^2}= k \frac{(\frac{q_1}{2})(2q_2)}{d^2}=k \frac{q_1 q_2}{d^2}=F

So the force has not changed.

2. F/4

In this case, q1 and q2 are unchanged. The distance between the charges is doubled to 2d.

So, we have:

q_1' = q_1\\q_2' = q_2\\d' = 2d

So, the new force is:

F'=k \frac{q_1' q_2'}{d'^2}= k \frac{q_1 q_2)}{(2d)^2}=\frac{1}{4} k \frac{q_1 q_2}{d^2}=\frac{F}{4}

So the force has decreased by a factor 4.

3. 6F

In this case, q1 is doubled and q2 is tripled. The distance between the charges remains d.

So, we have:

q_1' = 2 q_1\\q_2' = 3 q_2\\d' = d

So, the new force is:

F'=k \frac{q_1' q_2'}{d'^2}= k \frac{(2 q_1)(3 q_2)}{d^2}=6 k \frac{q_1 q_2}{d^2}=6F

So the force has increased by a factor 6.

8 0
3 years ago
Read 2 more answers
11. Which of the following phenomena is taking place when sound waves are reflected from a surface along parallel lines? A. Refr
allsm [11]
The Answer is C. Focusing             
4 0
3 years ago
How does adding more of a substance affect it's density?
fomenos
I’m going to use molasses as an example of a substance.

The mass and volume both change when changing the amount of molasses.
However, the density does not change. This is because the mass and volume increase at the same rate/proportion!

Even though there is more molasses (mass) in test tube A, the molasses also takes up more space (volume). Therefore, the spacing between those tiny particles that make up the molasses is constant (does not change).

The size or amount of a material/substance does not affect its density.
5 0
3 years ago
Read 2 more answers
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