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3 years ago

"Forensic entomologists study what to help solve crimes?"

Physics

2 answers:

Licemer1 [7]3 years ago

7 0

I believe the answer is flies since they study bugs, please get another opinion to be sure.

Inessa [10]3 years ago

7 0

Answer: Flies

Explanation:

Forensic entomology is the sub discipline of forensic science which use the insects and their life cycle to relate them with the decaying dead body. Insects found near the unknown dead body can be useful in establishing place of death, time since death, manner of death (poison or drug/alcohol toxicity).

On the basis of the above description, flies is the correct option because they are insects.

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An unstable particle at rest breaks up into two fragments of unequal mass. The mass of the lighter fragment is equal to 2.90 ✕ 1

motikmotik

Answer:

The speed of the heavier fragment is 0.335c.

Explanation:

Given that,

Mass of the lighter fragment $M_{l}=2.90\times10^{-28}\ kg$

Mass of the heavier fragment $M_{h}=1.62\times10^{-27}\ Kg$

Speed of lighter fragment = 0.893c

We need to calculate the speed of the heavier fragment

Let v is the speed of the second fragment after decay

Using conservation of relativistic momentum

$0=\drac{m_{1}v_{1}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}-\drac{m_{2}v_{2}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}$

$\drac{m_{1}v_{1}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}=\drac{m_{2}v_{2}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}$

$\dfrac{2.90\times10^{-28}\times0.893c}{\sqrt{1-(0.893)^2}}=\dfrac{1.62\times10^{-27}v_{2}}{\sqrt{1-\dfrac{v_{2}^2}{c^2}}}$

$\dfrac{v_{2}}{\sqrt{1-\dfrac{v_{2}^2}{c^2}}}=\dfrac{2.90\times10^{-28}\times0.893c}{1.62\times10^{-27}\times0.45}$

$\dfrac{v_{2}}{\sqrt{1-\dfrac{v_{2}^2}{c^2}}}=0.355c$

$\dfrac{v_{2}}{1-\dfrac{v_{2}^{2}}{c^2}}=(0.355c)^2$

$\dfrac{1-\dfrac{v_{2}^2}{c^2}}{v_{2}^2}=\dfrac{1}{(0.355c)}$

$\dfrac{1}{v_{2}^2}-\dfrac{1}{c^2}=\dfrac{1}{(0.355c)^2}$

$\dfrac{1}{v_{2}^2}=\dfrac{1}{c^2}+\dfrac{1}{0.126c^2}$

$\dfrac{1}{v_{2}^2}=\dfrac{1}{c^2}(1+\dfrac{1}{0.126})$

$\dfrac{1}{v_{2}^2}=\dfrac{8.93}{c^2}$

$v_{2}^2=\dfrac{c^2}{8.93}$

$v_{2}=0.335c$

Hence, The speed of the heavier fragment is 0.335c.

7 0

3 years ago

A 645 g block is released from rest at height h0 above a vertical spring with spring constant k = 530 N/m and negligible mass. T

Nina [5.8K]

Answer:

a)5.88J

b)-5.88J

c)0.78m

d)0.24m

Explanation:

a) W by the block on spring is given by

W= $\frac{1}{2}$ kx² = $\frac{1}{2}$ (530)(0.149)² = 5.88 J

b) Workdone by the spring = - Workdone by the block = -5.88J

c) Taking x = 0 at the contact point we have U top = U bottom

So, mg $h_o$ = $\frac{1}{2}$ kx² - mgx

And, $h_o$ = ( $\frac{1}{2}$ kx² - mgx )/(mg) = $[\frac{1}{2} (530)(0.149^2)-(0.645)(9.8)(0.149)$ ]/(0.645x9.8)

$h_o$ = 0.78m

d) Now, if the initial initial height of block is 3 $h_o$

$h_o$ = 3 x 0.78 = 2.34m

then, $\frac{1}{2}$ kx² - mgx - mg $h_o$ =0

$\frac{1}{2}$ (530)x² - [(0.645)(9.8)x] - [(0.645)(9.8)(2.34) = 0

265x² - 6.321x - 14.8 = 0

a=265

b=-6.321

c=-14.8

By using quadratic eq. formula, we'll have the roots

x= 0.24 or x=-0.225

Considering only positive root:

x= 0.24m (maximum compression of the spring)

4 0

3 years ago

A water tower is a familiar sight in many towns. The purpose of such a tower is to provide storage capacity and to provide suffi

Mademuasel [1]

The image of the water tower and the houses is in the attachment.

Answer: (a) P = 245kPa;

(b) P = 173.5 kPa

Explanation: <u>Gauge</u> <u>pressure</u> is the pressure relative to the atmospheric pressure and it is only dependent of the height of the liquid in the container.

The pressure is calculated as: P = hρg

where

ρ is the density of the liquid, in this case, water, which is ρ = 1000kg/m³;

When it is full the reservoir contains 5.25×10⁵ kg. So, knowing the density, you know the volume:

ρ = $\frac{m}{V}$

V = ρ/m

V = $\frac{5.25.10^{5}}{10^{3}}$

V = 525 m³

To know the height of the spherical reservoir, its diameter is needed and to determine it, find the radius:

V = $\frac{4}{3}.\pi.r^{3}$

$r = \sqrt[3]{ \frac{3}{4\pi } .V}$

r = $\sqrt[3]{\frac{525.3}{4\pi } }$

r = 5.005 m

diameter = 2*r = 10.01m

(a) Height for House A:

h = 15 + 10.01

h = 25.01

P = hρg

P = 25.01.10³.9.8

P = 245.10³ Pa or 245kPa

(b) h = 25 - 7.3

h = 17.71

P = hρg

P = 17.71.1000.9.8

P = 173.5.10³ Pa or 173.5 kPa

4 0

3 years ago

A first order reaction, A -> products, has a rate reaction of .00250 Ms-1 when [A] = . 484 M. (a) What is the rate constant,

tamaranim1 [39]

Answer: a) The rate constant, k, for this reaction is $0.00516s^{-1}$

b) No $t_{\frac{1}{2}}$ does not depend on concentration.

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

$A\rightarrow products$

Given: Order with respect to $A$ = 1

Thus rate law is:

a) $Rate=k[A]^1$

k= rate constant

$0.00250=k[0.484]^1$

$k=0.00516s^{-1}$

The rate constant, k, for this reaction is $0.00516s^{-1}$

b) Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant

t = age of sample

a = let initial amount of the reactant

a - x = amount left after decay process

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

$t_{\frac{1}{2}}=\frac{2.303}{k}\log\frac{100}{50}$

$t_{\frac{1}{2}}=\frac{0.69}{k}$

Thus $t_{\frac{1}{2}}$ does not depend on concentration.

8 0

3 years ago

What is one way that an electric field is different from a magnetic field?

wel

Answer:

The one of the major difference between the magnetic and electric field is that the electric field induces around the static charge particle which is either negative or positive, whereas the magnetic field produces around the poles (i.e., the north and south pole) of the magnet

Explanation:

4 0

3 years ago

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