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Semenov [28]
3 years ago
12

What would be the approximate PMI of a corpse whose body temperature is determined to be 26.3°C? Average body temperature is 37.

2*C.
PLEASE QUICKLY
Physics
1 answer:
lubasha [3.4K]3 years ago
4 0
Determining the time of death is both an art and a science and requires that the medical examiner use several techniques and observations to make his estimate. As a general rule, the sooner after death the body is examined, the more accurate this estimate will be.

Unfortunately, the changes that a body undergoes after death occur in widely variable ways and with unpredictable time frames. There is no single factor that will accurately indicate the time of physiological death. It is always a best guess. But when the principles are properly applied, the medical examiner can often estimate the physiologic time of death with some degree of accuracy.

One way we can estimate time of death is by measuring the body temperature. Average body temperate is 32.7c or 98.6 Fahrenheit .The current formula for heat loss is a loss of 1.5 degree per hour.


Hours since death = 98.6 – corpse core temperature / 1.5
(26.3 Celsius converted to Fahrenheit= 79.34 F)

98.6 F - 79.34 F / 1.5 = 45.7 hours since death
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Answer:

500km

Explanation:

Given parameters:

Speed  = 200km/hr

Time taken  = 2.5hrs

Unknown:

Distance  = ?

Solution:

To solve this problem, we use the speed, time and distance equation.

   Therefore;

  Distance  = Speed x time

So;

  Distance  = 200 x 2.5  = 500km

6 0
2 years ago
Use Eq. cosϕ=R/Z to show that the average power delivered by the source in an L−R−C series circuit is given by Pav = I^2rmsR .
Evgen [1.6K]

Answer:

Explanation:

In a L C R circuit, the average power is given by

P_{av}=V_{rms}I_{rms}Cos\phi

As given in the question

CosФ = R / Z

And we know that

V_{rms}=I_{rms}\times Z

So

P_{av}=I_{rms}\times Z\times I_{rms}\times \frac{R}{Z}

P_{av}=I_{rms}^{2}\times R

6 0
3 years ago
A 6.0-μF air-filled capacitor is connected across a 100-V potential source (a battery). After the battery fully charges the capa
Dovator [93]

Answer:

Change in Q = 2.1x 10^-3 C

Explanation:

We are given that

The Initialcapacitance C1 = 6.0μF

Initial charge oncapacitor

Q1 = C1 V

= 6.00 x 10^-6 x 100

= 6.00 x 10^-4 C

So the Final capacitance C2 will be

= K x C1 = 4.5 x 6.00 x 10^-6

= 2.7 x 10^ -5 F

So to get Finalcharge

We use Q2 = C2 x V

= 2.7 x 10^ - 5 x 100

= 27 x 10^ -4 C

So Charge flown in thecapacitor is change in Q

Which is = Q2 - Q1

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5 0
3 years ago
Which component acts a platform on which an application software runs.​
Alexandra [31]

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6 0
2 years ago
A single slit of width d = 0.08 mm is illuminated by light of two wavelengths, l = 446 nm and l = 662 nm. The diffraction patter
lorasvet [3.4K]

Answer

given,

width of slit, d = 0.08 mm

                   d = 8 x 10⁻⁵ m

light of two wavelength

I₁= 446 nm

I₂ = 662 nm

a)  angles at which the third dark fringe

    sin C = \dfrac{m\lambda}{d}

                m = 3  , I₁= 446 nm

    sin C = \dfrac{3\times 446 \times 10^{-9}}{8\times 10^{-5}}

                 C = 0.958°

                m = 3  , I₁= 662 nm

    sin C = \dfrac{3\times 662 \times 10^{-9}}{8\times 10^{-5}}

                 C = 1.423°

b)  angles at which the third dark fringe

    sin C = \dfrac{m\lambda}{d}

                m = 1  , I₁= 446 nm

    sin C = \dfrac{1\times 446 \times 10^{-9}}{8\times 10^{-5}}

                 C = 0.319°

                m = 1  , I₁= 662 nm

    sin C = \dfrac{1\times 662 \times 10^{-9}}{8\times 10^{-5}}

                C = 0.474°

4 0
2 years ago
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