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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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Pupils dilate and constrict in order to allow an adequate amount of light to pass through the retina and vision. If there is not enough light and the pupils do not dilate, a small amount of light will pass to the retina and the vision will be damaged.
6 0
3 years ago
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The pressure on a volume of liquid V = 1.0 mº at the surface is approximately equal to the atmospheric pressure Patm = 1.00 x 10
Alexus [3.1K]

Answer:-2.86*10⁻⁴

Explanation: Use the equation change in volume = (change in pressure * original volume) / Bulks Modulus. ΔV = (-Δp*V₀) / B

Plugging in your numbers, you should get ΔV = (-2.29*10⁷*1) / (8*10¹⁰) = -2.86*10⁻⁴

ΔP = P₂-P₁  ----> ΔP = 2.30*10⁷ - 1.00*10⁵ = 2.29*10⁷

3 0
3 years ago
Your little sister is building a radio from scratch. Plans call for a 500 μH inductor wound on a cardboard tube. She brings you
gayaneshka [121]

Answer:

N = 195 turns

Explanation:

The inductance of the inductor, L = 500 μH = 500 * 10⁻⁶H

The length of the tube, l = 12 cm = 0.12 m

The diameter of the tube, d = 4 cm = 0.04 m

Radius, r = 0.04/2 = 0.02 m

Area of the tube, A = πr² = 0.02²π = 0.0004π m²

\mu_{0} = 4\pi * 10^{-7}

The inductance of a solenoid is given by:

L = \frac{\mu_{0}N^{2} A }{l}

500 * 10^{-6} = \frac{4\pi *10^{-7}  N^{2} *4\pi  *10^{-4}  }{0.12}\\500 * 10^{-6} = 0.00000001316N^{2} \\N^{2} = \frac{500 * 10^{-6}}{0.00000001316}\\N^{2} = 37995.44\\N = \sqrt{37995.44} \\N = 194.92 turns

8 0
3 years ago
While tuning a string to the note C at 523 Hz, a piano tuner hears 2.00 beats/s between a reference oscillator and the string.
lara31 [8.8K]

Answer:

a)the possible frequencies are 521hz ,522hz, 523, 524hz,525hz

b) 526hz

c)0.989 or a 1.14% decrease in tension

Explanation:

a) While tuning a string at 523 Hz,piano tuner hears 2.00 beats/s between a reference oscillator and the string.

The possible frequencies of the string can be calculated by

fl=f' - B

where

fl= lower limit of the possible frequency

f'= frequency of the string

B= beat heard by the tuner

fl= 523hz + Or - (2beats/secs * 1hz/1beat per sc)

fl= 521hz or 525hz

So the possible frequencies are 521hz ,522hz, 523, 524hz,525hz

b)fl=f' - B

523hz= f' - 3

f'= 523 + 3= 526hz

c) The tension is directly proportional to the square of the frequencies

T1/T2 =f1^2/f2^2

523^2 / 526^2 = 0.989 or a 1.14% decrease in tensio

6 0
3 years ago
A unit of mass is ??
Effectus [21]
Hello,

Answer: kilogram

Further explaining: in science is used to measure weight of an object and used for accreditation.
Hope this helps!

3 0
3 years ago
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