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Elenna [48]
2 years ago
15

A naked, male corpse was found at 8AM on Tuesday, July 9.The air temperature wasalready 26.7˚C (81˚F). The body exhibited some s

tiffness in the face and eyelids and had a body temperature of 34.4˚C (93.9˚F). Livor mortis was not evident. Approximately how long ago did the man die? Justify your answer. Would clothing on the body have made a difference in determining the actual time of death in this case? Why or why not?
Physics
1 answer:
Slav-nsk [51]2 years ago
3 0

Answer:

x = 9.87 hours ago.

Further explanation is present below.

Explanation:

Data given:

Air Temperature = 26.7°C (81°F)

Body temperature = 34.4°C (93.9°F)

We are given that, Livor mortis was absent. Secondly, The body exhibited some stiffness in the face and eyelids.

So, it can be concluded that, body must have died recently, as Livor mortis was absent plus the presence of stiffness in the face and eyelids. Furthermore,

The normal body temperature = 37°C

Body temperature in our case = 34.4° C

Difference = 37 -34.4 = 2.6°C

And Air temperature = 26.7°C

Which is a very low difference, in addition, it has not acquired the air temperature so, all this evidence certifies that body died recently.

Now, we have to calculate, how recently it died.

Formula:

0.78 . x = difference of body temperature and air temperature

Where, x = number of hours = time required,

0.78 . x = (34.4 - 26.7)

0.78 . x = 7.7

x = 7.7/0.78

x = 9.87 hours ago.

It means that, naked male corpse was died 9.87 hours ago.

Now, we are asked to find out clothing on the body had made a difference or not.

So, Yes it would definitely make a difference. As body would have more warmer than being naked. and it would result in a calculation of a different time of death. (body temperature would have > 34.4°C)

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Bas_tet [7]

Answer:

v = 0.059 m/s

Explanation:

To find the final speed of Olaf and the ball you use the conservation momentum law. The momentum of Olaf and the ball before catches the ball is the same of the momentum of Olaf and the ball after. Then, you have:

mv_{1i}+Mv_{2i}=(m+M)v  (1)

m: mass of the ball = 0.400kg

M: mass of Olaf = 75.0 kg

v1i: initial velocity of the ball = 11.3m/s

v2i: initial velocity of Olaf = 0m/s

v: final velocity of Olaf and the ball

You solve the equation (1) for v and replace the values of all variables:

v=\frac{mv_{1i}}{m+M}=\frac{(0.400kg)(11.3m/s)}{0.400kg+75.0kg}=0.059\frac{m}{s}

Hence, after Olaf catches the ball, the velocity of Olaf and the ball is 0.059m/s

3 0
2 years ago
A 4.60 gg coin is placed 19.0 cmcm from the center of a turntable. The coin has static and kinetic coefficients of friction with
Komok [63]

Answer:

62.64 RPM.

Explanation:

Given that

m= 4.6 g

r= 19 cm

μs = 0.820

μk = 0.440.

The angular speed of the turntable = ω rad/s

Condition just before the slipping starts

The maximum value of the static friction force =Centripetal force

\mu_s\ m g=m\ \omega^2\ r\\ \omega^2=\dfrac{\mu_s\ m g}{m r}\\ \omega=\sqrt{\dfrac{\mu_s\  g}{ r}}\\ \omega=\sqrt{\dfrac{0.82\times 10}{ 0.19}}\ rad/s\\\omega= 6.56\ rad/s

\omega=\dfrac{2\pi N}{60}\\N=\dfrac{60\times \omega}{2\pi }\\N=\dfrac{60\times 6.56}{2\pi }\ RPM\\N=62.64\ RPM

Therefore the speed in RPM will be 62.64 RPM.

5 0
2 years ago
A football kicked in front of a goal post at an angle of 45 degree to the ground just clear the top by of the post 3m high. Calc
Firlakuza [10]

Answer:

A. 10.84 m/s

B. 1.56 s

Explanation:

From the question given above, the following data were obtained:

Angle of projection (θ) = 45°

Maximum height (H) = 3 m

Acceleration due to gravity (g) = 9.8 m/s²

Velocity of projection (u) =?

Time (T) taken to hit the ground again =?

A. Determination of the velocity of projection.

Angle of projection (θ) = 45°

Maximum height (H) = 3 m

Acceleration due to gravity (g) = 9.8 m/s²

Velocity of projection (u) =?

H = u²Sine²θ / 2g

3 = u²(Sine 45)² / 2 × 9.8

3 = u²(0.7071)² / 19.6

Cross multiply

3 × 19.6 = u²(0.7071)²

58.8 = u²(0.7071)²

Divide both side by (0.7071)²

u² = 58.8 / (0.7071)²

u² = 117.60

Take the square root of both side

u = √117.60

u = 10.84 m/s

Therefore, the velocity of projection is 10.84 m/s.

B. Determination of the time taken to hit the ground again.

Angle of projection (θ) = 45°

Velocity of projection (u) = 10.84 m/s

Time (T) taken to hit the ground again =?

T = 2uSine θ /g

T = 2 × 10.84 × Sine 45 / 9.8

T = 21.68 × 0.7071 / 9.8

T = 1.56 s

Therefore, the time taken to hit the ground again is 1.56 s.

7 0
3 years ago
Which of the following is an example of a properly written testable hypothesis? *
sveta [45]

Answer:

D. if it is dark, then an owl will find a mouse by the sound the mouse makes

3 0
3 years ago
What is the Gravitational Potential Energy of a 30 kg box lifted 1.5 meters off the ground?
juin [17]

Answer:

<h2>441 J</h2>

Explanation:

The potential energy of a body can be found by using the formula

PE = mgh

where

m is the mass

h is the height

g is the acceleration due to gravity which is 9.8 m/s²

From the question we have

PE = 30 × 9.8 × 1.5

We have the final answer as

<h3>441 J</h3>

Hope this helps you

6 0
2 years ago
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