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The doctor took the temperature of a dea...

The doctor took the temperature of a dead body at 11.30 Pm which was `94. 6^@F`. He took the temperature of the body again after one hour, which was `93. 4^@F`. If the temperature of the room was `70^@F` , estimate the time of death. Taking normal temperature of human body as `98. 6^@F`. [Given: `log(143/123)=0. 15066 ,log(123/117)=0. 05` ]

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To solve the problem, we will use Newton's Law of Cooling, which states that the rate of change of the temperature of an object is proportional to the difference between its temperature and the ambient temperature. ### Step-by-step Solution: 1. **Set Up the Differential Equation**: According to Newton's Law of Cooling, we have: \[ \frac{dT}{dt} = -\lambda (T - T_r) ...
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A Veterinary doctor was examining a sick cat brought by a pet lover. When it was brought to the hospital, it was already dead. The pet lover wanted to find its time of death. He took the temperature of the cat at 11.30 pm which was 94.6^(@)F . He took the temperature again after one hour, the temperature was lower than the first observation. It was 93.4^(@)F . The room in which the cat was put is always at 70^(@)F . The normal temperature of the cat is taken as 98.6^(@)F when it was alive. The doctor estimated the time of death using Newton law of cooling which is governed by the differential equation: (dT)/(dt) prop (T − 70), where 70^(@)F is the room temperature and T is the temperature of the object at time t. Substituting the two different observations of T and t made, in the solution of the differential equation (dT)/(dt) =k(T-70) where k is a constant of proportion, time of death is calculated. If the temperature was measured 2 hours after 11.30pm, will the time of death change? (Yes/No)

A Veterinary doctor was examining a sick cat brought by a pet lover. When it was brought to the hospital, it was already dead. The pet lover wanted to find its time of death. He took the temperature of the cat at 11.30 pm which was 94.6^(@)F . He took the temperature again after one hour, the temperature was lower than the first observation. It was 93.4^(@)F . The room in which the cat was put is always at 70^(@)F . The normal temperature of the cat is taken as 98.6^(@)F when it was alive. The doctor estimated the time of death using Newton law of cooling which is governed by the differential equation: (dT)/(dt) prop (T − 70), where 70^(@)F is the room temperature and T is the temperature of the object at time t. Substituting the two different observations of T and t made, in the solution of the differential equation (dT)/(dt) =k(T-70) where k is a constant of proportion, time of death is calculated. Which method of solving a differential equation helped in calculation of the time of death?

A Veterinary doctor was examining a sick cat brought by a pet lover. When it was brought to the hospital, it was already dead. The pet lover wanted to find its time of death. He took the temperature of the cat at 11.30 pm which was 94.6^(@)F . He took the temperature again after one hour, the temperature was lower than the first observation. It was 93.4^(@)F . The room in which the cat was put is always at 70^(@)F . The normal temperature of the cat is taken as 98.6^(@)F when it was alive. The doctor estimated the time of death using Newton law of cooling which is governed by the differential equation: (dT)/(dt) prop (T − 70), where 70^(@)F is the room temperature and T is the temperature of the object at time t. Substituting the two different observations of T and t made, in the solution of the differential equation (dT)/(dt) =k(T-70) where k is a constant of proportion, time of death is calculated. The solution of the differential equation (dT)/(dt)=k(T-70) is given by,

A Veterinary doctor was examining a sick cat brought by a pet lover. When it was brought to the hospital, it was already dead. The pet lover wanted to find its time of death. He took the temperature of the cat at 11.30 pm which was 94.6^(@)F . He took the temperature again after one hour, the temperature was lower than the first observation. It was 93.4^(@)F . The room in which the cat was put is always at 70^(@)F . The normal temperature of the cat is taken as 98.6^(@)F when it was alive. The doctor estimated the time of death using Newton law of cooling which is governed by the differential equation: (dT)/(dt) prop (T − 70), where 70^(@)F is the room temperature and T is the temperature of the object at time t. Substituting the two different observations of T and t made, in the solution of the differential equation (dT)/(dt) =k(T-70) where k is a constant of proportion, time of death is calculated. If t = 0 when T is 72, then the value of c is

A Veterinary doctor was examining a sick cat brought by a pet lover. When it was brought to the hospital, it was already dead. The pet lover wanted to find its time of death. He took the temperature of the cat at 11.30 pm which was 94.6^(@)F . He took the temperature again after one hour, the temperature was lower than the first observation. It was 93.4^(@)F . The room in which the cat was put is always at 70^(@)F . The normal temperature of the cat is taken as 98.6^(@)F when it was alive. The doctor estimated the time of death using Newton law of cooling which is governed by the differential equation: (dT)/(dt) prop (T − 70), where 70^(@)F is the room temperature and T is the temperature of the object at time t. Substituting the two different observations of T and t made, in the solution of the differential equation (dT)/(dt) =k(T-70) where k is a constant of proportion, time of death is calculated. State the degree of the above given differential equation.

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