A rich industrialist was found murdered in his house. The police arrived at the scene at 11:00 PM. The temperature of the corpse
1 answer:
Answer:
The dude was killed around 6:30PM
Explanation:
Newton's law of cooling states:
where,
= initial temp
= temp of room
= temp after t hours
= how fast the temp is changing
= time (hours)
because the body was initlally 31ºC when the police found it
because that was the room temp
because the body temp drop to 30ºC after 1 hour
t = 1 because that's the time it took for the body temp to drop to 30ºC
we don't know k so we must solve for this
rearrange the equation to solve for k
plug in the numbers to solve for k
Now that we know the value for k, we can find the moment the murder occur. A crucial information that the question left out is the temperature of a human body when they're still alive. A living human body is about 37ºC. We can use that as out initial temperature to solve this problem because we can assume that the freshly killed body will be around 37ºC.
because the body was 37ºC right after being killed
because that was the room temp
because the body temp when the police found it
we solved this earlier
t = ??? we don't know how long it took from the time of the murder to when the police found the body
Rearrange the equation to solve for t
plug in the values
t ≈ 4.337 hours from the time the body was killed to when the police found it.
The police found the body at 11:00PM so subtract 4.337 from that.
11 - 4.33 = 6.66 ≈ 6:30PM
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Answer:
a) Q = 251.758 kJ/mol
b) creep rate is
Explanation:
we know Arrhenius expression is given as
where
Q is activation energy
C is pre- exponential constant
At 700 degree C creep rate is% per hr
At 800 degree C creep rate is% per hr
activation energy for creep is =
solving for Q we get
Q = 251.758 kJ/mol
b) creep rate at 500 degree C
we know
Answer:
N_c = 3.03 N
F = 1.81 N
Explanation:
Given:
- The attachment missing from the question is given:
- The given expressions for the radial and θ direction of motion:
r = 0.5*θ
θ = 0.5*t^2 ...... (correction for the question)
- Mass of peg m = 0.5 kg
Find:
a) Determine the magnitude of the force of the rod on the peg at the instant t = 2 s.
b) Determine the magnitude of the normal force of the slot on the peg.
Solution:
- Determine the expressions for radial kinematics:
dr/dt = 0.5*dθ/dt
d^2r/dt^2 = 0.5*d^2θ/dt^2
- Similarly the expressions for θ direction kinematics:
dθ/dt = t
d^2θ/dt^2 = 1
- Evaluate each at time t = 2 s.
θ = 0.5*t^2 = 0.5*2^2 = 2 rad -----> 114.59°
r = 1 m , dr / dt = 1 m/s , d^2 r / dt^2 = 0.5 m/s^2
- Evaluate the angle ψ between radial and horizontal direction:
tan Ψ = r / (dr/dθ) = 1 / 0.5
Ψ = 63.43°
- Develop a free body diagram (attached) and the compute the radial and θ acceleration:
a_r = d^2r / dt^2 - r * dθ/dt
a_r = 0.5 - 1*(2)^2 = -3.5 m/s^2
a_θ = r * (d^2θ/dt^2) + 2 * (dr/dt) * (dθ/dt)
a_θ = 1(1) + 2*(1)*(2) = 5 m/s^2
- Using Newton's Second Law of motion to construct equations in both radial and θ directions as follows:
Radial direction: N_c * cos(26.57) - W*cos(24.59) = m*a_r
θ direction: F - N_c * sin(26.57) + W*sin(24.59) = m*a_θ
Where, F is the force on the peg by rod and N_c is the normal force on peg by the slot. W is the weight of the peg. Using radial equation:
N_c * cos(26.57) - 4.905*cos(24.59) = 0.5*-3.5
N_c = 3.03 N
F - 3.03 * sin(26.57) + 4.905*sin(24.59) = 0.5*5
F = 1.81 N
