Based on the trend produced by the dose - response graph, it would be best to evacuate the residents in other to prevent the increasing percentage of deaths due to the rising level of pollutant A.
- The curve shows that the pollutant level in mg/kg of pollutant A is still increasing, hence, groundwater monitoring alone won't be the best decision to make.
- Since the pollutant level is still increasing, then the spill level still need effective monitoring.
- Evacuation of residents seems to be the best decision that should be taken based on the information interpreted on the graph.
Therefore, Evacuating residents to prevent rising death percentage is required as the pollutant level is yet to subside.
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Answer:
1.3823 rad/s
20.7345 m/s
28.66129935 m/s²

2006.29095 N radially outward
Explanation:
r = Radius = 15 m
m = Mass of person = 70 kg
g = Acceleration due to gravity = 9.81 m/s²
Angular velocity is given by

Angular velocity is 1.3823 rad/s
Linear velocity is given by

The linear velocity is 20.7345 m/s
Centripetal acceleration is given by

The centripetal acceleration is 28.66129935 m/s²
Acceleration in terms of g


Centripetal force is given by

The centripetal force is 2006.29095 N radially outward
The torque will be experienced when the centrifuge is speeding up of slowing down i.e., when it is accelerating and decelerating.
Answer:
W = 9.93 10² N
Explanation:
To solve this exercise we must use the concept of density
ρ = m / V
the tabulated density of copper is rho = 8966 kg / m³
let's find the volume of the cylindrical tube
V = A L
V = π (R_ext ² - R_int ²) L
let's calculate
V = π (4² - 2²) 10⁻⁴ 3
V = 1.13 10⁻² m³
m = ρ V
m = 8966 1.13 10⁻²
m = 1.01 10² kg
the weight of the tube
W = mg
W = 1.01 10² 9.8
W = 9.93 10² N
The water will be cool and steam will be created by the hot and cold water reacting together
The springs stored energy is transferred to the cube as kinetic energy and then by the slop the KE is converted to height energy.
<span>0.5 . k . x^2 = 0.5 . m . v^2 = m . g . ∆h </span>
<span>0.5 . 50 . (0.1^2) = 0.05 . 9.8 . ∆h </span>
<span>∆h = 0.51 m = 51 cm </span>
<span>This is the height gained </span>
<span>Distance along the slope = ∆h / sin 60 = 0.589 = 59 cm </span>
<span>In the second case, the stored spring energy is converted into height energy AND frictional heat energy. </span>
<span>The height energy is m . g . d sin 60 where d is the distance the cube moves along the slope. </span>
<span>The Frictional energy converted is F . d </span>
<span>F ( the frictional force ) = µ . N </span>
<span>N ( the reaction to the component of the gravity force perpendicular to the surface of the slope ) = m . g . cos60 </span>
<span>Total energy converted </span>
<span>0.5 . k . x^2 = (m . g . dsin60) + (µ . m . g . cos60 . d ) </span>
<span>Solve for d </span>
<span>d = 0.528 = 53 cm</span>