Use the ideal-gas law, along with the knowledge that the pressure of the air above the fire is the same as that of the ambient air, to derive an expression for the acceleration a of an air parcel as a function of (t2/t1), where t1 is the absolute temperature of the air above the fire and t2 is the absolute temperature of the ambient air.
answer with explanation:
when dealing with heavy loads, one needs to take certain precautions in order to avoid or prevent from injuries.
for instance, one should avoid lifting above the shoulder level which can end up pulling your muscle.
also, you can either break the load in parts if possible or get some for lifting heavy bulks.
moreover, one should try to lift with the legs while keeping the back straight and not twisting it.
on/off components of a pid controlled accumulates the error over time and responds to system error after the error has been accumulated.
heat transfer through conduction is a slow process and both conduction and convection require material medium
answer with explanations:
we are given:
where t= time in seconds, and a(t) = acceleration as a function of time.
x(2) = -20 )
where x(t) = distance travelled as a function of time.
need to find x(4).
from (1), we express x(t) by integrating, twice.
velocity = v(t) = integral of (1) with respect to t
v(t) = 4t^3/3 - 2t + k1 )
where k1 is a constant, to be determined.
integrate (4) to find the displacement x(t) = integral of (4).
x(t) = integral of v(t) with respect to t
= (t^4)/3 - t^2 + (k1)t + k2 ) where k2 is another constant to be determined.
from (2) and (3)
we set up a system of two equations, with k1 and k2 as unknowns.
x(0) = 0 - 0 + 0 + k2 = -2 => k2 = 2 )
substitute (6) in (3)
x(2) = (2^4)/3 - (2^2) + k1(2) -2 = -20
16/3 -4 + 2k1 -2 = -20
2k1 = -20-16/3 +4 +2 = -58/3
k1 = -29/3 )
thus substituting (6) and (7) in (5), we get
x(t) = (t^4)/3 - t^2 - 29t/3 + 2 )
which, by putting t=4 in (8)
x(4) = (4^4)/3 - (4^2 - 29*4/3 +2
= 86/3, or
= 28 2/3, or
= 28.67 (to two places of decimal)