The distance a spring will stretch varies directly with how much weight is attached to the spring. If a spring stretches 5 inche
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Answer:
T = 3.967 C
Step-by-step explanation:
Density = mass / volume
Use the mass = 1kg and volume as the equation given V, we will come up with the following equation
D = 1 / 999.87−0.06426T+0.0085043T^2−0.0000679T^3
= (999.87−0.06426T+0.0085043T^2−0.0000679T^3)^-1
Find the first derivative of D with respect to temperature T
dD/dT =
Let dD/dT = 0 to find the critical value we will get
= 0
Using formula of quadratic, we get the roots:
T = 79.53 and T = 3.967
Since the temperature is only between 0 and 30, pick T = 3.967
Find 2nd derivative to check whether the equation will have maximum value:
Substituting the value with T=3.967,
d2D/dT2 = -1.54 x 10^(-8) a negative value. Hence It is a maximum value
Substitute T =3.967 into equation V, we get V = 0.001 i.e. the volume when the the density is the highest is at 0.001 m3 with density of
D = 1/0.001 = 1000 kg/m3
Therefore T = 3.967 C