Answer:
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Explanation:
Given:
diameter of sphere, d = 6 inches
radius of sphere, r =
= 3 inches
density,
= 493 lbm/ 
S.G = 1.0027
g = 9.8 m/
= 386.22 inch/ 
Solution:
Using the formula for terminal velocity,
=
(1)

where,
V = volume of sphere
= drag coefficient
Now,
Surface area of sphere, A = 
Volume of sphere, V = 
Using the above formulae in eqn (1):
= 
=
= 
Therefore, terminal velcity is given by:
=
inch/sec
Answer:
the percent increase in the velocity of air is 25.65%
Explanation:
Hello!
The first thing we must consider to solve this problem is the continuity equation that states that the amount of mass flow that enters a system is the same as what should come out.
m1=m2
Now remember that mass flow is given by the product of density, cross-sectional area and velocity
(α1)(V1)(A1)=(α2)(V2)(A2)
where
α=density
V=velocity
A=area
Now we can assume that the input and output areas are equal
(α1)(V1)=(α2)(V2)

Now we can use the equation that defines the percentage of increase, in this case for speed

Now we use the equation obtained in the previous step, and replace values

the percent increase in the velocity of air is 25.65%
Answer:
The material with the higher specific heat capacity would cause a more severe burn.
Explanation:
Quantity of heat (Q) = mass of material (m) × specific heat capacity (C) × temperature difference (∆T)
From the formula above, the relationship between Q and C is direct in which increase in one quantity (C) leads to a corresponding increase in the other quantity (Q)
The material with the higher specific heat capacity would produce more heat, thus cause a more severe burn.