Answer: ∆V for r = 10.1 to 10ft
∆V = 40πft^3 = 125.7ft^3
Approximate the change in the volume of a sphere When r changes from 10 ft to 10.1 ft, ΔV=_________
[v(r)=4/3Ï€r^3].
Step-by-step explanation:
Volume of a sphere is given by;
V = 4/3πr^3
Where r is the radius.
Change in Volume with respect to change in radius of a sphere is given by;
dV/dr = 4πr^2
V'(r) = 4πr^2
V'(10) = 400π
V'(10.1) - V'(10) ~= 0.1(400π) = 40π
Therefore change in Volume from r = 10 to 10.1 is
= 40πft^3
Of by direct substitution
∆V = 4/3π(R^3 - r^3)
Where R = 10.1ft and r = 10ft
∆V = 4/3π(10.1^3 - 10^3)
∆V = 40.4π ~= 40πft^3
And for R = 30ft to r = 10.1ft
∆V = 4/3π(30^3 - 10.1^3)
∆V = 34626.3πft^3
d = distance between the two cities
v₁ = average speed while going from chicago to kansas city = 440 knots
t₁ = time taken to travel distance going from chicago to kansas city
time taken to travel distance going from chicago to kansas city is given as
t₁ = d/v₁
t₁ = d/440 eq-1
v₂ = average speed while going from kansas city to chicago = 110 knots
t₂ = time taken to travel distance going from kansas city to chicago
time taken to travel distance going from kansas city to chicago is given as
t₂ = d/v₂
t₂ = d/110 eq-2
Given that :
t₂ = t₁ + 3
using eq-1 and eq-2
(d/110) = (d/440) + 3
d = 440
Answer:
1.110
Step-by-step explanation:
I hope that help's
