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
Step-by-step explanation:
Convert the speeds of Amir and Ryder to meters per second (m/s).
Amir:
8260 mm/s = 8.260 m/s
Ryder:
930 cm/s = 9.30 m/s
9.30 m/s > 8.260 m/s, so:
Ryder ran approximately 1 meter per second faster than Amir.
Answer:
45°
Step-by-step explanation:
There are sets of formulas you can refer to for different situations like this. In your case here, you need to apply the formula and solve for the unknown. My work is in the attachment.
