The volume of the balloon is given by:
V = 4πr³/3
V = volume, r = radius
Differentiate both sides with respect to time t:
dV/dt = 4πr²(dr/dt)
Isolate dr/dt:
dr/dt = (dV/dt)/(4πr²)
Given values:
dV/dt = 72ft³/min
r = 3ft
Plug in and solve for dr/dt:
dr/dt = 72/(4π(3)²)
dr/dt = 0.64ft/min
The radius is increasing at a rate of 0.64ft/min
The surface area of the balloon is given by:
A = 4πr²
A = surface area, r = radius
Differentiate both sides with respect to time t:
dA/dt = 8πr(dr/dt)
Given values:
r = 3ft
dr/dt = 0.64ft/min
Plug in and solve for dA/dt:
dA/dt = 8π(3)(0.64)
dA/dt = 48.25ft²/min
The surface area is changing at a rate of 48.25ft²/min
Answer:
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Answer:
The 60 Ohm bulb would be brighter than the 20 Ohm bulb.
Explanation:
Hi there!
Ohm's Law:
where V is voltage, I is current and R is resistance
Power equation: 
The greater the power of a bulb, the brighter it is.
The bulbs are connected in series, meaning the the current travelling through each will be the same. Plug Ohm's law into the power equation so we just have I and R:

Because the current stays the same, the bulb with greater resistance will have greater power, hence being brighter. Therefore, the 60 Ohm bulb would be brighter than the 20 Ohm bulb.
I hope this helps!