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
The answer I believe is shoes, tires and gloves.
As we know the time period of pendulum is given by the equation
so here
L = length of the pendulum
Now on increasing the temperature due to thermal expansion the length of the pendulum will increase and due to this the time period of the pendulum will also increase
Now due to increase in time period of the pendulum the clock will become slow and it will give incorrect time due to this increase in temperature
