When air expands adiabatically (without gaining or losing heat), its pressure P and volume V are related by the equation PV1.4=C where C is a constant. Suppose that at a certain instant the volume is 320 cm3, and the pressure is 95 kPa (kPa = kiloPascals) and is decreasing at a rate of 11 kPa/minute. At what rate is the volume increasing at this instant?

26.47 cm³/minute

Explanation:

We first differentiate the expression to obtain the rate of change of volume with time dV/dt

Now when V = 320 cm, P = 95 kPa and dP/dt = -11 kPa/minute since it is decreasing.

We substitute these values into the expression for dV/dt. So,

total distance travelled = 60 miles

total time taken =1.5 hours

average speed = total distance / total time

60 / 1.5

= 40

therefore, average speed = 40 m/h

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find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2