Ship a is 32 miles north of ship b and is sailing due south at 16 mph. ship b is sailing due east at 12 mph. at what rate is the distance between the ships changing an hour later? are they getting closer together or farther apart?

-5.6 mi/h

Explanation:

= Velocity of ship A = -16 mi/h (negative as the distance is decreasing)

Distance ship A travels in 1 hour = 16×1 = 16 mi

a = Distance remaining = 32-16 = 16 mi

= Velocity of ship B = 12 mi/h

b = Distance ship B travels in 1 hour = 18×1 = 12 mi

c = Distance between A and B after 1 hour = √(a²+b²) = √(16²+12²) = 20 mi

From Pythagoras theorem

a²+b² = c²

Now, differentiating with respect to time

∴ Rate at which distance between the ships is decreasing one hour later is 5.6 mi/h

the temperature in kelvin of 2.0 dm3 of an ideal gas is doubled and its pressure is increased by a factor of four. what is the final volume of the gas

force required to detach it is given by

charges are taken to be fundamental charge

now we have

so above is the minimum distance at which it will detach