Answer:
Explanation:
velocity of ship with respect to water = 6.5 m/s due north

velocity of water with respect to earth = 1.5 m/s at 40° north of east

velocity of ship with respect to water = velocity of ship with respect to earth - velocity of water with respect to earth



The magnitude of the velocity of ship relative to earth is
= 5.66 m/s
Answer:
I'm sorry I don't have a answer but I like your pfp
Answer:
The answer is "
"
Explanation:
Please find the complete question in the attached file.

pi = pressure only at two liquids' devices
PA = pressure atmosphere.
1 = oil density
2 = uncertain fluid density

The pressures would be proportional to the quantity
cm from below the surface at the interface between both the oil and the liquid.


A. compounds because the reactants of a double replacement reaction do not have charges