A little aluminum boat (mass of 14.50 g) has a volume of 450.00 cm3. The boat is place in a small pool of water and carefully filled with pennies. If each penny has a mass of 2.50 g, how many pennies can be added to the boat before it sinks?
Answers
Answer:
If the water level is at the top of the boat (just before it sinks), the buoyant force upwards from the water (the weight of water displaced) must be equal to the weight of the boat.
Suppose we place x pennies.
Mass of the boat in g - 14.50 + 2.50 x
Mass of the boat (kg) - (14.50 + 2.50x) / 1000
Weight of the boat (N) = mg = (14.50 + 2.50x) x 9.81 / 1000
Volume of water displaced = 450 cm3 = 0.45 l
I l water has mass 1 kg
So mass of water displaced = 0.45 kg
Weight of water displaced = 0.45 x 9.81
So (14.50 + 2.50 x ) x 9.81 / 1000 = 0.45 x 9.81
14.50 + 2.50 x = 1000 x 0.45 = 450
x = (450 - 14.50) / 2.50 = 174.2 pennies, so answer is 174.
Hope this answer is helpful for u.
Answer:
174
Explanation:
let the total mass of pennies that can be accommodated be m
buoyant force=volume of body×density of liquid×g
equate this to weight
450×1×g=(14.5+m)g
450-14.5=m
m=435.5
no of pennies=435.5/2.5=174.2