A hot–air balloon is moving at a speed of 10.0 meters/second in the +x–direction. The balloonist throws a brass ball in the +x–d
1 answer:
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Answer:
a) 3.37 x
b) 6.42kg/
Explanation:
a) Firstly we would calculate the volume of the metal using it`s weight in air and water , after finding the weight we would find the density .
Weight of metal in air = 50N = mg implies the mass of metal is 5kg.
Now the difference of weight of the metal in air and water = upthrust acting on it = volume (metal) p (liquid) g = V (1000)(10) = 14N. So volume of metal piece = 14 x
. So density of metal = mass of metal / volume of metal = 5 / 14 x
= 3.37 x
b) Water exerts a buoyant force to the metal which is 50−36 = 14N, which equals the weight of water displaced. The mass of water displaced is 14/10 = 1.4kg Since the density of water is 1kg/L, the volume displaced is 1.4L. Hence, we end up with 3.57kg/l. Moreover, the unknown liquid exerts a buoyant force of 9N. So the density of this liquid is 6.42kg/
Answer:
Approximately , assuming that this gas is an ideal gas.
Explanation:
- Let
and
denote the volume and pressure of this gas before the compression.
- Let
and
denote the volume and pressure of this gas after the compression.
By Boyle's Law, the pressure of a sealed ideal gas at constant temperature will be inversely proportional to its volume. Assume that this gas is ideal. By this ideal gas law:
.
Note that in Boyle's Law, is inversely proportional to
. Therefore, on the two sides of this equation, "final" and "initial" are on different sides of the fraction bar.
For this particular question:
.
.
.
- The pressure after compression,
Rearrange the equation to obtain:
.
Before doing any calculation, think whether the pressure of this gas will go up or down. Since the gas is compressed, collisions between its particles and the container will become more frequent. Hence, the pressure of this gas should increase.
.