Because hot air rises and cold air sinks. So when they heat the air in the balloon it becomes less dense or less heavy and that makes it rise ad the volume becomes more. Because hot air rises and cold air sinks. So when they heat the air in the balloon it becomes less dense or less heavy and that makes it rise and the volume becomes more.
Although volume and pressure are held constant, some of the gas escapes from the bottom of the balloon, thus decreasing the mass of the gas in the balloon. If mass is decreased, then density is also decreased because density=mass/volume. Such a decrease in density of the balloon causes it rise.
Nothing happens to the mass of the balloon. Mass is conserved, so the temperature of the balloon will not affect it's mass. Mass can be thought of the amount of "stuff" that makes up a balloon. It can be obtained by adding up the mass of all the molecules of rubber in the balloon. Obviously, putting the balloon in a warm room will not change the number of molecules in the balloon, therefore the mass stays constant. The volume of the balloon will probably increase. Because volume increases but mass remains constant, the density of the balloon would decrease. D = m/v
Yes. When you squeeze the balloon, you cause the volume to decrease. Since density is determined by dividing mass by volume, a decrease in volume will cause an increase in denisty.
Mass = Density x Volume Density = Mass ÷ Volume Volume = Mass ÷ Density
Density = (Mass) divided by (Volume) If you know the density and volume, then Mass = (Density) times (Volume)
Although volume and pressure are held constant, some of the gas escapes from the bottom of the balloon, thus decreasing the mass of the gas in the balloon. If mass is decreased, then density is also decreased because density=mass/volume. Such a decrease in density of the balloon causes it rise.
The density decreases by half. You find the answer by knowing that density is equal to mass divided by the volume. If the mass stays constants and the volume is doubled, then the density is halved.
The density decreases by half. You find the answer by knowing that density is equal to mass divided by the volume. If the mass stays constants and the volume is doubled, then the density is halved.
The density decreases by half. You find the answer by knowing that density is equal to mass divided by the volume. If the mass stays constants and the volume is doubled, then the density is halved.
Nothing happens to the mass of the balloon. Mass is conserved, so the temperature of the balloon will not affect it's mass. Mass can be thought of the amount of "stuff" that makes up a balloon. It can be obtained by adding up the mass of all the molecules of rubber in the balloon. Obviously, putting the balloon in a warm room will not change the number of molecules in the balloon, therefore the mass stays constant. The volume of the balloon will probably increase. Because volume increases but mass remains constant, the density of the balloon would decrease. D = m/v
as is, a rock, because density is mass over volume and a hot air balloon has a lesser/similar mass (stuff) but is spread out (volume) more than a rock.
Mass = Density x Volume Density = Mass/Volume Volume = Mass/Density
Yes. When you squeeze the balloon, you cause the volume to decrease. Since density is determined by dividing mass by volume, a decrease in volume will cause an increase in denisty.
It makes no difference. Density = mass / volume. You divide the mass by the volume. If the volume is greater than the mass your answer will necessarilybe less than one, but that is still the correct answer showing the density in terms of the units used.
Density = Mass / Volume Rearranging this gives: Volume = Mass / Density Mass = Density × Volume
Volume = mass / Density Mass = Volume * Density Density = Mass / Volume
Density = mass/volume Mass = (density) x (volume) Volume = mass/density