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2016-03-09T11:00:57+05:30

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                  Ascent rate of a balloon (assuming spherical symmetry) depends on the following forces:(1) The upward buoyant force FB=43πr3ρairgFB=43πr3ρairg
(2) The gravitational pull downwards: FG=43πr3ρgFG=43πr3ρg
(3) The drag force acting: $F_D\, =\, \tfrac12\, \rho\ \rho_{air}\, v^2\, C_D\, A$ On a first glance it might appear that the balloon soon reaches a terminal velocity. But the quantities involved in these equations aren't all independent of each other or remain constant. For example the density of air changes with altitude. And the atmospheric pressure drops as you ascent, causing the balloon to increase in volume, thereby increasing the drag on it. Thus to analyse the motion of the balloon carefully, one has to resort to numerical methods and computers. But if you are looking for an approximation the ascent rate could be taken as the terminal velocity and could be obtained by setting  ,FB=FG+FDFB=FG+FD leading to, v=8gr3CD(ρair−ρρair)−−−−−−−−−−−−−−v=8gr3CD(ρair−ρρair) Ascent rate of a balloon (assuming spherical symmetry) depends on the following forces:(1) The upward buoyant force $F_B=\frac{4}{3}\pi r^3\rho_{air}g$
(2) The gravitational pull downwards: $F_G=\frac{4}{3}\pi r^3\rho g$
(3) The drag force acting: $F_D\, =\, \tfrac12\, \rho\, v^2\, C_D\, A$ On a first glance it might appear that the balloon soon reaches a terminal velocity. But the quantities involved in these equations aren't all independent of each other or remain constant. For example the density of air changes with altitude. And the atmospheric pressure drops as you ascent, causing the balloon to increase in volume, thereby increasing the drag on it. Thus to analyse the motion of the balloon carefully, one has to resort to numerical methods and computers. But if you are looking for an approximation the ascent rate could be taken as the terminal velocity and could be obtained by setting   $$F_B=F_G+F_D$$ Ascent rate of a balloon (assuming spherical symmetry) depends on the following forces:(1) The upward buoyant force $F_B=\frac{4}{3}\pi r^3\rho_{air}g$
(2) The gravitational pull downwards: $F_G=\frac{4}{3}\pi r^3\rho g$
(3) The drag force acting: $F_D\, =\, \tfrac12\, \rho_{air}\, v^2\, C_D\, A$ On a first glance it might appear that the balloon soon reaches a terminal velocity. But the quantities involved in these equations aren't all independent of each other or remain constant. For example the density of air changes with altitude. And the atmospheric pressure drops as you ascent, causing the balloon to increase in volume, thereby increasing the drag on it. Thus to analyse the motion of the balloon carefully, one has to resort to numerical methods and computers. But if you are looking for an approximation the ascent rate could be taken as the terminal velocity and could be obtained by setting, $$F_B=F_G+F_D$$ leading to, $$v = \sqrt{\frac{8 g r}{3 C_D} \left( \frac{\rho_{air} - \rho}{\rho_{air}} \right)}$$ 1 source | link answered Jun 21 '11 at 19:55 Manu Ascent rate of a balloon (assuming spherical symmetry) depends on the following forces:(1) The upward buoyant force FB=43πr3ρairgFB=43πr3ρairg
(2) The gravitational pull downwards: FG=43πr3ρgFG=43πr3ρg
(3) The drag force acting: FD=12ρv2CDAFD=12ρv2CDA On a first glance it might appear that the balloon soon reaches a terminal velocity. But the quantities involved in these equations aren't all independent of each other or remain constant. For example the density of air changes with altitude. And the atmospheric pressure drops as you ascent, causing the balloon to increase in volume, thereby increasing the drag on it. Thus to analyse the motion of the balloon carefully, one has to resort to numerical methods and computers. But if you are looking for an approximation the ascent rate could be taken as the terminal velocity and could be obtained by setting FB=FG+FDFB=FG+FD

                               

On a first glance it might appear that the balloon soon reaches a terminal velocity. But the quantities involved in these equations aren't all independent of each other or remain constant. For example the density of air changes with altitude. And the atmospheric pressure drops as you ascent, causing the balloon to increase in volume, thereby increasing the drag on it. Thus to analyse the motion of the balloon carefully, one has to resort to numerical methods and computers. But if you are looking for an approximation the ascent rate could be taken as the terminal velocity and could be obtained by setting IT.



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2016-03-09T11:03:06+05:30
Ascent is the (h)height to which liquid rises in a capillary tube. ascent formula is h = 2Scosα/aβg  where β is the density of the liquid, g= acce. due to gravity.
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