V = 4/3*pi*r^3dV/dt = 4/3*pi*3*r^2*dr/dt = 4*pi*r^2*dr/dt
Therefore dV/dt = 5 => 4*pi*r^2*dr/dt = 5
therefore dr/dt = 5/(4*pi*r^2)
Diameter = 12 metres => radius = 6 metres
So dr/dt = 5/(4*pi*36) = 0.01105 metres per minute.
diameter = 2*radius => rate of change of diameter = 0.2210 metres per minute.
a water tank in the form of an inverted cone is being emptied at the rate 6 m3/min. the altitude of the cone is 24 m, and the base radius is 12 m. find how fast the water level is lowering when the water is 10 m deep.
A balloon payment calculator is not actually in the shape of a balloon. It is used to calculate a balloon payment; it is called a balloon payment because of its size.
To find the volume of an inflated balloon, you can measure its diameter using a ruler and then use the formula for the volume of a sphere, V = (4/3) * pi * r^3, where r is the radius (half of the diameter) of the balloon. Plug in the radius and calculate the volume. Alternatively, you can submerge the inflated balloon in a container of water and measure the water displacement to find the volume of the balloon.
a water tank in the form of an inverted cone is being emptied at the rate 6 m3/min. the altitude of the cone is 24 m, and the base radius is 12 m. find how fast the water level is lowering when the water is 10 m deep.
When a balloon is inflated, the pressure inside the balloon increases. This is because as the balloon expands, the air molecules become more tightly packed together, increasing the pressure. The pressure remains higher inside the balloon until the balloon is stretched to its maximum capacity.
There is not a PICC line with an inflated balloon for placement.
You did
Assuming the balloon is perfectly spherical and that the 12" you state refers to the diameter of the balloon then it is obviously 6". If you can't assume any of the above then your question cannot be answered.
There is no difference at all if the balloon is not inflated or inflated with air.
A deflated balloon is lighter than an inflated balloon because the air inside the inflated balloon adds weight to it. When the air is released, the balloon becomes lighter because it is no longer burdened by the added weight of the air molecules.
When the balloon is inflated, it is filled with air, which has less mass than the solid material of the deflated balloon. So overall, the balloon has less mass when inflated because the mass of the air inside it is lighter than the solid material of the balloon.
No, the mass of the deflated balloon is the same as the mass of the inflated balloon, as the only thing that changes is the volume and density of the air inside the balloon.
A 31-inch balloon would have a diameter of 31 inches when fully inflated. This means the width of the balloon, from one side to the other, would be approximately 31 inches.
The heat from the sun causes the air molecules inside the balloon to expand, increasing the pressure within the balloon. If the pressure becomes too high, the balloon will burst to release the excess pressure.