SOMEBODY ONCE TOLD ME THE WORLD WAS GONNA ROLL ME

A gasoline tank has the shape of an inverted right circular cone with base radius 4 meters and height 5 meters. Gasoline is bein

RSB [31]

Answer:

h'=0.25m/s

Explanation:

In order to solve this problem, we need to start by drawing a diagram of the given situation. (See attached image).

So, the problem talks about an inverted circular cone with a given height and radius. The problem also tells us that water is being pumped into the tank at a rate of $8m^{3}/s$ . As you may see, the problem is talking about a rate of volume over time. So we need to relate the volume, with the height of the cone with its radius. This relation is found on the volume of a cone formula:

$V_{cone}=\frac{1}{3} \pi r^{2}h$

notie the volume formula has two unknowns or variables, so we need to relate the radius with the height with an equation we can use to rewrite our volume formula in terms of either the radius or the height. Since in this case the problem wants us to find the rate of change over time of the height of the gasoline tank, we will need to rewrite our formula in terms of the height h.

If we take a look at a cross section of the cone, we can see that we can use similar triangles to find the equation we are looking for. When using similar triangles we get:

$\frac {r}{h}=\frac{4}{5}$

When solving for r, we get:

$r=\frac{4}{5}h$

so we can substitute this into our volume of a cone formula:

$V_{cone}=\frac{1}{3} \pi (\frac{4}{5}h)^{2}h$

which simplifies to:

$V_{cone}=\frac{1}{3} \pi (\frac{16}{25}h^{2})h$

$V_{cone}=\frac{16}{75} \pi h^{3}$

So now we can proceed and find the partial derivative over time of each of the sides of the equation, so we get:

$\frac{dV}{dt}= \frac{16}{75} \pi (3)h^{2} \frac{dh}{dt}$

Which simplifies to:

$\frac{dV}{dt}= \frac{16}{25} \pi h^{2} \frac{dh}{dt}$

So now I can solve the equation for dh/dt (the rate of height over time, the velocity at which height is increasing)

So we get:

$\frac{dh}{dt}= \frac{(dV/dt)(25)}{16 \pi h^{2}}$

Now we can substitute the provided values into our equation. So we get:

$\frac{dh}{dt}= \frac{(8m^{3}/s)(25)}{16 \pi (4m)^{2}}$

so:

$\frac{dh}{dt}=0.25m/s$

3 0

3 years ago

Which cannot be separated into simpler substances?

marin [14]

Answer:

An element is a pure substance that cannot be separated into simpler substances by chemical or physical means. There are about 117 elements.

Explanation:

6 0

3 years ago

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Where is the chart option found in the Excel application?

vodomira [7]

Answer:

its under the insert table

Explanation:

5 0

4 years ago

What is the relationship between ozone and smog?

Irina18 [472]

smog is a term used to describe air pollution that is a result of the interaction of sunlight with certain chemicals in the atmosphere. One of the primary components of photochemical smog is ozone. While ozone in the stratosphere protects earth from harmful UV radiation, ozone on the ground is hazardous to human health.

4 0

4 years ago

While working on her science fair project Venus connected a battery to a circuit that contained a light bulb. Venus decided to c

Dmitriy789 [7]

<h2>Answer:</h2>

<u>A) Increase the voltage by adding a bigger battery </u>

<h2>Explanation:</h2>

According to Ohm's law

V = IR

where V is voltage, I is current and R is the resistance. If we write the equation for resistance we would get

R= V / I

Here we can see that Voltage is directly proportional to Resistance so in order to keep the balance if we increase the resistance then we must increase the voltage to keep the current constant.

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3 years ago

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