State two advantages of thermal expansion

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

As a pole of a 2nd-order discrete-time system moves away from the origin in the z-plane, while its phase remains constant, the d

Rudik [331]

Answer:

False

Explanation:

When the location of the poles changes in the z-plane, the natural or resonant frequency (ω₀) changes which in turn changes the damped frequency (ωd) of the system.

As the poles of a 2nd-order discrete-time system moves away from the origin then natural frequency (ω₀) increases, which in turn increases damped oscillation frequency (ωd) of the system.

ωd = ω₀√(1 - ζ)

Where ζ is called damping ratio.

For small value of ζ

ωd ≈ ω₀

4 0

3 years ago

Which waves can u see

NemiM [27]

light waves. the other waves like gamma rays or infrared waves or radio waves are all not visible to the eye. light is the only thing out of those you can see.

3 0

3 years ago

Read 2 more answers

What are the two requirements of a circuit so that current will flow?

antoniya [11.8K]

Answer:

-The battery-the power source

-Closed conducting loop

Explanation:

-To produce an electric current, the following requirements must be met:

-A battery-This is the energy source than will do work on the charge thus moving from a low energy location to high energy location.

-Closed Conducting Loop-The loop is usually made of copper wires due to their high electric conductivity.

8 0

3 years ago

if you run around a circle at 4.5 m/s and the circle has a radius of 7.7 m, what is your centripetal acceleration?

madreJ [45]

Answer:

Centripetal acceleration,

$a_{c} =2.63\ m/s^{2} }$

Explanation:

Centripetal acceleration:

Centripetal acceleration is the idea that any object moving in a circle, in something called circular motion, will have an acceleration vector pointed towards the center of that circle.

Centripetal means towards the center.

Examples of centripetal acceleration (acceleration pointing towards the center of rotation) include such situations as cars moving on the cicular part of the road.

An acceleration is a change in velocity.

Formula for Centripetal acceleration:

$a_{c} =\frac{(velocity)^{2} }{radius}$

Given here,

Velocity = 4.5 m/s

radius = 7.7 m

To Find :

$a_{c} = ?$

Solution:

We have,

$a_{c} =\frac{(velocity)^{2} }{radius}$

Substituting given value in it we get

$a_{c} =\frac{(4.5)^{2}}{7.7} \\\\a_{c} =\frac{20.25}{7.7}\\\\a_{c} =2.629\ m/s^{2} \\\\\therefore a_{c} =2.63\ m/s^{2$

Centripetal acceleration,

$a_{c} =2.63\ m/s^{2} }$

7 0

3 years ago

State two advantages of thermal expansion​

State two advantages of thermal expansion