What's a beam of light?<br>

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

2 years ago

An empty oﬃce chair is at rest on a ﬂoor. Consider the following forces:. 1. A downward force due to gravity;. 2. An upward forc

Bumek [7]

4. 1 and 2 only.

1. the downward force is the force of gravity.

<span>2. The upward force exerted is the Normal reaction from the floor.</span>

8 0

3 years ago

A horizontal force of 400 N is exerted on a 2.0-kg ball as it rotates (at

frutty [35]

Answer:

the speed of the ball is 10 m/s

Explanation:

Given;

magnitude of exerted force, F = 400 N

mass of the ball, m = 2 kg

radius of the circle, r = 0.5

The speed of the ball is calculated by applying centripetal force formula;

$F = \frac{mv^2}{r} \\\\v^2 = \frac{Fr}{m}\\\\v = \sqrt{\frac{Fr}{m}}\\\\ v = \sqrt{\frac{400*0.5}{2}}\\\\v = 10 \ m/s$

Therefore, the speed of the ball is 10 m/s

6 0

3 years ago

Read 2 more answers

Faraday's Law states that the negative of the time rate of change of the flux of the magnetic field through a surface is equal t

MrRa [10]

Answer:

(C). The line integral of the magnetic field around a closed loop

Explanation:

Faraday's law states that induced emf is directly proportional to the time rate of change of magnetic flux.

This can be written mathematically as;

$EMF = -\frac{\delta \phi _B}{\delta t}$

$(\frac{\delta \phi _B}{\delta t} )$ is the rate of change of the magnetic flux through a surface bounded by the loop.

ΔФ = BA

where;

ΔФ is change in flux

B is the magnetic field

A is the area of the loop

Thus, according to Faraday's law of electric generators

∫BdL = $\frac{\delta \phi _B}{\delta t}$ = EMF

Therefore, the line integral of the magnetic field around a closed loop is equal to the negative of the rate of change of the magnetic flux through the area enclosed by the loop.

The correct option is "C"

(C). The line integral of the magnetic field around a closed loop

8 0

3 years ago

The three mountains task examines the development of

GalinKa [24]

Answer:

The Three Mountain Task was developed by Jean Piaget and Bärbel Inhelder in the 1940s to study children's ability to coordinate spatial perspectives. In the task, a child faced a display of three model mountains while a researcher placed a doll at different viewpoints of the display.

Explanation:

6 0

2 years ago

What's a beam of light?​

What's a beam of light?