The circulation system for mammals is very complex
Answer:
Explanation:
<u>Instant Velocity and Acceleration
</u>
Give the position of an object as a function of time y(x), the instant velocity can be obtained by

Where y'(x) is the first derivative of y respect to time x. The instant acceleration is given by

We are given the function for y

Note we have changed the last term to be quadratic, so the question has more sense.
The velocity is

And the acceleration is

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
. 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:

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:

When solving for r, we get:

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

which simplifies to:


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

Which simplifies to:

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:

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

so:

Answer:
54.6°
Explanation:
From law of reflection i=r.
So, construct the reflected ray at 55.7°degrees from the normal and let it fall on the other mirror.
Now draw the second normal at the point of incidence and again measure the angle of incidence, and draw the angle of reflection.
If you consider triangle AOB, one angle is ∠AOB=90°
and ∠OAB is 54.6°
From angle sum property third angle ie ∠ABO=180°-90°-54.6°=35.4°
So, the second incident angle will be 54.6°
Hence, the second reflected angle will be 54.6 degrees.
The correct answer is (A). The speed of light would increase to a speed larger than the maximum speed of light in vacuum.
The index of refraction is the ratio of speed of light in vacuum to the speed of light in a medium.
n=C/V
here, n is the index of refraction, c the speed of light in vacuum, v is speed of light in any medium.
Now if the value of index of refraction is less than one, than the value of speed of light would be greater than the speed of light in the vacuum.