Answer:
the answers the correct one is d
Explanation:
The speed of sound is constant so we can use the relations of uniform motion
v = x / t
x = v t
now let's calculate the distance for each person
t = 5s
x₁ = 300 5
x₁ = 1500 m
t = 6s
x₂ = 300 6
x₂ = 1800 m
therefore we have two possibilities
a) the two people are on the same side, therefore the distance between them is
Δx = x₂- x₁
Δx = 1800 - 1500
Δx = 300 m
let's reduce to km
Δx = 0.300 km
b) people are on opposite sides of the sound
Δx = x₂ + x₁
Δx = 1800 + 1500
Δx = 3300 m
Δx = 3.3 km
when checking the answers the correct one is d
The metals will start to rust lol. i think. because this messes up how the metals conduct the flow of the electricity.
Answer:
A. At warm tempetures, molecules move around more.
Explanation:
I'm at k12 and I just took the test got it right. Physical Science: Unit 2 Test
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:
