Sort the units based on the measurement system they're used in

Include units with all of your answers. A 1.84 kg bucket full of water is attached to a 0.76 m long string.

aleksandrvk [35]

<span>c. What is the magnitude of the tension in the string at the bottom of the circle if you are swinging it at 3.37 m/s? </span>

7 0

3 years ago

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

How do we calculate the value of work?

Katarina [22]

The percent complete is calculated by dividing the quantity of material progressed at a point in time by the total quantity required for the project. The resulting percent is multiplied by the current agreed committed value of the material item to obtain the VOWD for that item.

7 0

3 years ago

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Which tagline from a car advertisement has the BEST connotative appeal to the parents of a new teenage driver?

Nonamiya [84]

A. Advanced new safety features that get everyone home at night.

6 0

3 years ago

A vehicle starts to move from the rest gets an acceleration of 5 m/s2 within 2

likoan [24]

Answer:

When an object moves in a straight line with a constant acceleration, you can calculate its acceleration if you know how much its velocity changes and how long this takes.

The formula is,

Acceleration = change in velocity / time taken

The equation for acceleration can also be represented as:

a = (v-u) \ t

The change in velocity v – u = 5 – 0 = 5 m/s.

The acceleration = change in velocity ÷ time = 5 m/s ÷ 2 s = 2.5 m/s^2

3 0

2 years ago

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