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3 years ago

A tick is getting nutrients from a dog it has attached to. The dog in this scenario is the:

Physics

2 answers:

Daniel [21]3 years ago

8 0

The dog in this scenario is the host. A host is an object another organism is feeding off of.

Vesna [10]3 years ago

7 0

A tick is getting nutrients from a dog it has attached to. The dog in this scenario is the host.

The tick would be the parasite.

Hope this helps,

Davinia.

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There isnt enough sanitidzer(alcohol)what solution to be replaced

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For me: WASH OUR HANDS REGULARLY

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2 years ago

An acorn with a mass of .0300 kg is hanging from a branch in a tree it is 2.50 M off the ground what is the potential energy of

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According to another source this is what I got
0.735 J ( Ep-potential energy, m-mass,g-gravitational acceleration = 9.81m/s², h-height; Ep = m * g * h; Ep = 0.0300 kg * 9.81 m/s² * 2.5 m )
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Desmond builds a set of towers out of blocks. The table shows the number of blocks in each tower. Tower # 1 2 3 4 5 Number of bl

insens350 [35]

Answer:B

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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:

$\frac{dh}{dt}=0.25m/s$

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3 years ago

(15 points) (Asap!!) In what two ways can you increase the elastic potential energy of a spring?

adelina 88 [10]

Hello There

Answers: The elastic potential energy can be increased by: 

1) Getting a spring with a higher spring constant

2) Increasing the length at which the spring is compressed.

Reasons: Getting a stronger spring makes it stronger which equals more energy. While increasing the compression on the spring, increases the stored energy which makes it more powerful when its released

I hope this helps
-Chris

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