Do sponges have of appendages jointed, not Jointed, or absent

Which two surfaces would have the highest coefficient of friction?

scoray [572]

1. sandpaper against wood.
2. rubber mat against the floor of the shower stall.

# High coefficient of friction describes the degree of interaction between two surfaces. . A higher coefficient of friction indicates that two surfaces in contact have a greater resistance.

5 0

3 years ago

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Explain the meaning and use of the following acronyms: HELIOS, Exo-FMS, HAZMAT, NIRSpec.

adelina 88 [10]

Answer:

HELIOS- The god of the sun in

HAZMAT-Hazardous materials

NIRSpec- Near Infrared Spectrograph

Explanation:

4 0

2 years ago

Why are some substances dissolve both polar and nonpolar substances?

dlinn [17]

Answer:

Because water is polar and oil is nonpolar, their molecules are not attracted to each other. polar solvents dissolve polar solutes, and nonpolar solvents dissolve nonpolar solutes.

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use your brain :)

8 0

3 years ago

Why do astronauts float aboard the international space station?

lianna [129]

They’re falling toward earth & moving forward at about the same velocity. because the downward and forward forces are nearly equal, the astronauts are not pulled in any specific direction, so they float .

8 0

3 years ago

Water is leaking out of an inverted conical tank at a rate of 10,500 cm3/min at the same time that water is being pumped into th

satela [25.4K]

The tank has a volume of $\dfrac\pi3R^2H$ , where $H=6\,\rm m$ is its height and $R=\dfrac d2=2\,\rm m$ is its radius.

At any point, the water filling the tank and the tank itself form a pair of similar triangles (see the attached picture) from which we obtain the following relationship:

$\dfrac26=\dfrac rh\implies r=\dfrac h3$

The volume of water in the tank at any given time is

$V=\dfrac\pi3r^2h$

and can be expressed as a function of the water level alone:

$V=\dfrac\pi3\left(\frac h3\right)^2h=\dfrac\pi{27}h^3$

Implicity differentiating both sides with respect to time $t$ gives

$\dfrac{\mathrm dV}{\mathrm dt}=\dfrac\pi9h^2\,\dfrac{\mathrm dh}{\mathrm dt}$

We're told the water level rises at a rate of $\dfrac{\mathrm dh}{\mathrm dt}=20\,\frac{\rm cm}{\rm min}$ at the time when the water level is $h=2\,\mathrm m=200\,\mathrm{cm}$ , so the net change in the volume of water $\dfrac{\mathrm dV}{\mathrm dt}$ can be computed:

$\dfrac{\mathrm dV}{\mathrm dt}=\dfrac\pi9(200\,\mathrm{cm})^2\left(20\,\dfrac{\rm cm}{\rm min}\right)=\dfrac{800,000\pi}9\,\dfrac{\mathrm{cm}^3}{\rm min}$

The net rate of change in volume is the difference between the rate at which water is pumped into the tank and the rate at which it is leaking out:

$\dfrac{\mathrm dV}{\mathrm dt}=(\text{rate in})-(\text{rate out})$

We're told the water is leaking out at a rate of $10,500\,\frac{\mathrm{cm}^3}{\rm min}$ , so we find the rate at which it's being pumped in to be

$\dfrac{800,000\pi}9\,\dfrac{\mathrm{cm}^3}{\rm min}=(\text{rate in})-10,500\,\dfrac{\mathrm{cm}^3}{\rm min}$

$\implies\text{rate in}\approx289,753\,\dfrac{\mathrm{cm}^3}{\rm min}$

4 0

3 years ago