All categories

3 years ago

Which surface most likely has the least friction?

Physics

2 answers:

vitfil [10]3 years ago

5 0

The answer will be a water slide :)

elixir [45]3 years ago

3 0

Wood would likely have little friction.

You might be interested in

In a series circuit what effect will adding more resistors to the circuit have

zhuklara [117]

Answer:In a series circuit, adding more resistors increases total resistance and thus lowers current. But the opposite is true in a parallel circuit because adding more resistors in parallel creates more choices and lowers total resistance. If the same battery is connected to the resistors, current will increase.

Explanation:

8 0

3 years ago

Three beakers are of identical shape and size , one beaker is pained matt black, one is dull white and one is gloss white . The

JulsSmile [24]

Answer:

The beaker that is matt black heats more quickly because black attracts more heat.

Explanation:

5 0

3 years ago

[The illustration shows a lamp giving off light. What form of energy does the lamp use to produce light?

AnnZ [28]

Answer:

I think it's electrical energy

7 0

3 years ago

A mercury barometer reads 745.0 mm on the roof of a building and 760.0 mm on the ground. Assuming a constant value of 1.29 kg/m3

Ipatiy [6.2K]

Answer:

The height of the building is 158.140 meters.

Explanation:

A barometer is system that helps measuring atmospheric pressure. Manometric pressure is the difference between total and atmospheric pressures. Manometric pressure difference is directly proportional to fluid density and height difference. That is:

$\Delta P \propto \rho \cdot \Delta h$

$\Delta P = k \cdot \rho \cdot \Delta h$

Where:

$\Delta P$ - Manometric pressure difference, measured in kilopascals.

$\rho$ - Fluid density, measured in kilograms per cubic meter.

$\Delta h$ - Height difference, measured in meters.

Now, an equivalent height difference with a different fluid can be found by eliminating manometric pressure and proportionality constant:

$\rho_{air} \cdot \Delta h_{air} = \rho_{Hg} \cdot \Delta h_{Hg}$

$\Delta h_{air} = \frac{\rho_{Hg}}{\rho_{air}} \cdot \Delta h_{Hg}$

Where:

$\Delta h_{air}$ - Height difference of the air column, measured in meters.

$\Delta h_{Hg}$ - Height difference of the mercury column, measured in meters.

$\rho_{air}$ - Density of air, measured in kilograms per cubic meter.

$\rho_{Hg}$ - Density of mercury, measured in kilograms per cubic meter.

If $\Delta h_{Hg} = 0.015\,m$ , $\rho_{air} = 1.29\,\frac{kg}{m^{3}}$ and $\rho_{Hg} = 13600\,\frac{kg}{m^{3}}$ , the height difference of the air column is:

$\Delta h_{air} = \frac{13600\,\frac{kg}{m^{3}} }{1.29\,\frac{kg}{m^{3}} }\times (0.015\,m)$

$\Delta h_{air} = 158.140\,m$

The height of the building is 158.140 meters.

5 0

3 years ago

Read 2 more answers

Use the SI prefixes in Table 3 of this chapter to convert these hypothetical units of measure into appropriate quantities: a. 10

Amanda [17]

<span>The first response would be "deca," since this is a multiple of 10^1. B, since it's working off the thousands prefix (10^3), would be "kilo." The third, at 10^-6, would be "micro." Next, at 10^-9, would be "nano," and the final, 10^18, would have a prefix of "exa."</span>

6 0

4 years ago