Classify each model A-D as either an element, a compound, or a mixture. Explain your reasoning for each answer

How can liquid pressure phenomena be used in daily life. Give one example with explanation of working?

N76 [4]

Answer:

An example in which liquid pressure phenomena can be used in daily life is in Water blasting

Explanation:

Water blasting refers application of pressurized water to remove materials from the surface of objects.

There are different varieties of water blasting, including;

Hydrocleaning; Cleaning enabled by the use of high pressure water

Hydrodemolition; Demolition or removal of concrete using pressurized water

Hydrojetting; The spraying of water under pressure on surfaces in order to remove surface contaminants.

8 0

3 years ago

You drop a steel ball bearing, with a radius of 2.40 mm, into a beaker of honey. Note that honey has a viscosity of 6.00 Pa/s an

Stells [14]

Answer:

The “terminal speed” of the ball bearing is 5.609 m/s

Explanation:

Radius of the steel ball R = 2.40 mm

Viscosity of honey η = 6.0 Pa/s

$\text { Viscosity has Density } \sigma=1360 \mathrm{kg} / \mathrm{m}^{3}$

$\text { Steel has a density } \rho=7800 \mathrm{kg} / \mathrm{m}^{3}$

$\left.\mathrm{g}=9.8 \mathrm{m} / \mathrm{s}^{2} \text { (g is referred to as the acceleration of gravity. Its value is } 9.8 \mathrm{m} / \mathrm{s}^{2} \text { on Earth }\right)$

While calculating the terminal speed in liquids where density is high the stokes law is used for viscous force and buoyant force is taken into consideration for effective weight of the object. So the expression for terminal speed (Vt)

$V_{t}=\frac{2 \mathrm{R}^{2}(\rho-\sigma) \mathrm{g}}{9 \eta}$

Substitute the given values to find "terminal speed"

$\mathrm{V}_{\mathrm{t}}=\frac{2 \times 0.0024^{2}(7800-1360) 9.8}{9 \times 6}$

$\mathrm{V}_{\mathrm{t}}=\frac{0.0048 \times 6440 \times 9.8}{54}$

$\mathrm{V}_{\mathrm{t}}=\frac{302.9376}{54}$

$\mathrm{V}_{\mathrm{t}}=5.609 \mathrm{m} / \mathrm{s}$

The “terminal speed” of the ball bearing is 5.609 m/s

7 0

3 years ago

Consider two children sitting on a merry-go-round, with one closer to the outer edge and one closer to the center. show answer N

dolphi86 [110]

Answer:

They both have the same angular speed.

Explanation:

The mathematical formula for angular speed is:

$w=\frac{2\pi}{T}$

where $w$ is angular speed, $2\pi$ is a constant, and $T$ is the period (the time it takes the marry-go-round to complete a lap).

What we can see from the formula is that, since the $2\pi$ does not change its value, the angular speed depends only on the period T.

In this case for both the children closer to the outher edge and for the children closer to the center, the time to complete a lap is the same, because the time does not depend on where they are sitting in the marry go round. This means that the period for both is the same.

Thus, since the period for both is the same, the angular speed given by

$w=\frac{2\pi}{T}$ will also be the same

4 0

3 years ago

Which of the following is a terrestrial habitat?<br>A) Pond B) Garden C) Lake D) River

kenny6666 [7]

Answer:

garden

Explanation: All the other habitats are aquatic

8 0

3 years ago

If predators are increased, then the cricket frog population will (increase/decrease/stay the same) over the span of five years.

12345 [234]

With an increase of predators, then the population will most likely decrease over the span of 5 years.

4 0

3 years ago

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