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harkovskaia [24]
3 years ago
13

Describe the parts of the wheel and axle

Physics
2 answers:
suter [353]3 years ago
8 0

Answer:

A simple machine that may be used the most often is called the wheel and axle. The wheel and axle has two basic parts: wheel and axle.  It has two circular objects which includes a larger disc and a small cylinder both joined at the center.

Explanation:

Natalka [10]3 years ago
3 0

Explanation:

A simple machine that may be used the most often is called the wheel and axle. The wheel and axle has two basic parts: wheel and axle. They can be found everywhere. It has two circular objects which includes a larger disc and a small cylinder both joined at the center.The large disc is the wheel, and the small cylinder or rod is the axle. There may be two wheels attached to the axle.

There are two basic ways a wheel and axle can work together to help move things.

1. The Force is applied to the Wheel

For example, a screwdriver is an example of a wheel and axle. The handle is the wheel where the force is applied. It turns or spins and increases the force of the shaft or axle, which helps turn the screw.

Another example of force being applied to the wheel is when a doorknob is turned. The wheel (doorknob) is turned and the locking mechanism connected to the shaft turns and the door can then be opened.

2. The Force is applied to the Axle

A Ferris wheel is an example of force being applied to the axle. When the axle turns it results in the giant wheel turning. The wheel is much larger than the axle and covers more distance and area. A ceiling fan works the same way.

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What is the relative humidity if the dry-bulb temperature is 18°C and the wet-bulb temperature is 11°C?
Diano4ka-milaya [45]

Answer:

The amount of moisture air can hold is dependent on its temperature and pressure. The warmer the air the greater the quantity of water vapour it can contain. The air temperature is measured with a normal thermometer this is the Dry-Bulb reading. The actual amount of moisture known as the mixing ratio is measured in grams of water per kilogram of dry air. When air at a certain temperature is saturated it cannot hold any more moisture. The relative humidity of the air is the ratio of the actual amount of moisture in the air to the fully saturated amount.

We can use evaporation to measure the amount of moisture in the air. A wet cloth is placed over the bulb of a thermometer and then air is blown over the cloth causing the water to evaporate. Since evaporation takes up heat, the thermometer will cool to a lower temperature than a thermometer with a dry bulb at the same time and place. The depression in Wet-Bulb temperature allows the humidity to be calculated. If the air is fully saturated (100% relative humidity) the water cannot evaporate, so both the wet and dry bulb temperatures are the same. If the dry and wet bulb temperatures are set to the same value the calculator will show the saturation mixing ratio the amount of water in saturated air at that temperature.

If partly saturated air is cooled without changing its pressure or amount of water vapour, a point is reached when it becomes saturated. The moisture will be given up as dew or ice crystals. This temperature is the Dew Point. This is why condensation will form on a bottle of cold beer, as the air in the vicinity of the bottle is cooled below its dew point. Meteorological reports usually quote the temperature and dew point as well as the station pressure. Moisture content and relative humidity can be calculated from these figures.

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Explanation:

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4 years ago
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Answer:

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3 years ago
(m = 4.0 kg) the cockroach rides on the rim (at radius R) of a disk (M = 6.0 kg) that turns about its center like a merry-go-rou
kvasek [131]

Answer:

ω' = 2.5 rad/s

Explanation:

mass of cockroach, m = 4 kg

mass of disk, M = 6 kg

Radius of disc= R

initial angular velocity, ω = 2 rad/s

Let the final angular velocity is ω'

As no external torque is applied, so the angular momentum is constant.

Angular momentum = Moment of inertia x angular velocity

I ω = I' ω'

\frac{1}{2}\left ( M+m \right )R^{2}\times {2} = \left (\frac{1}{2}MR^{2}+m(0.5R)^{2})  \right )\omega '

10R^{2} = 4R^{2}\omega '

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Explanation:

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