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2 years ago

Assume you drink 250ml of soda out of a 1 liter soda bottle. how many liters are left?

Physics

1 answer:

charle [14.2K]2 years ago

5 0

The correct answer is 0.75 liters.

<h3>Briefing:</h3>

One liter of soda produced 250 ml of beverage.

1 liter is equal to 1000 milliliters.

250 ml was consumed, or 250/1000 or 0.25 liter.

There was 1 - 0.25 liters of soda left

0.75 liter of soda was left and the correct answer is 0.75 liter.

<h3>What is a liter of water?</h3>

Four glasses of water make up one liter, to be precise.

Let us understand this using the following explanation. Despite the fact that there is no set standard size for glasses, their capacity varies. In contrast, we calculate that a liter holds 32 ounces and a glass holds 8 ounces of water.

<h3>How are liters measured?</h3>

Convert the object's dimensions into centimeters before performing the volume calculation in litres. Next, determine a shape's volume using the volume formula. For instance, you could use Volume = length times width times height to determine the volume of a cube, and the result would be in cubic centimetres.

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There is the more role of the moon than the sun to occur tides in the oceans,why?

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Answer:

The ocean tides on earth are caused by both the moon's gravity and the sun's gravity. ... Even though the sun is much more massive and therefore has stronger overall gravity than the moon, the moon is closer to the earth so that its gravitational gradient is stronger than that of the sun.

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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

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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

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A shorter than the original source and in the researcher's words

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The summary is an abridged version of the original source and in the researcher's own very words.

Summaries gives an over-arching perspective and excludes implicit details from a given text.

A summary should not be detailed and must avoid overt illustrations. They must capture the true essence of piece leaving out flowery details.

A summary should be lesser in length than the original piece. Any third party reader should immediately be able to grab the details of the original piece from the summary.

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