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3 years ago

How many liters are equivalent to 500 centiliters

Physics

2 answers:

andreev551 [17]3 years ago

4 0

5 Liters are equal to 500 centiliters

PilotLPTM [1.2K]3 years ago

4 0

1000 centiliters = 1 liter.

500 centiliters = x liter

1000 / 500 = 1/x Cross multiply

1000x = 500 Divide by 1000

x = 500/1000

x = 1/2

500 centiliters = 1/2 or 0.5 of a liter.

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GalinKa [24]

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Explanation:

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3 years ago

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A triangular plate with a non-uniform areal density has a mass M=0.500 kg. It is suspended by a pivot at P and can oscillate as

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The period of the oscillations.T = 1.2042s

Opposition is the process of any quantity or measure fluctuating repeatedly about its equilibrium value throughout time. This process is referred to as oscillation. Oscillation, a periodic fluctuation of a substance, can also be described as alternating between two values or rotating around a central value.

Typically, the mathematical formula for the moment of inertia is

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Therefore, a moment of inertia

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I=0.054

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Read more about the period of the oscillations. brainly.com/question/14394641

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1 year ago

A point P1 is located by the vector A = (3.74)î + (1.64)ĵ and a point P2 is located by the vector B = (1.60)î + (3.66)ĵ. The vec

seropon [69]

Answer:

$\vec{C} = ( \ - 2.14 \ , \ 2.02 \ )$

Explanation:

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So:

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3 years ago

‼️‼️ Please help, urgent ‼️‼️ (check photo)

Alex787 [66]

Answer: The force constant k is 10600 kg/s^2

Step by step:

Use the law of energy conservation. When the elevator hits the spring, it has a certain kinetic and a potential energy. When the elevator reaches the point of still stand the kinetic and potential energies have been transformed to work performed by the elevator in the form of friction (brake clamp) and loading the spring.

Let us define the vertical height axis as having two points: h=2m at the point of elevator hitting the spring, and h=0m at the point of stopping.

The total energy at the point h=2m is:

$E_{tot}=E_{kin}+E_{pot}\\E_{tot}= \frac{1}{2}mv^2+mg\Delta h = \frac{1}{2}2000 kg 4^2\frac{m^2}{s^2}+2000kg\, 9.8\frac{m}{s^2}2m=55200\,kg\frac{m^2}{s^2}$

The total energy at the point h=0m is:

$E_{tot}=E_{kin}+E_{pot}+Work=0+0+ Work\\E_{tot} =F_{friction}\Delta h+\frac{1}{2}k (\Delta h)^2=17000N\cdot 2m+\frac{1}{2}k\cdot 2^2 m^2$

The two Energy values are to be equal (by law of energy conservation), which allows us to determine the only unknown, namely the force constant k:

$17000N\cdot 2m+\frac{1}{2}k\cdot 2^2 m^2 = 55200 \,kg\frac{m^2}{s^2}\\k = \frac{55200-34000}{2}\,\frac{kg}{s^2}=10600\frac{kg}{s^2}$

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3 years ago

What happens in the process of gravitational condensation?

Lerok [7]

Answer:

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