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ludmilkaskok [199]
2 years ago
6

What is Virtual Lab and what does people use it for and why they use it?

Physics
1 answer:
Dafna1 [17]2 years ago
7 0

The Virtual Laboratory is an interactive environment for creating and conducting simulated experiments: a playground for experimentation. It consists of domain-dependent simulation programs, experimental units called objects that encompass data files, tools that operate on these objects

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Why does the sun appear to rise in the east and set in the west every day?
anygoal [31]

Answer: even heating

Explanation:

the earth rotates

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2 years ago
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Pls help it’s easy ik but I want to make sure of them both I don’t have much time
LuckyWell [14K]

Answer:

To the right

Weight

Explanation:

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2 years ago
How do you convert mass to weight and visa versa
Darina [25.2K]
To change from mass to weight is Fw = 30 kg * 9.8 m/s^2 = 294 N. To change from weight to mass divide by gravity (9.8 m/s^2).
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3 years ago
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The vapor pressure of diethyl ether (ether) is 463.57 mm Hg at 25°C. How many grams of chlorophyll, C55H72MgN4O5, a nonvolatile,
mixer [17]

Answer:

29.4855 grams of chlorophyll

Explanation:

From Raoult's law

Mole fraction of solvent = vapor pressure of solution ÷ vapor pressure of solvent = 457.45 mmHg ÷ 463.57 mmHg = 0.987

Mass of solvent (diethyl ether) = 187.4 g

MW of diethyl ether (C2H5OC2H5) = 74 g/mol

Number of moles of solvent = mass/MW = 187.4/74 = 2.532 mol

Let the moles of solute (chlorophyll) be y

Total moles of solution = moles of solute + moles of solvent = (y + 2.532) mol

Mole fraction of solvent = moles of solvent/total moles of solution

0.987 = 2.532/(y + 2.532)

y + 2.532 = 2.532/0.987

y + 2.532 = 2.565

y = 2.565 - 2.532 = 0.033

Moles of solute (chlorophyll) = 0.033 mol

Mass of chlorophyll = moles of chlorophyll × MW = 0.033 × 893.5 = 29.4855 grams

8 0
2 years ago
A ball initially at rest rolls down a hill with an acceleration of 3.3 m/s2. If it accelerates for 7.5 s, how far will it move?
goldfiish [28.3K]

<u>Answer:</u>

  Ball will move 92.8125 meter along the cliff in 7.5 seconds.

<u>Explanation:</u>

We have equation of motion , s= ut+\frac{1}{2} at^2, s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

In this case initial velocity = 0 m/s, acceleration = 3.3 m/s^2, we need to calculate displacement when time = 7.5 seconds.

Substituting

  s=0*7.5+\frac{1}{2} *3.3*7.5^2\\ \\ =92.8125 meter

  So ball will move 92.8125 meter along the cliff in 7.5 seconds.

5 0
3 years ago
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