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guapka [62]
2 years ago
6

How many grams of sugar are in 0.5mol C6H12O6?

Physics
2 answers:
Anastasy [175]2 years ago
5 0

Answer:

90.07794 grams

Explanation:

Evgen [1.6K]2 years ago
3 0
Answer 90.07794
The answer to this question ^^^
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A rock with a mass of 8 kg falls straight down from a height of 7 m. What work is done?
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F=MA
F=(8 kg)(9.8 m/s)
F= 78.4 N
W=FD
W=(78.4 N)(7 m)
W=548.8 J
How this helps
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Dark matter may explain _____.
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Dark matter may explain <span>unexpected orbital velocities of stars in galaxies.</span>
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If the weight of the ruler is one Newton ,Gc cannot have a value more than 25cm
nevsk [136]

If the length of the ruler is 50 cm, the center of gravity cannot be greater than 25 cm.

The given parameters:

  • Weight of the ruler = 1 N

<h3>What is center of gravity (CG)?</h3>
  • Center of gravity is the point at which the weight of an object is concentrated.

Let the length of the ruler = L

The center of the gravity of the ruler is calculated as follows;

X_{CG} = \frac{W(L_0) + W(L -X_{CG})}{W} \\\\X_{CG} = \frac{1(0) + 1(L -X_{CG})}{1}\\\\X_{CG} = L - X_{CG}\\\\X_{CG } + X_{CG} = L\\\\2X_{CG} = L\\\\X_{CG} = \frac{L}{2} \\\\when , \ L = 50 \ cm\\\\X_{CG} = \frac{50}{2} \\\\X_{CG} = 25 \ cm

Thus, if the length of the ruler is 50 cm, the center of gravity cannot be greater than 25 cm. This may change if the length of the ruler changes because the center of gravity of uniform ruler depends on the length of the ruler.

Learn more about center of gravity here: brainly.com/question/6765179

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2 years ago
How long do you stay in school here in Auburn schools
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What does this supposed to mean?

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A package is dropped from an airplane flying horizontally with constant speed V in the positive xdirection. The package is relea
k0ka [10]

Answer:

D=V*\sqrt{\frac{2H}{g} } -\frac{H}{4}

Explanation:

From the vertical movement, we know that initial speed is 0, and initial height is H, so:

Y_{f}=Y_{o}-g*\frac{t^{2}}{2}

0=H-g*\frac{t^{2}}{2}    solving for t:

t=\sqrt{\frac{2H}{g} }

Now, from the horizontal movement, we know that initial speed is V and the acceleration is -g/4:

X_{f}=X_{o}+V*t+a*\frac{t^{2}}{2}   Replacing values:

D=V*\sqrt{\frac{2H}{g} }-\frac{g}{4}*\frac{1}{2} *(\sqrt{\frac{2H}{g} })^{2}

Simplifying:

D=V*\sqrt{\frac{2H}{g} } -\frac{H}{4}

4 0
3 years ago
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