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koban [17]
3 years ago
8

Your code returns a number of 99.123456789 +0.00455679 for your calculation. How should you report it in your lab write-up?

Physics
1 answer:
Aneli [31]3 years ago
6 0

Answer: Your code returns a number of 99.123456789 +0.00455679

Ok, you must see where the error starts to affect your number.

In this case, is in the third decimal.

So you will write 99.123 +- 0.004 da da da.

But you must round your results. In the number you can see that after the 3 comes a 4, so the 3 stays as it is.

in the error, after the 4 comes a 5, so it rounds up.

So the final presentation will be 99.123 +- 0.005

you are discarding all the other decimals because the error "domains" them.

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Answer:

distance between the dime and the mirror, u = 0.30 m

Given:

Radius of curvature, r = 0.40 m

magnification, m = - 2 (since,inverted image)

Solution:

Focal length is half the radius of curvature, f = \frac{r}{2}

f = \frac{0.40}{2} = 0.20 m

Now,

m = - \frac{v}{u}

- 2 = -\frac{v}{u}

\frac{v}{u} = 2                  (2)

Now, by lens maker formula:

\frac{1}{f} = \frac{1}{u} + \frac{1}{v}

\frac{1}{v} = \frac{1}{f} - \frac{1}{u}

v = \frac{uf}{u - f}            (3)

From eqn (2):

v = 2u

put v = 2u in eqn (3):

2u = \frac{uf}{u - f}

2 = \frac{f}{u - f}

2(u - 0.20) = 0.20

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What controls how fast an object falls?
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Read 2 more answers
Glycerin (C3H8O3) is a nonvolatile liquid. What is the vapor pressure of a solution made by adding 154 g of glycerin to 316 mL o
garri49 [273]

Answer:

P_{sol}=50.4\ mm.Hg

Explanation:

According to given:

  • molecular mass of glycerin, M_g=3\times 12+8+3\times 16=92\ g.mol^{-1}
  • molecular mass of water, M_w=2+16=18\ g.mol^{-1}
  • ∵Density of water is 0.992\ g.cm^{-3}= 0.992\ g.mL^{-1}
  • ∴mass of water in 316 mL, m_w=316\times 0.992=313.5 g
  • mass of glycerin, m_g=154\ g
  • pressure of mixture, P_x=55.32\ torr= 55.32\ mm.Hg
  • temperature of mixture, T_x=40^{\circ}C

<em>Upon the formation of solution the vapour pressure will be reduced since we have one component of solution as non-volatile.</em>

<u>moles of water in the given quantity:</u>

n_w=\frac{m_w}{M_w}

n_w=\frac{313.5}{18}

n_w=17.42 moles

<u>moles of glycerin in the given quantity:</u>

n_g=\frac{m_g}{M_g}

n_g=\frac{154}{92}

n_g=1.674 moles

<u>Now the mole fraction of water:</u>

X_w=\frac{n_w}{n_w+n_g}

X_w=\frac{17.42}{17.42+1.674}

X_w=0.912

<em>Since glycerin is non-volatile in nature so the vapor pressure of the resulting solution will be due to water only.</em>

\therefore P_{sol}=X_w\times P_x

\therefore P_{sol}=0.912\times 55.32

P_{sol}=50.4\ mm.Hg

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