All categories

3 years ago

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.

You might be interested in

3. Will a plant grow higher with more light?

sleet_krkn [62]

Answer:

When we analyze the sentence we see that this is a hypotype with the growth of plants must behave and it has a prediction included.

Therefore the correct answer is a

Explanation:

In this exercise you are asked to identify the given sentence with a specific part of the scientific method.

Among the parts of the method we have.

* Independent variable . The controlled variable in research

*Dependent variable. The magnitude measured in the experiment

* Control variable. The magnitude that is not controlled

*Experiment. It is the design of the procedure to evaluate the hypothesis

* Hypothesis. It is the assumption with which scientific work begins

* Prediction. It is a consequence of work if the mortgage is correct.

When we analyze the sentence we see that this is a hypotype with the growth of plants must behave and it has a prediction included.

Therefore the correct answer is a

6 0

2 years ago

When a body is moving with a uniform velocity, the acceleration is ___?

denis23 [38]

According to the formula :
a=ΔV / ΔT
When a body is moving with a uniform velocity, the acceleration is zero. That's it. You should remember, that velocity is not constant whereas speed is constant.

4 0

2 years ago

(4.3 x 10-2 )4.950 x 105) O 21.285 x 103 O 2.13 x 104 O 21 x 103 O 2.1 x 104

Natalka [10]

Answer:

The answer is:

It's 21.285 × 103

7 0

2 years ago

The energy from 0.015 moles of octane was used to heat 250 grams of water. The temperature of the water rose from 293.0 K to 371

arsen [322]

Answer : The correct option is, (B) -5448 kJ/mol

Explanation :

First we have to calculate the heat required by water.

$q=m\times c\times (T_2-T_1)$

where,

q = heat required by water = ?

m = mass of water = 250 g

c = specific heat capacity of water = $4.18J/g.K$

$T_1$ = initial temperature of water = 293.0 K

$T_2$ = final temperature of water = 371.2 K

Now put all the given values in the above formula, we get:

$q=250g\times 4.18J/g.K\times (371.2-293.0)K$

$q=81719J$

Now we have to calculate the enthalpy of combustion of octane.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy of combustion of octane = ?

q = heat released = -81719 J

n = moles of octane = 0.015 moles

Now put all the given values in the above formula, we get:

$\Delta H=\frac{-81719J}{0.015mole}$

$\Delta H=-5447933.333J/mol=-5447.9kJ/mol\approx -5448kJ/mol$

Therefore, the enthalpy of combustion of octane is -5448 kJ/mol.

5 0

3 years ago

A positive point charge Q is located at x=a and a negative point charge −Q is at x=−a. A positive charge q can be placed anywher

Scilla [17]

Answer:

$\vec{F}_x = \frac{2KQqa}{(a^2 + y^2)^{3/2}} \^x$

Explanation:

The Coulomb's Law gives the force by the charges:

$\vec{F} = K\frac{q_1q_2}{r^2}\^r$

Let us denote the positon of the charge q on the y-axis as 'y'.

The force between 'Q' and'q' is

$F_1 = K\frac{Qq}{x^2 + y^2}\\F_1_x = F_1\cos(\theta)$

where Θ is the angle between $F_1$ and x-axis.

$F_1_x = K\frac{Qq}{x^2 + y^2}(\frac{x}{\sqrt{x^2 + y^2}}) = \frac{KQqa}{(a^2 + y^2)^{3/2}}$

whereas

$F_2_x = K\frac{-Qq}{a^2 + y^2}(-\frac{a}{\sqrt{a^2 + y^2}}) = \frac{KQqa}{(a^2 + y^2)^{3/2}}$

Finally, the x-component of the net force is

$\vec{F}_x = \frac{2KQqa}{(a^2 + y^2)^{3/2}} \^x$

8 0

3 years ago