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Lana71 [14]
3 years ago
6

What is one advantage of doing a feild experiment instead of a lobortory expirament

Physics
1 answer:
Scrat [10]3 years ago
5 0

Answer:

- High internal validity as there is good control over all variables

- Allows for precise control over extraneous variables and IV

- Easy to replicate due to standardised procedure

you can choose one of them

hope this can help you

You might be interested in
What is the element with the lowest electronegativity value?
Lerok [7]
<h2>Answer: Francium </h2>

Let's start by explaining that electronegativity is a term coined by Linus Pauling and is determined by the <em>ability of an atom of a certain element to attract electrons when chemically combined with another atom. </em>

So, the more electronegative an element is, the more electrons it will attract.

It should be noted that this value can not be measured directly by experiments, but it can be determined indirectly by means of calculations from other atomic or molecular properties of the element. That is why the scale created by Pauling is an arbitrary scale, where the maximum value of electronegativity is 4, assigned to Fluorine (F) and the <u>lowest is 0.7, assigned to Francium (Fr).</u>

7 0
3 years ago
An object is dropped from rest from a 70.6 m tower. Air resistance is negligible. After 0.32 seconds, what is magnitude and dire
dem82 [27]

Answer:

<em>1,378.9ms²</em>

Explanation:

Given the following

Distance S = 70.6m

Time t = 0.32secs

Initial velocity = 0m/s

Required

Acceleration

Using the equation of motion

S = ut+1/2at²

Substitute

70.6 = 0+1/2a(0.32)²

70.6 = 0.0512a

a = 70.6/0.0512

a = 1,378.9

<em>Hence the acceleration is 1,378.9ms²</em>

7 0
2 years ago
What is the equivalent resistance between points A and C if R1=1430, R2=1350, R3=1100, R4=1350, and R5=1150.
Marianna [84]

R1 + R4 = 1430 + 1350 = 2780 = R14    series combination of R1 & R4

R2 + R5 = 1350 + 1150 = 2500 = R25

The circuit has been reduced to 3 resistors in parallel

R314 = 2780 * 1100 / (2780 + 1100) = 788  this is the resistance of the parallel combination of R14 and R3

R31425 = 2500 * 788 / (2500 + 788) = 599 which is the equivalent of the circuit  - you can also use the formula for 3 resistors in parallel but this seems simpler

7 0
3 years ago
4. A family leaves from New York City and is flying to Los Angles which is 2800mi away. It takes
Blizzard [7]

The average speed of the whole travel is equal to <u>400 mph</u>.

Why?

From the statement, we know that whole travel is divided into three parts. For the first part (traveling from New York to Chicago), we have that it was 3.25 hours and the covered distance was half of the total distance (1400mi). For the second part, we have that it was 1 hour (layover time), and the covered no distance. For the third part (traveling from Chicago to Los Angeles), we have that it was 2.75 hours, and it took the other half of the total distance (1400mi).

We can calculate the average speed of the whol travel using the following formula:

AverageSpeed=\frac{distance_{1}+distance_{2}+distance_{3}}{time_{1}+time{2}+time_{3}}

Now, substituting and calculating, we have:

AverageSpeed=\frac{1400mi+0mi+1400mi}{3.25h+1h+2.75h}

AverageSpeed=\frac{2800mi}{7h}=400mph

Hence, we have the average speed of the whole travel is equal to 400 mph.

Have a nice day!

7 0
3 years ago
Consider the hydrogen atom. How does the energy difference between adjacent orbit radii change as the principal quantum number i
Kisachek [45]

Answer:

the energy difference between adjacent levels decreases as the quantum number increases

Explanation:

The energy levels of the hydrogen atom are given by the following formula:

E=-E_0 \frac{1}{n^2}

where

E_0 = 13.6 eV is a constant

n is the level number

We can write therefore the energy difference between adjacent levels as

\Delta E=-13.6 eV (\frac{1}{n^2}-\frac{1}{(n+1)^2})

We see that this difference decreases as the level number (n) increases. For example, the difference between the levels n=1 and n=2 is

\Delta E=-13.6 eV(\frac{1}{1^2}-\frac{1}{2^2})=-13.6 eV(1-\frac{1}{4})=-13.6 eV(\frac{3}{4})=-10.2 eV

While the difference between the levels n=2 and n=3 is

\Delta E=-13.6 eV(\frac{1}{2^2}-\frac{1}{3^2})=-13.6 eV(\frac{1}{4}-\frac{1}{9})=-13.6 eV(\frac{5}{36})=-1.9 eV

And so on.

So, the energy difference between adjacent levels decreases as the quantum number increases.

5 0
3 years ago
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