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

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Acceleration of 1.5 ms expressed in km /hr2?

V125BC [204]

You’re answer is 5 because !! :)

8 0

3 years ago

A 50n box is lifted 2 meters in 3 seconds.

anygoal [31]

F=G=50 N

L=F*d=50*2=100 J

is the work

P=L/t=100/3=33.33 W

4 0

3 years ago

The electric potential at the origin of an xy-coordinate system is 40 V. A -8.0-μC charge is brought from x = +∞ to that point.

vredina [299]

Answer:

-320 μJ.

Explanation:

Consider a point with an electrical charge of $q$ . Assume that $V$ is the electrical potential at the position of that charge. The electrical potential of that point charge will be equal to:

$\text{Potential Energy} = q \cdot V$ .

Keep in mind that since both $q$ and $V$ might not be positive, the size of the electrical potential energy might not be positive, either.

For this point charge,

$q = \rm -8.0\; \mu C$ ; (that's -8.0 microjoules, which equals to $\rm -8.0\times 10^{-6}\; J$ )
$V = \rm 40\; V$ .

Hence its electrical potential energy:

$\text{Potential Energy} = q\cdot V = \rm (-8.0\; \mu C) \times 40\; V = -320\; \mu J$ .

Why is this value negative? The electrical potential energy of a charge is equal to the work needed to bring that charge from infinitely far away all the way to its current position. Also, negative charges are attracted towards regions of high electrical potential. Bringing this $\rm -8.0\; \mu C$ negative charge to the origin will not require any external work. Instead, this process will release 320 μJ of energy. As a result, the electrical potential energy is a negative value.

7 0

3 years ago

The output voltage of a power supply is normally distributed with mean 5 V and standard deviation 0.02 V. If the lower and upper

-BARSIC- [3]

To solve this problem we will apply the normal distribution, with which we will obtain the probability that the given event will occur. Concepts such as the mean and standard deviation will be present throughout the solution of the problem. Increasing or decreasing the average would change the location or center point of the curve. The change in the standard deviation would lead to the change in the dispersion of the data. As the standard deviation increases, the curve would become flatter.

Let X be the output voltage of power supply

X∼N $(5,0.02^2)$