Which of the following is true about organizing items and information during scientific investigations?

Physics

2 answers:

LekaFEV [45]3 years ago

5 0

There are usually multiple methods for organizing items and information in a scientific investigation.

frez [133]3 years ago

3 0

Answer:

There are usually multiple methods for organizing items and information in a scientific investigation.

Explanation:

<em>There are usually multiple methods for organizing items and information in a scientific investigation.</em> For example, a scientist might keep separate records for each trial he or she runs in an experiment. Alternatively, records for all trials could be kept on one data sheet. Records might be organized by date. In other cases, they might be organized based on patterns in experimental results.

Scientists also use multiple methods for displaying and analyzing data. For example, collected data in one investigation might best be displayed in a data table. In other cases, plotting data on a graph or chart might help the scientists better understand the data.

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2. A Se lanza un electrón con rapidez inicial v0 = 1.60×106 m/s hacia el interior de un campo uniforme entre las placas paralela

Vanyuwa [196]

Answer:

A) E = 145.6 N / C , B) y= 2,8 10-7 m with a downward direction

C) he shape of the trajectory of the two particles is to simulate a parabola,

D) F_{e} /F_{g} = 10³⁴

Explanation:

A) For this exercise we use Newton's second law to find the acceleration of the electron, where the force is electric

F = m a

- e E = m a

a = - e E / m

with the field directed downward, the acceleration is in the vertical upward direction.

We look for how much the electron moves with kinematics, in the x direction there is no acceleration,

x axis (parallel to plates)

x = v₀ t

t = x / v₀

y axis (perpendicular to plates)

y = y₀ + $v_{oy}$ t + ½ a t²

Let's take the zero of the system in the middle of the plates y₀ = 0, also the initial vertical velocity is zero (v_{oy} = 0) the width of the plate is known

y = ½ a t²

we substitute

y = ½ (e E /m) (x / v₀)²

y = ½ e x2 /m v₀² E

we look for the electric field

E = 2 m y v₀² / e x²

where to use this expression the length and width of the condenser must be known, suppose that the length is x = l = 1 cm = 1 10⁻² m and the width is y = 0.5 mm = 0.5 10⁻³ m

let's calculate

E = 2 9.1 10⁻³¹ 0.5 10⁻³ (1.6 10⁶)² / (1.6 10⁻¹⁹ (1 10⁻²)²)

E = 145.6 N / C

B) The electron is exchanged for a proton

Let's look for the vertical displacement, in this case as the proton has a positive charge it moves towards the bottom of the plates

y = ½ e x² / m v₀² E

y = ½ 1.6 10⁻¹⁹ 1 10⁻⁴ / (1.67 10⁻²⁷ (1.6 10⁶)² 145.6

y = 28.4375 10⁻⁸ m

since the distance between the plates is 0.5 10-3 m, the proton passes the condensate because its deflection is very small

In summary, its displacement is y= 2,8 10-7 m and with a downward direction (the same direction of the electric field)

C) The shape of the trajectory of the two particles is to simulate a parabola, but one for having a negative charge (electron) the force is upwards and the other for having a positive charge (proton) the trajectory is downwards

D) The force of gravity

$F_{g}$ = G m M / R²