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3 years ago

What is the unit of measurement used for atomic particles?

Physics

2 answers:

dalvyx [7]3 years ago

8 0

Atomic mass unit (amu)

Wittaler [7]3 years ago

7 0

That's going to depend on exactly what you want to measure about the particles. If it's the particle's mass then the unit is related to the kilogram. If your measuring the size of the particle then you will use a unit related to the meter. If you want to measure the age of the particle then you'll use a unit related to the second.

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Un ladrillo se le imparte una velocidad inicial de 6m/s en su trayectoria hacia abajo. ¿cual sera su velocidad final despues de

marshall27 [118]

Saludos!

Respuesta:

28,64 m/s.

Explicación:

Datos:

Altura o distancia recorrida: 40 m
Vo: 6 m/s
Aceleración de la gravedad: 9,81 m/s²

El ejercicio puede ser resuelto facilmente utilizando la siguiente formula, sin embargo es posible realizarlo utilizando formulas diferentes.

Entonces tenemos que:

$Vf ^{2} -Vo ^{2} =2 x g x h$

Es importante saber que al estar lanzando el ladrillo hacia abajo, el sentido del movimiento sigue el sentido de la gravedad, es decir es necesario que tomes el valor de la gravedad como positivo (+) y no negativo (-) como normalmente se usa.

Sustituyendo tenemos que:

$Vf x^{2} =(6 m/s) ^{2} +(2)x(9,81 m/s ^{2})x(40m) \\ Vf ^{2} =820,8 m^{2} /s ^{2} \\ \sqrt{Vf ^{2} } = \sqrt{820,8m^{2}/s ^{2} } \\ Vf=28,64 m/s$

Que tengas un buen día!

6 0

3 years ago

A ball rolls off the end of a horizontal table that is 4 meters off the ground. It is measured that the ball lands 3 meters away

ElenaW [278]

Answer:

The speed at which the ball rolled off the end of the table is 3.3 m/s

Explanation:

Hi there!

Please, see the attached figure for a graphical description of the problem. Notice that the origin of the frame of reference is located at the edge of the table.

The position vector of the ball can be calculated as follows:

r = (x0 + v0x · t, y0 + v0y · t + 1/2 · g · t²)

Where:

r = position vector.

x0 = initial horizontal position.

v0x = initial horizontal velocity.

t = time.

y0 = initial vertical position.

v0y = initial vertical velocity.

g = acceleration due to gravity.

When the ball reaches the ground, its position will be:

r final = (3, -4)

Then:

3 = x0 + v0x · t

-4 = y0 + v0y · t + 1/2 · g · t²

Since the origin of the frame of reference is located at the edge of the table, x0 and y0 = 0. v0y is also 0 ( see the initial velocity vector in the figure to elucidate why). Then:

3 m = v0x · t

-4 m = 1/2 · g · t²

We can solve for "t" in the equation of the y-component and use it in the equation of the x-component to obtain v0x:

-4 m = 1/2 · g · t²

-4 m = -1/2 · 9.8 m/s² · t²

8 m / 9.8 m/s² = t²

t = 0.9 s

Then:

3 m = v0x · 0.9s

3 m/ 0.9 s = v0x

v0x = 3.3 m/s

The speed at which the ball roll off the end of the table is 3.3 m/s

8 0

3 years ago

Joe moved to a new town and was having a hard time making friends. Recently he met some people he really liked at a party. They

siniylev [52]

Joey should say no to drugs next time he is offered to smoke weed

3 0

3 years ago

In the diagram, q1 = -6.39*10^-9 C and q2 = +3.22*10^-9 C. What is the electric field at point P? pls help

Alexxx [7]

Answer:

Below

Explanation:

First draw the vectors that represent both electric fields.

E1 is the elictric field created by q1, E2 is the one created by q2.

● q1 is negative so E1 will point from P.

● q2 is positive so E2 will point out of P

(Picture below)

■■■■■■■■■■■■■■■■■■■■■■■■■■

The resulting electric field is equal to the sum of the two fields since both vectors are colinear.

Let E be the total field.

● E = E1 + E2

The formula of the electric field intensity is:

● E = K ×(q/d^2)

-K is Coulomb's constant

-d is the distance between the charge and the object ( here P)

-q is the charge

■■■■■■■■■■■■■■■■■■■■■■■■■■

● E1 = K × (q1/d1^2)

The distance between q1 and P is the qum of 0.15 m 0.25 m. (0.4 m)

Coulombs constant is 9×10^9 m^2/C^2

● E1 = 9×10^9 ×[-6.39 × 10^(-9)/ 0.4^2]

● E1 = -359.43 N/C

■■■■■■■■■■■■■■■■■■■■■■■■■■

● E2 = K ×(q2/d^2)

The distance between q2 and P is 0.25 m.

● E2 = 9×10^9×[3.22×10^(-9) /0.25^2]

● E2 = 463.68 N/C

■■■■■■■■■■■■■■■■■■■■■■■■■■

● E = E1 + E2

● E = -359.43+463.68

● E = 105.25 N/C

4 0

3 years ago

Read 2 more answers

How many centimeters are in 3.50 feet?

Kazeer [188]

106.68 centimetres are in 3.50 feet

4 0

3 years ago