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3 years ago

Why is it important to have only one set of chemical symbols in the world ?

Physics

1 answer:

Pavlova-9 [17]3 years ago

3 0

Answer:

As you learned, scientists standardized the short-hand way we represent elements, by their chemical symbol. One of the main reasons this was developed was because using letters was the easiest way to represent the elements. Another reason that we use chemical symbols is to allow us to write chemical formulas easily

One set of chemical symbols means that every science student and scientist -- no matter what languages they speak -- can identify these chemicals.

I hope this helps

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S When an uncharged conducting sphere of radius a is placed at the origin of an x y z coordinate system that lies in an initiall

telo118 [61]

The sphere has a constant potential. It is the electric field.

$E = V_{0} = 0$

In the sphere, then

$E_{x} = 0, E_{y}=0, E_{z}=0$

Outside the sphere, then

$V = V_{0} - E_{0}z + \frac{E_{0}a^{3}z}{(x^{2} +y^{2} + z^{2})^{3/2} }$

The elements of the electric field include

$E_{x} =\frac{3E_{0}a^{3}xy}{(x^{2} +y^{2} +z^{2})^{5/2}}\\E_{y} = \frac{3E_{0}a^{3}xz}{(x^{2} +y^{2}+z^{2})^{5/2}}$

Which becomes,

$=E_{0} (1-\frac{a^{3}}{x^{2} +y^{2}+z^{2})^{3/2}}+\frac{3a^{3}z^{2}}{(x^{2} +y^{2}+z^{2})^{5/2}})$

<h3>In a consistent electric field, is force constant?</h3>

Similar to an ordinary object in the uniform gravitational field near the Earth's surface, a charged item in a uniform electric field experiences a constant force and consequently experiences a uniform acceleration. The vector cross product of p and E determines the torque's direction.

If the charge is positive, the force either moves in the same direction as E or in the opposite direction (if charge is negative).

A torque is experienced by an electric dipole (p) in an even electric field (E). The vector cross product of p and E determines the torque's direction.

To learn more about uniform electric field, visit

brainly.com/question/17426130

#SPJ4

5 0

1 year ago

18. Un avión de rescate de animales que vuela hacia el este a 36.0 m/s deja caer una paca de

Alona [7]

Answer:

Definimos momento como el producto entre la masa y la velocidad

P = m*v

(tener en cuenta que la velocidad es un vector, por lo que el momento también será un vector)

Sabemos que el peso de la paca de heno es 175N, y el peso es masa por aceleración gravitatoria, entonces.

Peso = m*9.8m/s^2 = 175N

m = (175N)/(9.8m/s^2) = 17.9 kg

Ahora debemos calcular la velocidad de la paca justo antes de tocar el suelo.

Sabemos que la velocidad horizontal será la misma que tenía el avión, que es:

Vx = 36m/s

Mientras que para la velocidad vertical, usamos la conservación de la energía:

E = U + K

Apenas se suelta la caja, esta tiene velocidad cero, entonces su energía cinética será cero y la caja solo tendrá energía potencial (Si bien la caja tiene velocidad horizontal en este punto, por la superposición lineal podemos separar el problema en un caso horizontal y en un caso vertical, y en el caso vertical no hay velocidad inicial)

Entonces al principio solo hay energía potencial:

U = m*g*h

donde:

m = masa

g = aceleración gravitatoria

h = altura

Sabemos que la altura inicial es 60m, entonces la energía potencial es:

U = 175N*60m = 10,500 N

Cuando la paca esta próxima a golpear el suelo, la altura h tiende a cero, por lo que la energía potencial se hace cero, y en este punto solo tendremos energía cinética, entonces:

10,500N = (m/2)*v^2

De acá podemos despejar la velocidad vertical justo antes de golpear el suelo.

√(10,500N*(2/ 17.9 kg)) = 34.25 m/s

La velocidad vertical es 34.25 m/s

Entonces el vector velocidad se podrá escribir como:

V = (36 m/s, -34.25 m/s)

Donde el signo menos en la velocidad vertical es porque la velocidad vertical es hacia abajo.

Reemplazando esto en la ecuación del momento obtenemos:

P = 17.9kg*(36 m/s, -34.25 m/s)

P = (644.4 N, -613.075 N)

6 0

3 years ago

How often do the earth's magnetic poles switch?

RSB [31]

Reversals are the rule, not the exception. Earth has settled in the last 20 million years into a pattern of a pole<span> reversal about every 200,000 to 300,000 years, although it has been more than twice that </span>long<span> since the last reversal.</span>

3 0

3 years ago

3. A car has a mass of 2.50 x 10^3 kg. If the force acting on the car is 7.65 x 10^3 N to the

const2013 [10]

Answer:

3.06m/s² to the east

Explanation:

Given parameters:

Mass of car = 2.5 x 10³kg

Force acting on the car = 7.65 x 10³N

Unknown:

Acceleration of the car = ?

Solution:

From Newton's second law of motion:

Force = mass x acceleration

Acceleration = $\frac{Force }{mass}$ = $\frac{7.65 x 10^{3} }{2.5 x 10^{3} }$ = 3.06m/s² to the east

6 0

3 years ago

A body moves in a straight line. At time, it's acceleration is given by a = 12t + 1. When t =0, the velocity of the body v is 2

madreJ [45]

Answer:

v = 6t² + t + 2, s = 2t³ + ½ t² + 2t

59 m/s, 64.5 m

Explanation:

a = 12t + 1

v = ∫ a dt

v = 6t² + t + C

At t = 0, v = 2.

2 = 6(0)² + (0) + C

2 = C

Therefore, v = 6t² + t + 2.

s = ∫ v dt

s = 2t³ + ½ t² + 2t + C

At t = 0, s = 0.

0 = 2(0)³ + ½ (0)² + 2(0) + C

0 = C

Therefore, s = 2t³ + ½ t² + 2t.

At t = 3:

v = 6(3)² + (3) + 2 = 59

s = 2(3)³ + ½ (3)² + 2(3) = 64.5

5 0

2 years ago