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4 years ago

The "opposite" of ionization energy is

1 answer:

6 0

The opposite of ionization energy is B, electronegativity. In short, we can define ionization energy as the amount of energy required to remove one electron from the outermost valence shell from an atom. Conversely, we can define electronegativity as an atom's "willingness" to accept one electron.

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Un cuerpo de 480 g de masa es atraído con una fuerza de 3.9 E-6 N por otro cuerpo de 196 g de masa. Calcula la distancia a la qu

Soloha48 [4]

Answer:

La distancia entre los dos cuerpos es aproximadamente 1.269 milímetros.

Explanation:

Asumamos que ambos cuerpos son partículas, la fuerza de atracción ( $F$ ), en newtons, entre ambos cuerpos se define mediante la Ley de Newton de la Gravitación Universal, cuya ecuación es:

$F = G \cdot \frac{m_{1}\cdot m_{2}}{r^{2}}$ (1)

Donde:

$G$ - Constante de la gravitación universal, en metros cúbicos por kilogramo-segundo cuadrado.

$m_{1}, m_{2}$ - Masas de los cuerpos, en kilogramos.

$r$ - Distancia entre los cuerpos, en metros.

Si sabemos que $G = 6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}}$ , $m_{1} = 0.48\,kg$ , $m_{2} = 0.196\,kg$ y $F = 3.9\times 10^{-6}\,N$ , entonces la distancia entre los dos cuerpos es:

$r = \sqrt{\frac{G\cdot m_{1}\cdot m_{2}}{F} }$

$r \approx 1.269\times 10^{-3}\,m$

Es decir, la distancia entre los dos cuerpos es aproximadamente 1.269 milímetros.

5 0

3 years ago

Step by step process on how to solve this problem

rusak2 [61]

Step-1: Remember or look up the formula for the force of gravity between two objects. F = G · m₁ · m₂ / R²

Step-2: Remember or look up the value of G. G = 6.67 x 10⁻¹¹ m³·kg/s²

Step-3: Write the numbers you know into the formula.

(1.989 x 10²⁰ Newtons) =

(6.67 x 10⁻¹¹ mtr³·kg/s²) · (5.9742 x 10²⁴ kg) · (moon mass) / (3.84 x 10⁸ mtr)²

Step-4: Sit back, relax, take your time, look this mess over, carefully, end-to-end, and decide how to solve it for (moon mass) .

Step-5: Divide each side by (6.67x10⁻¹¹mtr³·kg/s²)·(5.9742x10²⁴kg)/(3.84x10⁸mtr)²:

Moon mass =

(1.989x10²⁰ Newtons)·(3.84x10⁸ mtr)²/(6.67x10⁻¹¹ mtr³·kg/s²)·(5.9742x10²⁴ kg)

Step-6: Crunch the numbers. Be careful to KEEP all the units as you go along. When you're done, the units of your answer will be the first instant indication if you made a mistake. You're looking for the MASS of the moon. If the answer doesn't have units of 'kg', then that'll be an immediate red flag, telling you that there's been a mistake somewhere.

Moon mass =

(1.989x10²⁰ Newtons)·(3.84x10⁸ mtr)²/(6.67x10⁻¹¹ mtr³·kg/s²)·(5.9742x10²⁴ kg)

Collect the numbers, and collect the units:

Moon mass = (1.989x10²⁰ · (3.84x10⁸)² / (6.67x10⁻¹¹ · 5.9742x10²⁴)

(kg-mtr/s² · mtr²)/(mtr³·kg/s² · kg)

Moon mass = (1.989 · 3.84² x 10³⁶) / (6.67 · 5.9742 x 10¹³)

(kg-mtr/s² · mtr²)/(mtr³·kg/s² · kg)

Moon mass = 0.736 x 10²³ kg

Moon mass = 7.36 x 10²² kg

Step 7: Look it up in a book or online. See if you're anywhere close.

When I search "moon mass" on Floogle, the first hit says

" 7.348 x 10²² kg " .

yay ! I'm satisfied.

6 0

3 years ago

Problem 1: Spherical mirrorConsider a spherical mirror of radius 2 m, and rays which go parallel to the optic axis. What is thep

SIZIF [17.4K]

Answer:

1) iii i= 1m, 2) iii and iv, 3) i = f₂ (L-f₁) / (L - (f₁ + f₂))

Explanation:

Problem 1

For this problem we use two equations the equations of the focal distance in mirrors

f = r / 2

f = 2/2

f = 1 m

The builder's equation

1 / f = 1 / o + 1 / i

Where f is the focal length, "o and i" are the distance to the object and the image respectively.

For a ray to arrive parallel to the surface it must come from infinity, whereby o = ∞ and 1 / o = 0

1 / f = 0 + 1 / i

i = f

i = 1 m

The image is formed at the focal point

The correct answer is iii

Problem 2

For this problem we have two possibilities the lens is convergent or divergent, in both cases the back face (R₂) must be flat

Case 1 Flat lens - convex (convergent)

R₂ = infinity

R₁ > 0

Cas2 Flat-concave (divergent) lens

R₂ = infinity

R₁ <0

Why the correct answers are iii and iv

Problem 3

For a thick lens the rays parallel to the first surface fall in their focal length (f₁), this is the exit point for the second surface whereby the distance to the object is o = L –f₁, let's apply the constructor equation to this second surface

1 / f₂ = 1 / (L-f₁) + 1 / i

1 / i = 1 / f₂ - 1 / (L-f₁)

1 / i = (L-f₁-f₂) / f₂ (L-f₁)

i = f₂ (L-f₁) / (L - (f₁ + f₂))

This is the image of the rays that enter parallel to the first surface

6 0

4 years ago

A ramp is 2.8 m long and 1.2 m high. How much power is needed to push a box up the ramp in 4.6 s with a force of 96 N?

tino4ka555 [31]

Power = work /time

and W = force * dis

W = 96 * 3.046 J = 292.44 J

P = 292.44 / 4.6 = 63.575 watt

6 0

4 years ago

An astronaut of mass 65kg in training rides in a seat that is moved in uniform circular motion by a radial arm 5.10 meters long.

ikadub [295]

<h3>Answer;</h3>

= 2868 Newtons

<h3>Explanation;</h3>

Centripetal force is a force that acts on an object or a body in circular path and is directed towards the center of the circular path.

Centripetal force is given by the formula;

mv²/r ; where m is the mass of the body, r is the radius of the circular path and v is the velocity of a body;

mass = 65 kg, velocity = 15 m/s and r = 5.1 m

Therefore;

Centripetal force = (65 × 15²)/ 5.10

= 2867.65 Newtons

= 2868 N

7 0

4 years ago