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Furkat [3]
3 years ago
14

What force opposes gravity and keeps objects from falling to the center of the earth?

Chemistry
1 answer:
stiv31 [10]3 years ago
7 0

La fuerza de la gravedad depende de la masa (el peso) de cada objeto. La fuerza con que se atraen dos objetos es proporcional a su masa y disminuye rápidamente en el momento en que los separamos. De hecho, nosotros también atraemos objetos con ‘nuestra’ fuerza gravitatoria, pero pesamos tan poco que no podemos percibirlo. En cambio, el Sol es tan grande que es capaz de mantenernos girando a su alrededor a pesar de estar muy lejos. La Luna también ejerce su propia fuerza gravitatoria, pero, como es más pequeña y ligera que la Tierra, si nos pesásemos sobre su superficie veríamos que pesamos unas seis veces menos que en la Tierra.

Podríamos preguntarnos por qué la Luna no cae sobre la Tierra al igual que una manzana cae del árbol. La razón es que nuestro satélite nunca está quieto. Se mueve constantemente a nuestro alrededor. Sin la fuerza de atracción terrestre, se alejaría flotando en el espacio. Gracias a esta combinación de velocidad y distancia de nuestro planeta, la Luna siempre está en equilibrio, ni cae ni se aleja. Si se moviera más rápido, se alejaría, si se moviera con más lentitud, ¡caería!

Hemos dicho que la fuerza de la gravedad también depende de la distancia. Si nos alejásemos lo suficiente de la Tierra, escaparíamos a su fuerza de atracción. Y eso es lo que tratamos de hacer con las naves espaciales. Necesitamos superar la llamada ‘velocidad de escape’, que es aproximadamente 11,2 km/s (a esa velocidad, podríamos viajar de Londres a Nueva York ¡en tan solo 10 minutos!). Cuando un cohete alcanza esa velocidad, ya es libre para viajar por el sistema solar.

Dentro de una nave en órbita, no sentimos la fuerza de la gravedad terrestre. Los objetos no caen, sino que flotan, así que si saltas, no regresas al suelo. Es lo que les ocurre a los astronautas cuando están a bordo de una estación espacial que orbita alrededor de la Tierra.

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Physical properties of Ammonium <br> Sulphate
Sliva [168]

the physical properties of ammonium  sulphate are colorless gas, less dense than air, pungent shell, very soluble in water, alkaline gas

6 0
2 years ago
To achieve the highest return during recrystallization of a given solid, one must: Group of answer choices Add the minimum amoun
TiliK225 [7]

Answer:

Option A

Explanation:

Addition of too much of solvent will make the solution dilute due to which the crystals will not form. Hence option D is incorrect

On the other hand adding a minimum amount of boiling solvent will give a saturated solution for recrystallization. Hence, option A is incorrect

Addition of cold solvent will lower the rate of formation of crystals. Hence, both option B and C are incorrect

4 0
3 years ago
How many grams of sodium bromide must be dissolved in 400.0 g of water to produce a 0.500 molar solution?
kiruha [24]

Answer:

D. 20.6g

Explanation:

Molarity is defined as ratio between moles of solute (sodium bromide, NaBr) per liter of water.

As density of water is 1g/mL; volume of 400.0g of water is 400.0mL = 0.4000L.

That means 0.400L of 0.500M solution contains:

0.400L × (0.500mol / 1L) = <em>0.200moles of sodium bromide</em>.

In mass (NaBr = 102.9g/mol):

0.200mol NaBr × (102.9g/mol) = 20.6g of NaBr

Right answer is:

<em>D. 20.6g</em>

8 0
3 years ago
How many liters are needed to make a 4. 8M solution of KBr if 4 mole of KBr are available for use?
Inessa [10]

4.8 M = \frac{4 mol}{x}

x = 0.83 L

4 0
2 years ago
The side chain of tyrosine has a pKa of about 10. What percent of tyrosine side chains would be deprotonated at pH 8.5?
bezimeni [28]

Answer:

Explanation:

Let the tyrosine molecule be represented by TH . It will ionise in water as follows

TH ⇄ T⁻ + H⁺

Let C be the concentration of undissociated TH and α be the degree of dissociation

TH       ⇄       T⁻ + H⁺

c                    0       0   ( before )

c( 1-α )            αc      αc ( after ionisation)

Ka = α²c² / c( 1-α )

= α²c  ( neglect α in the denominator as it is very small )

pKa = 10

Ka  = 10⁻¹⁰

pH = 8.5

H⁺ = 10⁻⁸°⁵

αc = 10⁻⁸°⁵

α²c =Ka = 10⁻¹⁰

α x10⁻⁸°⁵ = 10⁻¹⁰

α = 10⁻¹⁰⁺⁸°⁵

= 10⁻¹°⁵ = 1 / 31.62

Percentage of dissociation = 100 / 31.62

= 3.16 %

percent of  tyrosine side chains   deprotonated

3 0
3 years ago
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