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Alexus [3.1K]
3 years ago
9

What 4 things affect how great the change in temperature for a substance is?

Physics
2 answers:
yawa3891 [41]3 years ago
7 0

Explanation:

The heat Q transferred to cause a temperature change depends on the magnitude of the temperature change, the mass of the system, and the substance and phase involved.

Fiesta28 [93]3 years ago
5 0

Answer: Usually, increasing the temperature increases the solubility of solids and liquids. Increasing the temperature always decreases the solubility of gases.

Explanation:

When you add a solute to a solvent, the kinetic energy of the solvent molecules overcomes the attractive forces among solute particles.

The solute particles leave the surface of the solid and move into the dissolved (aqueous) phase. In the image below the mass of grey (-) balls and green (+) balls represent a salt crystal. As the salt dissolves, the positive and negative ions are pulled apart and become surrounded by water molecules.

If we heat the solvent, the average kinetic energies of its molecules increases. Hence, the solvent is able to dislodge more particles from the surface of the solute.

Thus, increasing the temperature increases the solubilities of substances. For example, sugar and salt are more soluble in water at higher temperatures.

But, as the temperature increases, the solubility of a gas in a liquid decreases. As temperature increases, the average kinetic energy of the gas molecules increases.

As a result, the gas molecules dissolved in the liquid are more likely to escape to the gas phase and not return.

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Suppose that Hubble's constant were H0 = 51 km/s/Mly (which is not its actual value). What would the approximate age of the univ
bija089 [108]

Given the Hubble's constant, the approximate age of the universe is 5.88 × 10⁹ Years.

Given the data in the question;

Hubble's constant; H_0 = 51km/s/Mly

Age of the universe; t = \ ?

We know that, the reciprocal of the Hubble's constant ( H_0 ) gives an estimate of the age of the universe ( t ). It is expressed as:

Age\ of\ Universe; t = \frac{1}{H_0}

Now,

Hubble's constant; H_0 = 51km/s/Mly

We know that;

1\ light\ years = 9.46*10^{15}m

so

1\ Million\ light\ years = [9.46 * 10^{15}m] * 10^6 = 9.46 * 10^{21}m

Therefore;

H_0 = 51\frac{km}{\frac{s}{Mly} } = 51000\frac{m}{s\ *\ Mly}  \\\\H_0 = 51000\frac{m}{s\ *\ (9.46*10^{21}m)} \\\\H_0 =  5.39 *10^{-18}s^{-1}\\

Now, we input this Hubble's constant value into our equation;

Age\ of\ Universe; t = \frac{1}{H_0}\\\\t = \frac{1}{ 5.39 *10^{-18}s^{-1}} \\\\t = 1.855 * 10^{17}s\\\\We\ convert\ to\ years\\\\t =  \frac{ 1.855 * 10^{17}}{60*60*24*365}yrs \\\\t = \frac{ 1.855 * 10^{17}}{31536000}yrs\\\\t = 5.88 *10^9 years

Therefore, given the Hubble's constant, the approximate age of the universe is 5.88 × 10⁹ Years.

Learn more: brainly.com/question/14019680

6 0
3 years ago
How does 'g' vary from place to place?​
r-ruslan [8.4K]

Explanation:

The acceleration g varies by about 1/2 of 1 percent with position on Earth's surface, from about 9.78 metres per second per second at the Equator to approximately 9.83 metres per second per second at the poles.

8 0
2 years ago
The largest building in the world by volume is the boeing 747 plant in Everett, Washington. It measures approximately 632 m long
Schach [20]

Explanation:

Given that,

The dimensions of the largest building in the world is 632 m long, 710 yards wide, and 112 ft high. It basically forms a cuboid. The volume of a cuboidal shape is given by :

Since,

1 meter = 3.28084 feet

632 m = 2073.49 feet

1 yard= 3 feet

710 yards = 2130 feet

V = lbh

V=2073.49 \ ft\times 2130\ ft\times 112\ ft

V=494651774.4\ ft^3

V=4.94\times 10^8\ ft^3

Also,

V=(4.94\times 10^8\ ft^3)(\dfrac{1\ m}{3.281})^3

V=1.39\times 10^7\ m^3

Hence, this is the required solution.

3 0
3 years ago
An object on a vertical spring oscillates up and down in simple harmonic motion with an angular frequency of 25.7 rad/s. Calcula
Sergio [31]

Answer:

The distance by which the spring stretches is 1.48 cm.

Explanation:

Given that,

An object on a vertical spring oscillates up and down in simple harmonic motion with an angular frequency of 25.7 rad/s, \omega=25.7\ rad/s

We know that angular frequency in SHM is given by :

\omega=\sqrt{\dfrac{k}{m}} \\\\\omega^2=\dfrac{k}{m}\\\\\dfrac{k}{m}=(25.7)^2.............(1)

When the object is allowed to hang stationary from it, the force due to spring is balanced by its weight. Such that :

kd=mg\\\\d=\dfrac{g}{(k/m)}

From equation (1) :

d=\dfrac{9.8}{(25.7)^2}\\\\d=0.0148\ m\\\\d=1.48\ cm

So, the distance by which the spring stretches is 1.48 cm.

6 0
3 years ago
Uh yeah- answer please and ill give brain list to whoever answers it fully
gulaghasi [49]

Answer:

1. A

2. D

3. 1

Explanation:

5 0
2 years ago
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