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Viktor [21]
3 years ago
11

Vapor pressure is related to the temperature of the liquid. user: in an open system, the vapor pressure is equal to the _____. i

nside air pressure outside air pressure inside temperature outside temperature
Physics
2 answers:
Naddik [55]3 years ago
8 0
In an open system the vapor pressure of a liquid, if it is not boliling is less thatn the atmospheric pressure.

As the temperature of the liquid increases, the vapor pressure increases. When the vapor pressure equals the atmospheric pressure the molecules of liquid escape and the liquid boils. The temperature at which this happens is the boiling point.
iVinArrow [24]3 years ago
6 0

Answer

-Directly;  outside air pressure

Vapor pressure is directly related to the temperature of the liquid. user: in an open system, the vapor pressure is equal to the outside air pressure.

Explanation;

-As the temperature of a system increases, the average kinetic energy of the molecules increases in both the liquid and gas phases.

-A higher average kinetic energy facilitates the escape of molecules from the liquid phase into the gas phase. At the same time, the rate of return of gas phase molecules to the liquid also increases. A new equilibrium point is reached at a higher gaseous vapor pressure. The increase in vapor pressure with temperature is exponential.


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Answer:

0.00583 seconds

Explanation:

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The product of 14 and a cubed
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14 × a^3

Explanation:

Product means multiplication

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1.Which statement is true when two objects are 20 miles apart but one is much bigger?
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1. C.  Gravitational attraction exists between the two objects.

Explanation:

Gravitational attraction is always exerted between two objects which have mass, and its magnitude is given by:

F=G\frac{m_1 m_2}{r^2}

where G is the gravitational constant, m1 and m2 the masses of the two objects, and r the separation between them. Since the two objects have for sure non-zero masses m1 and m2, even if they are 20 miles apart, the value of the gravitational attraction F is non-zero, so the correct answer is C.


2. D.  Two atoms come together to form a molecule.

Explanation:

this outcome is actually caused by the electrostatic forces between the two atoms, not by gravitational force. In fact, gravitational force becomes relevant only when the masses of the two objects involved are large enough: this is the case for planets, stars, galaxies, and objects in the universe. However, two atoms have very small masses, so the gravitational force between them is really negligible. On this smaller scales, the electrostatic force is much stronger than the gravitational force, so the electrostatic force is the real responsible for the formation of bonds between atoms.

7 0
3 years ago
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A worker drives a 0.562 kg spike into a rail tie with a 2.26 kg sledgehammer. The hammer hits the spike with a speed of 64.4 m/s
zubka84 [21]

Answer:

Explanation:

Given that,

Mass of sledge hammer;

Mh =2.26 kg

Hammer speed;

Vh = 64.4 m/s

The expression fot the kinetic energy of the hammer is,

K.E(hammer) = ½Mh•Vh²

K.E(hammer) = ½ × 2.26 × 64.4²

K.E ( hammer) = 4686.52 J

If one forth of the kinetic energy is converted into internal energy, then

ΔU = ¼ × K.E(hammer)

∆U = ¼ × 4686.52

∆U = 1171.63 J

Thus, the increase in total internal energy will be 1171.63 J.

4 0
3 years ago
Jupiter's satellite Europa orbits Jupiter with a period of 3.55 d at an orbital radius of 6.71 108 m (assume the orbit to be cir
yawa3891 [41]

Answer:

(a)

M = 1.898 x 10^27 kg

(b)

v = 13.74 km/s

(c) E = 0.28 N/kg

Explanation:

Time period, T = 3.55 days = 3.55 x 24 x 3600 second = 306720 s

Radius, r = 6.71 x 10^8 m

G = 6.67 x 10^-11 Nm^2/kg^2

(a) T=2\pi \sqrt{\frac{r^{3}}{GM}}

M=\frac{4\pi ^{2}r^{3}}{GT^{2}}

M=\frac{4\times3.14^{2}\times 6.71^{3}\times 10^{24}}{6.67\times 10^{-11}\times 306720^{2}}

M = 1.898 x 10^27 kg

(b) Let v be the orbital velocity

v=\frac{2\pi r}{T}

v=\frac{2\times 3.14\times 6.71\times 10^{8}}{306720}

v = 13739.5 m/s

v = 13.74 km/s

(b) The gravitational field E is given by

E = \frac{GM}{r^{2}}

E = \frac{6.67\times10^{-11}\times 1.898\times 10^{27}}{6.71^{2}\times 10^{16}}

E = 0.28 N/kg

6 0
3 years ago
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