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nikdorinn [45]
4 years ago
8

A 0.0434-m3 container is initially evacuated. Then, 4.19 g of water is placed in the container, and, after some time, all of the

water evaporates. If the temperature of the water vapor is 417 K, what is its pressure?Number units in Pa
Physics
1 answer:
adell [148]4 years ago
6 0

Answer:

P =18760.5 Pa

Explanation:

Given that

Volume ,V= 0.0434 m³

Mass ,m= 4.19 g = 0.00419 kg

T= 417 K

If we assume that water vapor is behaving like a ideal gas ,then we can use ideal gas equation

Ideal gas equation    P V = m R T

p=Pressure ,V = Volume ,m =mass

T=Temperature ,R=Universal gas constant

Now by putting the values

P V = m R T

For water R= 0.466 KJ/kgK

P x 0.0434 = 0.00419 x 0.466 x 417

P =18.7605 KPa

P =18760.5 Pa

Therefore the answer is 18760.5 Pa

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A particle travels along the curve from A to B in 5 s. It takes 8 s for it to go from B to C and then 10 s to go from C to A. De
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5.1 m/s

Explanation:

The figure is missing: find it in attachment.

In order to find the average speed, we have to calculate the length of the total path, and divide it by the total time elapsed.

The curve from A to B is a quarter of a circle with a radius of r = 20 m, so its length is:

AB=\frac{\pi r}{2}=\frac{\pi (20)}{2}=31.4 m

The path BC is the hypothenuse of a right triangle with sides equal to 20 m and 30 m, so its length is

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Finally, the length of the path AC is the sum of the side of 30 m and the radius of the curve, so

AC=30 + 20 = 50 m

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t=5 s + 8 s + 10 s =23 s

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6 0
3 years ago
A 200g of iron at 120 degrees and a 150 g piece of copper at -50 degrees are dropped into an insulated beaker containing 300 g o
kodGreya [7K]

Answer:

T = 15.03°C

Explanation:

given data:

copper specific heat = Sc = 0.385 J/g °C

iron specific iron = Si = 0.450 J/g °C

specific heat of ethanol = Se = 2.46 J/g °C

net heat loss is equal to zero

(m*S*\Delta T)_{copper} +(m*S*\Delta T)_ {iron} +(m*S*\Delta T)_ {ethanol} = 0

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57.75( T - (-50)) + 0.90(T - 120) +738(T -20) = 0

57.75T + 2887.5 + 0.90T - 108 + 738T - 14760 = 0

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4 0
3 years ago
A uniform solid sphere rolls down an incline without slipping. if the linear acceleration of the center of mass of the sphere is
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3 years ago
If the entire population of Earth were transferred to the Moon, how far would the center of mass of the Earth-Moon-population sy
Genrish500 [490]

The question is incomplete. The complete question is :

If the entire population of Earth were transferred to the Moon, how far would the center of mass of the Earth-Moon population system move? Assume the population is 7 billion, the average human has a mass of 65 kg, and that the population is evenly distributed over both the Earth and the Moon. The mass of the Earth is 5.97×1024 kg and that of the Moon is 7.34×1022 kg. The radius of the Moon’s orbit is about 3.84×105 m.

Solution :

Given :

Mass of earth, $M_e = 5.97 \times 10^{24} \ kg$

Mass of moon, $M_m = 7.34 \times 10^{22} \ kg$

Mass of each human, $m_p =65 \ kg$

Therefore mass of total population, $M_p = 65  \times 7 \times 10^{9} \ kg$

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Let the earth is at the origin of the coordinate system. Then,

Since $M_e>> M_p$

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$ 4.68 \times 10^3 \ km$ from the earth.

         

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