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jekas [21]
4 years ago
8

A 0.025 g sample of o is injected into a 1.00 l flask at 30.0c. what forms of water will be present at equilibrium? vapor pressu

re of water at 30.0c is 31.8 mm hg.
Physics
1 answer:
marysya [2.9K]4 years ago
5 0
The question asked is only the forms of water that will be present at equilibrium. Since the temperature of the system is 30 C, water will be in a form of liquid. Assuming that equilibrium will be achieved in the container, some liquid water will vaporize and mix with the sample. So, the forms of water that will be present are liquid and water vapor. 
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A semicircular plate with radius 6 m is submerged vertically in water so that the top is 2 m above the surface. Express the hydr
dalvyx [7]

Answer: 313920

Explanation:First, we’re going to assume that the top of the circular plate surface is 2 meters under the water. Next, we will set up the axis system so that the origin of the axis system is at the center of the plate.

Finally, we will again split up the plate into n horizontal strips each of width Δy and we’ll choose a point y∗ from each strip. Attached to this is a sketch of the set up.

The water’s surface is shown at the top of the sketch. Below the water’s surface is the circular plate and a standard xy-axis system is superimposed on the circle with the center of the circle at the origin of the axis system. It is shown that the distance from the water’s surface and the top of the plate is 6 meters and the distance from the water’s surface to the x-axis (and hence the center of the plate) is 8 meters.

The depth below the water surface of each strip is,

di = 8 − yi

and that in turn gives us the pressure on the strip,

Pi =ρgdi = 9810 (8−yi)

The area of each strip is,

Ai = 2√4− (yi) 2Δy

The hydrostatic force on each strip is,

Fi = Pi Ai=9810 (8−yi) (2) √4−(yi)² Δy

The total force on the plate is found on the attached image.

5 0
3 years ago
What is the maximum magnitude of charge that can be placed on each plate if the electric field in the region between the plates
34kurt

The given question is incomplete. The complete question is as follows.

A parallel-plate capacitor has capacitance C_{0} = 8.50 pF when there is air between the plates. The separation between the plates is 1.00 mm.

What is the maximum magnitude of charge that can be placed on each plate if the electric field in the region between the plates is not to exceed 3.00 \times 10^{4} V/m?

Explanation:

It is known that relation between electric field and the voltage is as follows.

             V = Ed

Now,  

              Q = CV

or,           Q = C \times Ed

Therefore, substitute the values into the above formula as follows.

              Q = C \times Ed

                  = 8.50 pF \times (\frac{10^{-12} F}{1 pF})(3 \times 10^{4} m/s)(1 mm)(\frac{10^{-3} m}{1 mm})

                  = 2.55 \times 10^{-10} C

Hence, we can conclude that the maximum magnitude of charge that can be placed on each given plate is 2.55 \times 10^{-10} C.

3 0
4 years ago
Magnet A doesn't have its poles labeled, but Magnet B has a clearly labeled north and south pole. If the
dusya [7]

Answer:

D:   The side of Magnet A that's attracted to Magnet B's south pole must be Magnet A's north pole

Explanation:

D:  The side of Magnet A that's attracted to Magnet B's south pole must be Magnet A's north pole because

1) opposite poles attract each other

2) similar poles repel each other

3)magnetic lines of force start at the north pole and end at the south pole

4 0
3 years ago
Read 2 more answers
In the 2016 Olympics in Rio, after the 50 mm freestyle competition, a problem with the pool was found. In lane 1 there was a gen
gogolik [260]

Answer:

Explanation:

still water speed is 50 m / 25.0 s = 2.00 m/s or 200 cm/s

In lane 1 the effective speed would be 201.2 cm/s

5000 cm / 201.2 cm/s = 24.85 s

The change is 25.00 - 24.85 = 0.15 s decrease in time

In lane 8, the effective speed would be 198.8 cm/s

5000 cm / 198.8 cm/s = 25.15 s

The change is 25.00 - 25.15 = 0.15 s increase in time

6 0
3 years ago
What was the runoff from precipitation on the property for areas with vegetation and roofed or paved areas over the past four mo
puteri [66]

Answer:

Higher percentage of runoff water is absorbed by the vegetation whereas 100 % of water runoff on the roofed area.

Explanation:

Higher percentage of runoff water is absorbed by the vegetation and very little amount of runoff water is received on the end of vegetative cover while on the other hand, 100 percent of precipitation water runoff on the surface of roofed or paved area because no water is absorbed by the roof due to its solid surface. The percentage of runoff depends on the factors such as rainfall intensity, slope, soil water storage capacity, and infiltration rate.

5 0
3 years ago
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