Current = charge per second
2 Coulombs per second = 2 Amperes
Potential difference = (current)x(resistance) in volts.
That's (2 Amperes) x (2 ohms).
That's how to do it.
I think you can find the answer now.
That's two different things it depends on:
-- surface area exposed to the air
AND
-- vapor already present in the surrounding air.
Here's what I have in mind for an experiment to show those two dependencies:
-- a closed box with a wall down the middle, separating it into two closed sections;
-- a little round hole in the east outer wall, another one in the west outer wall,
and another one in the wall between the sections;
So that if you wanted to, you could carefully stick a soda straw straight into one side,
through one section, through the wall, through the other section, and out the other wall.
-- a tiny fan that blows air through a tube into the hole in one outer wall.
<u>Experiment A:</u>
-- Pour 1 ounce of water into a narrow dish, with a small surface area.
-- Set the dish in the second section of the box ... the one the air passes through
just before it leaves the box.
-- Start the fan.
-- Count the amount of time it takes for the 1 ounce of water to completely evaporate.
=============================
-- Pour 1 ounce of water into a wide dish, with a large surface area.
-- Set the dish in the second section of the box ... the one the air passes through
just before it leaves the box.
-- Start the fan.
-- Count the amount of time it takes for the 1 ounce of water to completely evaporate.
=============================
<span><em>Show that the 1 ounce of water evaporated faster </em>
<em>when it had more surface area.</em></span>
============================================
============================================
<u>Experiment B:</u>
-- Again, pour 1 ounce of water into the wide dish with the large surface area.
-- Again, set the dish in the second half of the box ... the one the air passes
through just before it leaves the box.
-- This time, place another wide dish full of water in the <em>first section </em>of the box,
so that the air has to pass over it before it gets through the wall to the wide dish
in the second section. Now, the air that's evaporating water from the dish in the
second section already has vapor in it before it does the job.
-- Start the fan.
-- Count the amount of time it takes for the 1 ounce of water to completely evaporate.
==========================================
<em>Show that it took longer to evaporate when the air </em>
<em>blowing over it was already loaded with vapor.</em>
==========================================
Answer:
a) v = 21.34 m/s
b) v = 21.34 m/s
c) v = 21.34 m/s
Explanation:
Mass of the snowball, m = 0.560 kg
Height of the cliff, h = 14.2 m
Initial velocity of the ball, u = 13.3 m/s
θ = 26°
The speed of the slow ball as it reaches the ground, v = ?
The initial Kinetic energy of the snow ball, ![KE_{0} = 0.5 mu^{2}](https://tex.z-dn.net/?f=KE_%7B0%7D%20%20%3D%200.5%20mu%5E%7B2%7D)
Potential energy of the snow ball at the given height, PE = mgh
Final Kinetic energy of the ball as it reaches the ground, ![KE_{f} = 0.5mv^{2}](https://tex.z-dn.net/?f=KE_%7Bf%7D%20%3D%200.5mv%5E%7B2%7D)
a) Using the principle of energy conservation,
![KE_{0} + PE = KE_{f} \\0.5mu^{2} + mgh = 0.5mv^{2}\\v^{2} =2( 0.5u^{2} + gh)\\v^{2} =u^{2} + 2gh\\v = \sqrt{u^{2} + 2gh} \\v = \sqrt{13.3^{2} + 2*9.8*14.2}\\v = 21.34 m/s](https://tex.z-dn.net/?f=KE_%7B0%7D%20%2B%20PE%20%3D%20KE_%7Bf%7D%20%5C%5C0.5mu%5E%7B2%7D%20%2B%20mgh%20%3D%200.5mv%5E%7B2%7D%5C%5Cv%5E%7B2%7D%20%3D2%28%200.5u%5E%7B2%7D%20%2B%20gh%29%5C%5Cv%5E%7B2%7D%20%3Du%5E%7B2%7D%20%2B%202gh%5C%5Cv%20%3D%20%5Csqrt%7Bu%5E%7B2%7D%20%2B%202gh%7D%20%5C%5Cv%20%3D%20%5Csqrt%7B13.3%5E%7B2%7D%20%2B%202%2A9.8%2A14.2%7D%5C%5Cv%20%3D%2021.34%20m%2Fs)
b) The speed remains v = 21.34 m/s since it is not a function of the angle of launch
c)The principle of energy conservation used cancels out the mass of the object, therefore the speed is not dependent on mass
v = 21. 34 m/s
Gold’s molar mass is about 196 while aluminum is about 27, thus 50cm of gold has more mass
Answer:
<u>B. the stars of spectral type A and F are considered reasonably to have habitable planets but much less likely to have planets with complex plant - or animal - like life.</u>
Explanation:
The appropriate spectral range for habitable stars is considered to be "late F" or "G", to "mid-K" or even late "A". <em>This corresponds to temperatures of a little more than 7,000 K down to a little less than 4,000 K</em> (6,700 °C to 3,700 °C); the Sun, a G2 star at 5,777 K, is well within these bounds. "Middle-class" stars (late A, late F, G , mid K )of this sort have a number of characteristics considered important to planetary habitability:
• They live at least a few billion years, allowing life a chance to evolve. <em>More luminous main-sequence stars of the "O", "B", and "A" classes usually live less than a billion years and in exceptional cases less than 10 million.</em>
• They emit enough high-frequency ultraviolet radiation to trigger important atmospheric dynamics such as ozone formation, but not so much that ionisation destroys incipient life.
• They emit sufficient radiation at wavelengths conducive to photosynthesis.
• Liquid water may exist on the surface of planets orbiting them at a distance that does not induce tidal locking.
<u><em>Thus , the stars of spectral type A and F are considered reasonably to have habitable planets but much less likely to have planets with complex plant - or animak - like life.</em></u>