Heat will always flow from High temperature to low temperature
here we know that the temperature inside the room is 60 degree while out side temperature is 80 degree
so the flow of heat is from outside to inside the room as heat flow is always from higher temperature to lower temperature
So after some time due to heat flow from outside the temperature of room will increase
And after long time the temperature of room and outside will be same and heat transfer will become in equilibrium
Answer: Choice B
There are triple bonds between the carbon (C) and oxygen (O) atoms. Then there are 2 dots on either side
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Explanation:
When it comes to interaction and chemistry, all that matters is the valence shell or valence electrons. This is the outermost shell. This is because various elements do not interact with the inner electrons (they're locked in place so to speak and don't move to other elements).
Carbon has 6 protons, which is what uniquely makes up this element. This means there are 6 electrons. The inner shell has 2 electrons and the valence shell has 4 electrons. Two electrons are shown as the two blue dots on the left side of the C. The other two electrons form two of the lines, or the bonds, between the C and O.
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Oxygen has 8 protons and 8 electrons. It has 2 electrons in the inner shell and 6 electrons in the valence shell. Two of those electrons are the red dots on the right side of the O. The other 4 electrons are shared to form the bonds with the carbon atom.
This is where things get a bit tricky. I've shown a diagram below indicating that one of the oxygen electrons (red dot) is passed to the carbon, as this carbon atom is pulling on the oxygen electron. But the oxygen atom is pulling on it as well, which forms one of the triple bonds.
So this is why diagram B is the final answer. This is something you can logically determine (remembering the rules of how each electron shell is formed), or it's something you'll need to memorize. In the real world, it's easy to forget a lot of info like this, so that's why having it handy as a lookup reference is preferable.
Let's be clear: The plane's "395 km/hr" is speed relative to the
air, and the wind's "55 km/hr" is speed relative to the ground.
Before the wind hits, the plane moves east at 395 km/hr relative
to both the air AND the ground.
After the wind hits, the plane still maintains the same air-speed.
That is, its velocity relative to the air is still 395 km/hr east.
But the wind vector is added to the air-speed vector, and the
plane's velocity <span>relative to the ground drops to 340 km/hr east</span>.
Answer:
a) 35.75 ft/s
b) 45 ft
Explanation:
<u>Given </u>
Weight W = 100 lbf
mass(m) = 100*32.174/32.2=99.92 lb
decrease in kinetic energy ΔKE = -500 ft.lbf
increase in kinetic energy ΔPE= 1500 ft.lbf
initial velocity V_1 = 40 ft/s
initial height h_1 = 30 ft/s
The gravitational acceleration g = 32.2 ft/s2 Required
(a) Final velocity V_2 (a) Final elevation h_2
<u>Solution </u>
Change in kinetic energy is defined by
ΔKE = .5*m *( V_2 ^2-V_1^2)
Change in potential energy is defined by
ΔKE = W *( h_2 -h_1 )
Then,
-500=.5*99.92*1/32.174*(V_2 ^2-40^2)
V_2=35.75 ft/s
1500 = 100 x (h_2 — 30)
h_2= 45 ft