Answer: a) E= 6.63x10^-19J
E= 3.97×10^2KJ/mol
b) E = 3.31×10^-19J
E= 18.8×10^4 KJ/mol
C) E = 1.32×10^-33J
E= 8.01×10^-10KJ/mol
Explanation:
a) E = h ×f
h= planks constant= 6.626×10^-34
E=(6.626×10^-34)×(1.0×10^15)
E=6.63×10^-19J
1mole =6.02×10^23
E=( 6.63×10^-19)×(6.02×10^23)
E=3.97×10^2KJ/mol
b) E =(6.626×10^-34)/(1.0×10^15)
E=3.13×10^-19J
E= 3.13×10^-19) ×(6.02×10^23)
E= 18.8×10^3KJ/MOL
c) E= (6.626×10^-34) /0.5
E= 1.33×10^-33J
E= (1.33×10^-33) ×(6.02×10^23)
E= 8.01×10^-10KJ/mol
The answers are -2.4 for the distance. 7.2 for the height. Concave for the type
Answer:
V = 20 miles /sec
Explanation:
We have remaining distance = d = 96 miles
Lets call Pascal velocity V in miles per hour
Now if he increases his velocity by 50 % (equivalent to multiply by 1.5 ) he will need a time t₁ to arrive then as V = d/t
1.5* V = d/ t₁ ⇒ 1.5 * V = 96 /t₁
And in the case of reducing his velocity
(V / 4) = d/ (t₁ + 16 ) ⇒ V * (t₁ + 16 ) = 4*d ⇒ V*t₁ + 16*V = 384
So we a 2 equation system with two uknown variables
1.5*V = 96/t₁ (1)
V*t₁ + 16*V = 384 (2)
We solve from equation (1) t₁ = 64/V
And by substitution in equation (2)
V * (64/V) + 16* V = 384
64 + 16 *V = 384 ⇒ 16*V = 320 ⇒ V= 320/16
V = 20 miles /sec
Great experiment ! Everybody should try it if they can get the equipment.
It demonstrates a lot of things that are very hard to explain in words.
I hope the students remembered to tilt the axis of the globe. If they didn't,
and instead kept it straight up and down, then each city had pretty much
the same amount of bulb-light all the way around, and there were no seasons.
If the axis of the globe was tilted, then City-D had the least variation in
seasons. City-D is only 2° from the equator, so the sun is more direct
there all year around than it is at any of the others.
