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I need someone to check my answers and help me with the questions I couldn't answer.
What is the max. # of electrons located in the groud state of
a) an orbital 2
b) d sublevel 10
c) Be atom...
The atom that is used as the standard for the atomic mass scale is the Carbon atom with an atomic number of 6 and a mass number of 12 and this carbon atom is equal to 12 unified atomic mass units. Could another atom be used as the standard, and if so, how would this be accomplished??
Where is the fifth energy level in relation to the fourth energy level? Well, it shows that energy level five is at n= 5 and energy level four is at n=4 away from the nucleus, so that means that the fifth energy level is at the 16th orbital and energy level four is at the 25th orbital?? I am not...
55.8 ft/sec - 45 ft/sec = 10.8 ft/sec
10.8ft/sec/6.30ft/sec^2 = 1.714 sec. = 1.7 seconds
d) determine how much of the 300 ft displacement is due to the initial velocity and how much is due to the acceleration
e) find the time required for traveling the first 150 ft.
(76ft/sec + 45 ft/sec.)/2 = 121/2 = 60.5 ft/ sec.
300 ft / 60.5 ft/ sec. = 4.9 seconds??
acceleration = m = a = change in velocity divided by change in time = (76ft/sec - 45 ft/sec)/4.9 sec. = 31ft/sec/4.9 sec. = 6.3 ft/sec^2
c) find the time when the speed of the car will be 55.8 ft/sec. ...
Initially you are driving at 45 ft/sec on a straight road. You accelerate the car at a steady rate and increase the speed to 76ft/sec. During this acceleration process you travel a distance of 300 ft.
a) Calculate the time required for the 300 ft travel...
But the instantaneous speed appears to be the same number as the average speed, which is 22.5 mi/hr. Now what about: what about the time when the speedometer will read the average speed you calculated for part a) as the instantaneous speed??
Show that Newton's Second Law is NOT valid in a reference frame moving past the laboratory frame of problem 1 with a constant acceleration?
Problem 1: In a laboratory frame of reference, an observer notes that Newton's Second Law is valid. Show that it is also valid for an...
9 mi/hr/sec. times 2.5 seconds = 22.5 mi/hr because the seconds cancel??
So I get the instantaneous from multiplying the acceleration times the time? which would be 9 mi/hr/sec. times 5 seconds = 45 mi/hr?? I am confused.