Always stop as soon as you can at a safe place and shut the engine off when the warning/gauge light goes on in a vehicle.
Answer: P= W/t so P=50/20 =2.5 W
What it looks to be that you found in A was the "initial"...b/c the question asks:
<span>"how much energy does the electron have 'initially' in the n=4 excited state?" </span>
<span>"final" would be where it 'finally' ends up at, ie. its last stop...as for this question...the 'ground state' as in its lowest energy level. </span>
The answer comes to: <span>−1.36×10^−19 J</span>
You use the same equation for the second part as for part a.
<span>just have to subract the 2 as in the only diff for part 2 is that you use 1squared rather than 4squared & subract "final -initial" & you should get -2.05*10^-18 as your answer. </span>
Answer:
A) 1.122 m
B) 5.4 hertz
C) 6.06 m/s ( towards negative x axis )
D) 0.22
E) max speed = amplitude * 34
= 0.22 * 34 = 7.48
minimum speed ( speed at rest ) = 0
Explanation:
The travelling wave = D = 0.22sin ( 5.6x + 34t)
this wave is represented in the form : D = Asin ( kx + wt)
A) wavelength
k = 2
/ v
v = wavelength
k = 5.6
therefore wavelength ( v ) from the equation = 2
/ 5.6 = 1.122 m
B ) frequency
w = 2
f
w = 34
f = frequency
therefore f = w / 2
= 34 / 2
= 5.4 hertz
C ) velocity
speed = f v
v = 5.4 * 1.122 = 6.06 m/s ( towards negative x axis )
D) amplitude
A = 0.22
E ) maximum and minimum speeds of particles on the chord
max speed = amplitude * 34
= 0.22 * 34 = 7.48
minimum speed ( speed at rest ) = 0