Answer:
the energy difference between adjacent levels decreases as the quantum number increases
Explanation:
The energy levels of the hydrogen atom are given by the following formula:
where
is a constant
n is the level number
We can write therefore the energy difference between adjacent levels as
We see that this difference decreases as the level number (n) increases. For example, the difference between the levels n=1 and n=2 is
While the difference between the levels n=2 and n=3 is
And so on.
So, the energy difference between adjacent levels decreases as the quantum number increases.
Answer:
4.96 × 10⁵ Pa
Explanation:
F = mg
This force is evenly distributed on the three leg
radius, r = d/2
= 2.8 / 2
= 1.4 cm = 0.014 m
total cross sectional area of the three legs, A = 3*pi*r^2
Pressure due to weight,
P = Weight/A
P = 4.96 × 10⁵ Pa
Answer:
5.1*10^3 J/m^3
Explanation:
Using E = q/A*eo
And
q =75*10^-6 C
A = 0.25
eo = 8.85*10^-12
Energy density = 1/2*eo*(E^2) = 1/2*eo*(q/A*eo)^2 = [q^2] / [2*(A^2)*eo]
= [(75*10^-6)^2] / [2*(0.25)^2*8.85*10^-12]
= 5.1*10^3 J/m^3
Answer:cart B
Explanation:
For cart A speed is constant therefore there is no acceleration because acceleration is rate of change of velocity
thus there is no net force
For cart B there is change in velocity in the left direction , so there is net acceleration towards left
so there is net force in the left direction
For cart C there is decrease in velocity i.e. negative acceleration or deceleration . Therefore there is a net force towards right which opposes the motion
-- We know that the y-component of acceleration is the derivative of the
y-component of velocity.
-- We know that the y-component of velocity is the derivative of the
y-component of position.
-- We're given the y-component of position as a function of time.
So, finding the velocity and acceleration is simply a matter of differentiating
the position function ... twice.
Now, the position function may look big and ugly in the picture. But with the
exception of 't' , everything else in the formula is constants, so we don't even
need any fancy processes of differentiation. The toughest part of this is going
to be trying to write it out, given the text-formatting capabilities of the wonderful
envelope-pushing website we're working on here.
From the picture . . . . . y (t) = (1/2) (a₀ - g) t² - (a₀ / 30t₀⁴ ) t⁶
First derivative . . . y' (t) = (a₀ - g) t - 6 (a₀ / 30t₀⁴ ) t⁵ = (a₀ - g) t - (a₀ / 5t₀⁴ ) t⁵
There's your velocity . . . /\ .
Second derivative . . . y'' (t) = (a₀ - g) - 5 (a₀ / 5t₀⁴ ) t⁴ = (a₀ - g) - (a₀ /t₀⁴ ) t⁴
and there's your acceleration . . . /\ .
That's the one you're supposed to graph.
a₀ is the acceleration due to the model rocket engine thrust
combined with the mass of the model rocket
'g' is the acceleration of gravity ... 9.8 m/s² or 32.2 ft/sec²
t₀ is how long the model rocket engine burns
Pick, or look up, some reasonable figures for a₀ and t₀
and you're in business.
The big name in model rocketry is Estes. Their website will give you
all the real numbers for thrust and burn-time of their engines, if you
want to follow it that far.
