When two ions form a bond, the overall charge of that compound will ALWAYS become..
1 answer:
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Lets se
And
So
If spring constant is doubled mass must be doubled
Answer:
q = q₀ sin (wt)
Explanation:
In your statement it is not clear the type of circuit you are referring to, there are two possibilities.
1) The circuit of this problem is a system formed by an Ac voltage source and a capacitor, in this case all the voltage of the source is equal to the voltage at the terminals of the capacitor
ΔV = Δ
we assume that the source has a voltage of the form
ΔV = ΔV₀o sin wt
The capacitance of a capacitor is
C = q / ΔV
q = C ΔV sin wt
the current in the circuit is
i = dq / dt
i = c ΔV₀ w cos wt
if we use
cos wt = sin (wt + π / 2)
we make this change by being a resonant oscillation
we substitute
i = w C ΔV₀ sin (wt + π/2)
With this answer we see that the current in capacitor has a phase factor of π/2 with respect to the current
2) Another possible circuit is an LC circuit.
In this case the voltage alternates between the inductor and the capacitor
V_{L} + V_{C} = 0
L di / dt + q / C = 0
the current is
i = dq / dt
they ask us for a solution so that
L d²q / dt² + 1 / C q = 0
d²q / dt² + 1 / LC q = 0
this is a quadratic differential equation with solution of the form
q = A sin (wt + Ф)
to find the constant we derive the proposed solution and enter it into the equation
di / dt = Aw cos (wt + Ф)
d²i / dt²= - A w² sin (wt + Ф)
- A w² + 1 /LC A = 0
w = √ (1 / LC)
To find the phase factor, for this we use the initial conditions for t = 0
in the case of condensate for t = or the charge is zero
0 = A sin Ф
Ф = 0
q = q₀ sin (wt)
Answer:
h = 10.4 m
R = 22.48 m
v= 16,2 m/s , α = 61.7°, below the horizontal
v = (7.7)i + (-14.3)j in meters per second (i horizontal, j downward)
Explanation:
The ball describes a parabolic path, and the equations of the movement are:
Equation of the uniform rectilinear motion (horizontal ) :
x = vx*t :
Equations of the uniformly accelerated rectilinear motion of upward (vertical ).
y = (v₀y)*t - (1/2)*g*t² Equation (2)
vfy² = v₀y² -2gy Equation (3)
vfy = v₀y -gt Equation (4)
Where:
x: horizontal position in meters (m)
t : time (s)
vx: horizontal velocity in m
y: vertical position in meters (m)
v₀y: initial vertical velocity in m/s
vfy: final vertical velocity in m/s
g: acceleration due to gravity in m/s²
Known data
y= 8.8 m
v = ( (7.7)i + (5.7)j ) m/s : vx= 7.7 m/s , vy= 5.7 m/s
g = 9.8 m/s²
Calculation of the initial vertical velocity ( v₀y)
We apply Equation (3) with the known data
(vfy)² = (v₀y)² -2*g*y
(5.7)² = (v₀y)²- (2)*(9.8)*(8.8)
(5.7)²+ 172.48 = (v₀y)²
v₀y = 14.3 m/s
Calculation of the maximum height the ball rise (h)
In the maximum height vfy=0
We apply the Equation (3) :
(vfy)² = (v₀y)² -2*g*y
0 = (14.3)² - 2*98*h
h = (14.3)² / 19.6
h = 10.4 m
Calculation of the time it takes for the ball to the maximum height
We apply the Equation (4) :
vfy = v₀y -gt
0 = v₀y -gt
gt = v₀y
t = v₀y/g
t = 14.3/9.8
t= 1.46 s
Flight time = 2t = 2.92 s
Total horizontal distance traveled by the ball (R)
We replace data in the equation (1)
x =vx*t vx= 7.7 m/s , t =2.92 s (Flight time)
R = (7.7)* (2.92) = 22.48 m
Velocity of the ball (magnitude (v) and direction (α)) the instant before it hits the ground
vx = 7.7 m/s
vy = v₀y -gt = 14.3 - 9.8* (2.92) = -14.3 m/s
v= 16,2 m/s
α = -61.7°
α = 61.7°, below the horizontal
i- j components of the v
v = (7.7)i + (-14.3)j in meters per second (i horizontal, j downward)