Recall that
where and
are the initial and final velocities, respecitvely;
is the acceleration; and
is the change in position.
So we have
(Normally, this equation has two solutions, but we omit the negative one because the car is moving in one direction.)
Answer:
they meet from the girl's original position at: 2.37 (meters)
Explanation:
We need to use the Newton's law, exactly the second law that relate force, mass and acceleration as: with this we can get both accelerations; solving for acceleration
. Now
and
. Then knowing that they both travel at the same time and assuming that the distance among the girl and the sled is: 15.0-x, so,
and
, solving for the time we get:
and
with this equations we solving for the x that is the distance between the girl and the sled after the apply the force and we get:
. Finally we get:
and replacing the values we have got:
so
so x=2.37 (meters).
Answer:
The time is 1.8s
Explanation:
The ball droped, will freely fall under gravity.
Hence we use free fall formula to calculate the time by the ball to hit the ground
Where h is the height from which the ball is droped, g is the acceleration due to gravity that acted on the ball, and t is the time taken by the ball to hit the ground.
From the question,
h=16m
Also, let take
By substitution we obtain,
Diving through by 9.8
square root both sides, we obtain
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)