Answer:
a) k = 120 N / m
, b) f = 0.851 Hz
, c) v = 1,069 m / s
, d) x = 0
, e) a = 5.71 m / s²
, f) x = 0.200 m
, g) Em = 2.4 J
, h) v = -1.01 m / s
Explanation:
a) Hooke's law is
F = k x
k = F / x
k = 24.0 / 0.200
k = 120 N / m
b) the angular velocity of the simple harmonic movement is
w = √ k / m
w = √ (120 / 4.2)
w = 5,345 rad / s
Angular velocity and frequency are related.
w = 2π f
f = w / 2π
f = 5.345 / 2π
f = 0.851 Hz
c) the equation that describes the movement is
x = A cos (wt + Ф)
As the body is released without initial velocity, Ф = 0
x = 0.2 cos wt
Speed is
v = dx / dt
v = -A w sin wt
The speed is maximum for sin wt = ±1
v = A w
v = 0.200 5.345
v = 1,069 m / s
d) when the function sin wt = -1 the function cos wt = 0, whereby the position for maximum speed is
x = A cos wt = 0
x = 0
e) the acceleration is
a = d²x / dt² = dv / dt
a = - Aw² cos wt
The acceleration is maximum when cos wt = ± 1
a = A w²
a = 0.2 5.345
a = 5.71 m / s²
f) the position for this acceleration is
x = A cos wt
x = A
x = 0.200 m
g) Mechanical energy is
Em = ½ k A²
Em = ½ 120 0.2²
Em = 2.4 J
h) the position is
x = 1/3 A
Let's calculate the time to reach this point
x = A cos wt
1/3 A = A cos 5.345t
t = 1 / w cos⁻¹(1/3)
The angles are in radians
t = 1.23 / 5,345
t = 0.2301 s
Speed is
v = -A w sin wt
v = -0.2 5.345 sin (5.345 0.2301)
v = -1.01 m / s
i) acceleration
a = -A w² sin wt
a = - 0.2 5.345² cos (5.345 0.2301)
a = -1.91 m / s²
Answer:
Part 1) Voltage in secondary windings is 61.08 Volts
Part 2) Current in secondary windings is 0.53 Amperes
Explanation:
The potential developed in the primary and secondary winding of a transformer are related as
where
Np no of turns in primary coil
Ns no of turns in secondary coil
Vp Voltage of turns in primary coil
Vs Voltage of turns in secondary coil
Applying values in the formula we get
Part 2)
Using Ohm's law the current is given by
Answer:
what if I do and b then someone else c I don't have enough time pls
A. Impulse is simply the product of Force and time.
Therefore,
I = F * t --->
1
where I is impulse, F is force, t is time
However another formula for solving impulse is:
I = m vf – m vi --->
2
where m is mass, vf is final velocity and vi is initial
velocity
Therefore using equation 2 to solve for impulse I:
I = 2000kg (0) – 2000kg (77 m/s)
I = -154,000 kg m/s
B. By conservation of momentum, we also know that Impulse
is conserved. That means that increasing the time by a factor of 3 would still
result in an impuse of -154,000 kg m/s. So,
I = F’ * (3 t) = -154,000 kg m/s
Since t is multiplied by 3, therefore this only means
that Force is decreased by a factor of 3 to keep the impulse constant,
therefore:
(F/3) (3t) = -154,000 kg m/s
Summary of Answers:
A. I = -154,000 kg m/s
B. Force is decreased by factor of 3
