Answer:
a) T = 1,467 s
, b) A = 0.495 m
, c) v = 4.97 10⁻² m / s
Explanation:
The simple harmonic movement is described by the expression
x = A cos (wt + Ф)
Where the angular velocity is
w = √ k / m
a) Ask the period
Angular velocity, frequency and period are related
w = 2π f = 2π / T
T = 2π / w
T = 2pi √ m / k
T = 2π √ (1.2 / 22)
T = 1,467 s
f = 1 / T
f = 0.68 Hz
b) ask the amplitude
The mechanical energy of a harmonic oscillator
E = ½ k A²
A = √2 E / k
A = √ (2 2.7 / 22)
A = 0.495 m
c) the mass changes to 8.0 kg
As released from rest Ф = 0, the equation remains
x = A cos wt
w = √ (22/8)
w = 1,658
x = 3.0 cos (1,658 t)
Speed is
v = dx / dt
v = -A w sin wt
The speed is maximum when without wt = ±1
v = Aw
v = 0.03 1,658
v = 4.97 10⁻² m / s
To solve this problem it is necessary to apply the concepts related to transformers, that is to say passive electrical device that transfers electrical energy from one electrical circuit to one or more circuits.
From the mathematical definition we have that the relationship between the voltage of the first coil and the second coil is proportional to the number of loops of the first and second loop, that is:

Where
input voltage on the primary coil.
input voltage on the secondary coil.
number of turns of wire on the primary coil.
number of turns of wire on the secondary coil.
Replacing our values we have:



Replacing,


From the same relations of number of turns and the voltage of the first and second coil we also have the relation of electricity and voltage whereby:

Where
= Current Primary Coil
= Current secundary Coil
Therefore:



Therefore the maximum values for the secondary coil of the voltage is 410.56V and Current is 1.87A
B4 the tackle:
<span>The linebacker's momentum = 115 x 8.5 = 977.5 kg m/s north </span>
<span>and the halfback's momentum = 89 x 6.7 = 596.3 kg m/s east </span>
<span>After the tackle they move together with a momentum equal to the vector sum of their separate momentums b4 the tackle </span>
<span>The vector triangle is right angled: </span>
<span>magnitude of final momentum = √(977.5² + 596.3²) = 1145.034 kg m/s </span>
<span>so (115 + 89)v(f) = 1145.034 ←←[b/c p = mv] </span>
<span>v(f) = 5.6 m/s (to 2 sig figs) </span>
<span>direction of v(f) is the same as the direction of the final momentum </span>
<span>so direction of v(f) = arctan (596.3 / 977.5) = N 31° E (to 2 sig figs) </span>
<span>so the velocity of the two players after the tackle is 5.6 m/s in the direction N 31° E </span>
<span>btw ... The direction can be given heaps of different ways ... N 31° E is probably the easiest way to express it when using the vector triangle to find it</span>
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