Answer:
Magnitude of displacement = 2.07 km
Magnitude of average velocity = 1.17 kmph
Explanation:
Let east represent positive x axis and north represent positive y axis.
A bird watcher meanders through the woods, walking 1.93 km due east, 1.03 km due south, and 3.84 km in a direction 52.8 ° north of west.
1.93 km due wast
s ₁ = 1.93 i km
1.03 km due south
s₂ = -1.03 j km
3.84 km in a direction 52.8 ° north of west
s₃ = -3.84 cos 52.8 i + 3.84 sin 52.8 j = -2.32 i + 3.06 j km
Total displacement
s = s ₁+ s₂+ s₃ = 1.93 i - 1.03 j -2.32 i + 3.06 j = -0.39 i + 2.03 j
Magnitude of displacement, 
Time taken = 1.771 hour
Magnitude of average velocity, 
Answer:
a) A = 4.0 m
, b) w = 3.0 rad / s
, c) f = 0.477 Hz
, d) T = 20.94 s
Explanation:
The equation that describes the oscillatory motion is
x = A cos (wt + fi)
In the exercise we are told that the expression is
x = 4.0 cos (3.0 t + 0.10)
let's answer the different questions
a) the amplitude is
A = 4.0 m
b) the frequency or angular velocity
w = 3.0 rad / s
c) angular velocity and frequency are related
w = 2π f
f = w / 2π
f = 3 / 2π
f = 0.477 Hz
d) the period
frequency and period are related
T = 1 / f
T = 1 / 0.477
T = 20.94 s
e) the phase constant
Ф = 0.10 rad
f) velocity is defined by
v = dx / dt
v = - A w sin (wt + Ф)
speed is maximum when sine is + -1
v = A w
v = 4 3
v = 12 m / s
g) the angular velocity is
w² = k / m
k = m w²
k = 1.2 3²
k = 10.8 N / m
h) the total energy of the oscillator is
Em = ½ k A²
Em = ½ 10.8 4²
Em = 43.2 J
i) the potential energy is
Ke = ½ k x²
for t = 0 x = 4 cos (0 + 0.1)
x = 3.98 m
j) kinetic energy
K = ½ m v²
for t = 00.1
²
v = A w sin 0.10
v = 4 3 sin 0.10
v = 1.98 m / s
So we want to know what is the purpose of a lanyard attached to a safety switch. So in case the operator falls overboard a safety switch is installed and connected to the operators hand or waist. Which ever is more practical. This safety switch turns off the motor.
Answer:
3.82 Ns
Explanation:
Time varying horizontal Force is given as
F(t) = A t⁴ + B t²
F(t) = 4.50 t⁴ + 8.75 t²
Impulse imparted is given as
