Answer:
v = 3.81 m/s
Explanation:
First, we will find out the radius (r)
sinΘ = r / L
or
r = LsinΘ
on substituting the values, we get
r = 1.2 × sin18°
or
r = 0.370 m
Now,
The tension (T) can be calculated as
ΣF(y) = 0
or
Tcos18° - mg = 0
or
T = mg / cos18.8°
on substituting the values, we have
T = (2kg x 9.8m/s²) / cos18°
or
T = 20.60 N
Now, applying the equilibrium condition in horizontal direction. we have
also, the centripetal acceleration 
on substituting the values, we get
m(v² / r) = Tsin18°
v = (rTsin18° / m)
v = [(1.2m)(20.60 × sin18°) / 2kg]
v = 3.81 m/s
<span>Assuming the speeds of both the stick and Ingrid are taken from a stationary point, the velocities of both Ingrid and the stick will be subtracted to see how much faster the stick is going.
16 - 9
= 7 km/h is the speed of the stick relative to Ingrid</span>
The pressure at a certain depth underwater is:
P = ρgh
P = pressure, ρ = sea water density, g = gravitational acceleration near Earth, h = depth
The pressure exerted on the submarine window is:
P = F/A
P = pressure, F = force, A = area
The area of the circular submarine window is:
A = π(d/2)²
A = area, d = diameter
Set the expressions for the pressure equal to each other:
F/A = ρgh
Substitute A:
F/(π(d/2)²) = ρgh
Isolate h:
h = F/(ρgπ(d/2)²)
Given values:
F = 1.1×10⁶N
ρ = 1030kg/m³ (pulled from a Google search)
g = 9.81m/s²
d = 30×10⁻²m
Plug in and solve for h:
h = 1.1×10⁶/(1030(9.81)π(30×10⁻²/2)²)
h = 1540m
Of the cliff?
Projectile motion
In the problem we are asked to find a height of certain cliff when a motorcycle stunt driver zoom out horizontally at the end the cliff at an initial velocity. So we will use one of the kinematics equation for projectile motion,
y
=
v
o
y
t
+
1
2
g
t
where
v
o
y
is just equal to zero since we can assume that the driver zooms out horizontally,
g
=
9.8
m
/
s
2
and
t
is time after
