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
Answer:
(a) θ = 33.86°
(b) Ay = 49.92 N
Explanation:
You have that the magnitude of a vector is A = 89.6 N
The x component of such a vector is Ax = 74.4 N
(a) To find the angle between the vector and the x axis you use the following formula for the calculation of the x component of a vector:
(1)
Ax: x component of vector A
A: magnitude of vector A
θ: angle between vector A and the x axis
You solve the equation (1) for θ, by using the inverse of cosine function:
the angle between the A vector and the x axis is 33.86°
(b) The y component of the vector is given by:
the y comonent of the vecor is Ay = 49.92 N
Explanation:
formula for force is:
force=mass × acceleration
but in case of friction
force =coefficient of friction × Normal Reaction
F. = u × R
U = F/R
but when placed horizontally
R= M×G
M=mass=60kg
G=Gravity(10m/s or 9.8m/s)
F=140N
U=140/60×10
U=140/600
U=0.2333333333
approximately to 3 significant figures
U=0.233
if i am correct rate it 5 star
For purposes of completing our calculations, we're going to assume that
the experiment takes place on or near the surface of the Earth.
The acceleration of gravity on Earth is about 9.8 m/s², directed toward the
center of the planet. That means that the downward speed of a falling object
increases by 9.8 m/s for every second that it falls.
3 seconds after being dropped, a stone is falling at (3 x 9.8) = 29.4 m/s.
That's the vertical component of its velocity. The horizontal component is
the same as it was at the instant of the drop, provided there is no horizontal
force on the stone during its fall.
