I believe I seen on google if you go to Mather
Answer:
10.58 ft
Explanation:
Force, F = 1.4 N
Diameter of membrane = 7.4 mm
radius of membrane, r = 7.4 / 2 = 3.7 mm = 3.7 x 10^-3 m
Area, A = 3.14 x (3.7 x 10^-3)^2 = 4.3 x 10^-5 m^2
Density, d = 1.03 x 10^3 kg/m^3
Pressure at depth, P = h x d x g
Let h be the depth.
Pressure = force / Area
h x 1.03 x 10^3 x 9.8 = 1.4 / (4.3 x 10^-5)
h = 3.225 m = 10.58 ft
Thus, the depth of water is 10.58 ft.
Question in proper order
The rotational kinetic energy term is often called the <em>kinetic energy </em><em>in</em> the center of mass, while the translational kinetic energy term is called the <em>kinetic energy </em><em>of</em> the center of mass.
You found that the total kinetic energy is the sum of the kinetic energy in the center of mass plus the kinetic energy of the center of mass. A similar decomposition exists for angular and linear momentum. There are also related decompositions that work for systems of masses, not just rigid bodies like a dumbbell.
It is important to understand the applicability of the formula

Which of the following conditions are necessary for the formula to be valid?
a. The velocity vector
must be perpendicular to the axis of rotation
b.The velocity vector
must be perpendicular or parallel to the axis of rotation
c. The moment of inertial must be taken about an axis through the center of mass
Answer:
Option c
Explanation:

The first two conditions are untrue, this is because, you can have rotation in any direction and translation in any direction of any collection of masses. Rotational and translational velocities of masses do not depend on each other
The last statement is true because by definition, the moment of inertia, which is a measure of reluctance, is usually taken about a reference point which is the center of mass
Usain bolt is 43km/h
Deer 3km/3.75min because 45sec/60sec = .75min
All we need to do is convert
3km/3.75min * 60min/1h = 48km/h
Usain bolt 43km/h - deer 48km/h
= 5km/h
Answer:
0.5 m/s
Explanation:
From Newton's Third law of motion,
Momentum of the cannon = momentum of the flatcar.
mv = MV........................ Equation 1
Where m = mass of the cannon, v = velocity of the cannon, M = mass of the flatcar, V = Velocity of the flat car.
make V the subject of the equation
V = mv/M................... Equation 2
Given: m = 10 kg, v = 50 m/s, M = 1000 kg
Substitute into equation 2
V = 10×50/1000
V = 500/1000
V = 0.5 m/s
Hence the speed of the flatcar = 0.5 m/s