To solve this problem it is necessary to apply the continuity equations in the fluid and the kinematic equation for the description of the displacement, velocity and acceleration.
By definition the movement of the Fluid under the terms of Speed, acceleration and displacement is,
Where,
Velocity in each state
g= Gravity
h = Height
Our values are given as,
Replacing at the kinetic equation to find 
we have,
Applying the concepts of continuity,
We need to find A_2 then,
So the cross sectional area of the water stream at a point 0.11 m below the faucet is
Therefore the cross-sectional area of the water stream at a point 0.11 m below the faucet is 
It is most likely vegetable oil
Sky diving involves free fall under gravity along with the drag due to air molecules pushing against the body slowing the rate of fall of a body. This is actually a significant amount of force. The drag force depends on the contact surface area and weight of the body. More the surface area in contact, more would be the drag. The sitting position of the skydiver would experience less drag than the chest down position. This is because of the less contact surface area of the body with the air molecules while in the former case. Since no two persons have identical body shape and weight, the rate of fall can be made nearly equal but not exactly equal. This is would be possible when they are having same body position.
Answer:
Explanation:
Kinetic energy generated = work done by force = force x displacement
= 3 x 7 = 21 J
Answer:
v = 120 m/s
Explanation:
We are given;
earth's radius; r = 6.37 × 10^(6) m
Angular speed; ω = 2π/(24 × 3600) = 7.27 × 10^(-5) rad/s
Now, we want to find the speed of a point on the earth's surface located at 3/4 of the length of the arc between the equator and the pole, measured from equator.
The angle will be;
θ = ¾ × 90
θ = 67.5
¾ is multiplied by 90° because the angular distance from the pole is 90 degrees.
The speed of a point on the earth's surface located at 3/4 of the length of the arc between the equator and the pole, measured from equator will be:
v = r(cos θ) × ω
v = 6.37 × 10^(6) × cos 67.5 × 7.27 × 10^(-5)
v = 117.22 m/s
Approximation to 2 sig. figures gives;
v = 120 m/s
