Answer:
Explanation:
Drag force is given by
C = Drag coefficient is constant
A = Area is constant
= Velocity of the passenger jet = 1200 km/h =
= Velocity of the prop plane =
= Density of the air where the jet was flying =
= Density of the air where the prop plane was flying =
The ratio of the drag forces is .
Answer:
A. you would want to use the 12 ound ball and the 18 mph speed in order to transfer the most kinetic energy
B. KE= 1/2 MV^2 so you would want to maximize both your Mass(weight of the ball) and Velocity(speed of the roll) in order to achive the greatest KE.
C. see image
Explanation:
Full disclosure not totally sure what you are being asked in the last one but I put togethere a rough sketch that is my best guess.
Answer:
The height above the waterline that the stream reaches at the horizontal position of the insect is 15 cm.
Explanation:
Please, see the attached figure for a description of the problem.
The motion is parabolic and this is the equation that describes the position of an object in such a motion:
r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)
Where:
r = position vector
x0 = initial horizontal position
v0 = initial velocity
t = time
α = angle of the stream with the waterline
y0 = initial vertical position
g = acceleration due to gravity
First, let´s calculate how much time it takes the stream to reach the horizontal position of 0.27 m. For this, we will use the equation of the x-component of the vector position:
x = x0 + v0 · t · cos α
Since the origin of the reference system is located at the mouth of the fish, x0 = 0. Then:
0.27 m = 3.7 m/s · t · cos 35º
t = 0.27 m /(3.7 m/s · cos 35º)
t = 0.089 s
Now, with this time, we can calulate the vertical position (height) of the stream using the equation for the y-component of the vector "r":
y = y0 + v0 · t · sin α + 1/2 · g · t²
y = 0 m + 3.7 m/s · 0.089 s · sin 35º + 1/2 · (-9.8 m/s²) · (0.089s)²
y = 0.15 m
when the stream reaches 27 cm horizontally, it will reach 15 cm vertically and hit the insect!
