Answer:
Explanation:
Given data
Terminal velocity for spread eagle position vt=130 km/h
Terminal velocity for nosedive position vt=326 km/h
The terminal speed of the diver is given by
Therefore the area is given by
Since everything else is constant in the two dives except for the terminal velocity, the ratio between the area in the slow position to the area in the fast position is
Answer:
speed and time are Vf = 4.43 m/s and t = 0.45 s
Explanation:
This is a problem of free fall, we have the equations of kinematics
Vf² = Vo² + 2g x
As the object is released the initial velocity is zero, let's look at the final velocity with the equation
Vf = √( 2 g X)
Vf = √(2 9.8 1)
Vf = 4.43 m/s
This is the speed with which it reaches the ground
Having the final speed we can find the time
Vf = Vo + g t
t = Vf / g
t = 4.43 / 9.8
t = 0.45 s
This is the time of fall of the body to touch the ground
Answer:
A) x4
Explanation:
Magnification is equal to image size divided by the actual size, or M = I/A.
The image size is the student's drawing, which is 28.8 cm, and the actual size is 7.2 cm. Divide them, and cancel out the units, and you should get:
28.8 cm/7.2 cm = 4
