Answer:
The terminal speed of this object is 12.6 m/s
Explanation:
It is given that,
Mass of the object, m = 80 kg
The magnitude of drag force is,

The terminal speed of an object is attained when the gravitational force is balanced by the gravitational force.



On solving the above quadratic equation, we get two values of v as :
v = 12.58 m/s
v = -15.58 m/s (not possible)
So, the terminal speed of this object is 12.6 m/s. Hence, this is the required solution.
Answer:
Distance 20 km and Displacement 0 km
His displaceent is 0 km because he ends his walk where he started. The total distance of his walk is 20 km because he walks 10 km to the store + 10km back home.
Answer:
The velocity of the man is 0.144 m/s
Explanation:
This is a case of conservation of momentum.
The momentum of the moving ball before it was caught must equal the momentum of the man and the ball after he catches the ball.
Mass of ball = 0.65 kg
Mass of the man = 54 kg
Velocity of the ball = 12.1 m/s
Before collision, momentum of the ball = mass x velocity
= 0.65 x 12.1 = 7.865 kg-m/s
After collision the momentum of the man and ball system is
(0.65 + 54)Vf = 54.65Vf
Where Vf is their final common velocity.
Equating the initial and final momentum,
7.865 = 54.65Vf
Vf = 7.865/54.65 = 0.144 m/s
Answer:
0.2cm towards the retina.
Explanation:
the focal length of the frog eye is
(1/f) = (1/10) + (1/0.8)
f = 0.74cm
Since the distance of the object is 15cm Hence
(1/0.74) = (1/15) + (1/V)
V = 0.78cm
Therefore the distance the retina is to move is
0.78cm - 0.8cm = 0.02cm towards the retina.