<span>a thin fibrous cartilage between the surfaces of some joints, e.g., the knee.</span>
Answer:
aₓ = 0
, ay = -6.8125 m / s²
Explanation:
This is an exercise that we can solve with kinematics equations.
Initially the rabbit moves on the x axis with a speed of 1.10 m / s and after seeing the predator acceleration on the y axis, therefore its speed on the x axis remains constant.
x axis
vₓ = v₀ₓ = 1.10 m / s
aₓ = 0
y axis
initially it has no speed, so v₀_y = 0 and when I see the predator it accelerates, until it reaches the speed of 10.6 m / s in a time of t = 1.60 s. let's calculate the acceleration
= v_{oy} -ay t
ay = (v_{oy} -v_{y}) / t
ay = (0 -10.9) / 1.6
ay = -6.8125 m / s²
the sign indicates that the acceleration goes in the negative direction of the y axis
Answer:
(a) v = 3..6 m/s
(b) The rain falling downward has been able to affect the horizontal motion of the car by reducing it's velocity from 4 m/s to 3.6 m/s.
Explanation:
from the question we have the following:
mass of the car (Mc) = 24,000 kg
initial velocity of the car (u) = 4 m/s
mass of water (Mw) = 3000 kg
final velocity of the car (v) = ?
(a) we can calculate the final momentum of the car by applying the conservation of momentum where
initial momentum = final momentum
Mc x U = (Mc + Mw) x V
24000 x 4 = (24000 + 3000) x v
96,000 = 27000v
v =3.6 m/s
(b) The rain falling downward has been able to affect the horizontal motion of the car by reducing it's velocity from 4 m/s to 3.6 m/s.
Answer:
The minimum coefficient of friction is 0.544
Solution:
As per the question:
Radius of the curve, R = 48 m
Speed of the car, v = 16 m/s
To calculate the minimum coefficient of static friction:
The centrifugal force on the box is in the outward direction and is given by:

where
= coefficient of static friction
The net force on the box is zero, since, the box is stationary and is given by: