m = mass of the person = 82 kg
g = acceleration due to gravity acting on the person = 9.8 m/s²
F = normal force by the surface on the person
f = kinetic frictional force acting on the person by the surface
μ = Coefficient of kinetic friction = 0.45
The normal force by the surface in upward direction balances the weight of the person in down direction , hence
F = mg eq-1
kinetic frictional force on the person acting is given as
f = μ F
using eq-1
f = μ mg
inserting the values
f = (0.45) (82) (9.8)
f = 361.6 N
Answer:
static coefficient = 0,203 & kinetic coefficient = 0,14
Explanation:
There are two (2) conditions, when the desk is about to move and when the desk is moving. In the attachements you can see the two free body diagram for each condition.
In the first condition, there is no movement and the force is 12 N, in the image we can see the total forces are equal to 0 and by the definition of the friction force we can get the static friction coefficient.
In the second condition there is movement in the direction of the force which is equal to 8 N, again by the definition of the friction force we can get the kinetic friction coefficient. Since the desk is moving with constant velocity there is not acceleration.
what causes a star to shine brightly:
by squeezing atoms together in its core
Answer:
a = - 1.987 × 10⁶ ft/s²
t = 6.84 × 10⁻⁴ s
Explanation:
v₀ = 910 ft/s
x = 5 in.
relation v = v₀ - k x
v = 0 as body comes to rest
0 = 900 - 5k/12
k = 2184 s⁻¹
acceleration

where
(A) a = -k × v
at v= 910 ft/s
a = - 1.987 × 10⁶ ft/s²
(B) at x = 3.9 in.
v = 910 - 3.9(2184)/12
v = 200.2 m/s




t = 6.84 × 10⁻⁴ s