Becuse your weighting with chalk that has pigment
        
             
        
        
        
Answer:
s = 20 m
Explanation:
given,
mass of the roller blader = 60 Kg
length = 10 m
inclines at = 30°
coefficient of friction = 0.25
using conservation of energy
 
 
 
 
 
 
u = 9.89 m/s
Using second law of motion  
ma =μ mg
a = μ g
a = 0.25 x 9.8
a = 2.45 m/s²
Using third equation of motion ,  
v² - u² = 2 a s
0² - 9.89² = 2 x 2.45 x s
s = 20 m
the distance moved before stopping is 20 m
 
        
             
        
        
        
Answer:
V=4.7m/s
Explanations:
Let Ma mass of cat A=7kg
Va velocity of cat A=7m/s
Mb mass of cat b=6.1kg
VB velocity of cat b=2m/s
From conservation of linear momentum
MaVa+MbVb=(Ma+Mb)V
7*7+6.1*2=(7+6.1)V
61.2=13.1V
V=4.7m/s
 
        
             
        
        
        
Answer:
hello your question has some missing values attached below is the complete question with the missing values
answer :
a) 0.083 secs
b) 0.33 secs
c)  3e^-4/3
Explanation:
Given that 
g = 32 ft/s^2 ,  spring constant ( k ) = 2 Ib/ft
initial displacement = 1 ft above equilibrium 
mass = weight / g = 4/32 = 1/8 
damping force = instanteous velocity  hence  β = 1
a<u>)Calculate the time at which the mass passes through the equilibrium position.</u>
time mass passes through equilibrium = 1/12 seconds = 0.083
<u>b) Calculate the time at which the mass attains its extreme displacement </u>
time when mass attains extreme displacement = 1/3 seconds = 0.33 secs
<u>c) What is the position of the mass at this instant</u>
position = 3e^-4/3
attached below is the detailed solution to the given problem