Answer:
v=0.94 m/s
Explanation:
Given that
M= 5.67 kg
k= 150 N/m
m=1 kg
μ = 0.45
The maximum acceleration of upper block can be μ g.
a= μ g ( g = 10 m/s²)
The maximum acceleration of system will ω²X.
ω = natural frequency
X=maximum displacement
For top stop slipping
μ g =ω²X
We know for spring mass system natural frequency given as
By putting the values
ω = 4.47 rad/s
μ g =ω²X
By putting the values
0.45 x 10 = 4.47² X
X = 0.2 m
From energy conservation
150 x 0.2²=6.67 v²
v=0.94 m/s
This is the maximum speed of system.
Answer:
R = 9880 yd * 3 ft/yd / 5280 ft/mi = 5.61 mi
If you do it in steps
R = 9880 yd * 3 ft/yd = 29640 ft
R = 29640 ft / 5280 ft/mi = 5.61 mi
When object reached the terminal speed then its acceleration is zero
So as per Newton's II law we can say
now in that case we can say that net force is zero so here weight of the object is counter balanced by the drag force when it will reach at terminal speed
so we can write
so here we are given that
so terminal speed will be nearly 2 m/s
Explanation:
Total surface area of cylinder:
sheet of metal required = 120.95 cm^2
