Answer:
v = 4.25 m / s
Explanation:
To solve this exercise we must use Newton's second law, let's set a reference system that is horizontal and vertical
Axis y
- W = 0
X axis
Tₓ = m a
The acceleration is centripetal
a = v² / r
Tₓ = m v² / r
v² = Tₓ r / m (1)
Let's use trigonometry to find the tension components,
sin θ = Tₓ / T
cos θ =
/ T
= T cos θ
Tₓ = T sin θ
Let's look angle for the maximum tension 103 N
= T cos θ = W
θ = cos⁻¹ W / T
θ = cos⁻¹ (5.77 9.8 / 103)
θ = 56.7°
Now let's find the radius of the circle
sin 56.7 = r / L
r = L sin 56.7
We substitute in the speed equation (1)
v² = T sin 56.7 L sin 56.7 / m
v = √ T L sin² 56.7 / m
Let's calculate
v = √ (103 1.45 sin² 56.7 /5.77)
v = 4. 25 m / s
Answer:
1.176m
Explanation:
Local ambient pressure(P1) = 100 kPa
Absolute pressure(P2)=260kPa
Net pressure=absolute pressure-local ambient absolute pressure
Net pressure=P1(absolute pressure)-P2(local ambient absolute pressure)
Net pressure=260-100=160kPa
Pressure= ρgh
160kPa=13600*10*h
h=
h=1.176m
<em>The gravitational force between two objects is inversely proportional to the square of the distance between the two objects.</em>
The gravitational force between two objects is proportional to the product of the masses of the two objects.
The gravitational force between two objects is proportional to the square of the distance between the two objects. <em> no</em>
The gravitational force between two objects is inversely proportional to the distance between the two objects. <em> no</em>
The gravitational force between two objects is proportional to the distance between the two objects. <em> no</em>
The gravitational force between two objects is inversely proportional to the product of the masses of the two objects. <em> no</em>
Given:
k = 100 lb/ft, m = 1 lb / (32.2 ft/s) = 0.03106 slugs
Solution:
F = -kx
mx" = -kx
x" + (k/m)x = 0
characteristic equation:
r^2 + k/m = 0
r = i*sqrt(k/m)
x = Asin(sqrt(k/m)t) + Bcos(sqrt(k/m)t)
ω = sqrt(k/m)
2π/T = sqrt(k/m)
T = 2π*sqrt(m/k)
T = 2π*sqrt(0.03106 slugs / 100 lb/ft)
T = 0.1107 s (period)
x(0) = 1/12 ft = 0.08333 ft
x'(0) = 0
1/12 = Asin(0) + Bcos(0)
B = 1/12 = 0.08333 ft
x' = Aω*cos(ωt) - Bω*sin(ωt)
0 = Aω*cos(0) - (1/12)ω*sin(0)
0 = Aω
A = 0
So B would be the amplitude. Therefore, the equation of motion would be x
= 0.08333*cos[(2π/0.1107)t]