Answer:
2.23 Hz
Explanation:
From the attached diagram below; there exists a diagrammatic representation of the equilibrium position of the cylinder.
The equilibrium position of the spring is expressed as:
mg = K
where;
m = mass of the object
g = acceleration due to gravity
K = spring constant
= static deflection of the string
Given that:
m = 30 kg
g = 9.81 m/s²
= 50 mm = 50 × 
= 0.05 m
Then;
From here; let us find the angular velocity which will be needed to determine the natural frequency aftewards.
The angular velocity of the cylinder can be expressed by the formula:
Finally; the natural frequency 
can be calculated by using the equation
= 2.229305729
≅ 2.23 Hz
Thus; the resulting natural frequency of the vertical vibration of the cylinder = 2.23 Hz
Answer:
d) 15.12 N
e) 15.12 N
Explanation:
Draw a free body diagram of the each block.
Block A has three forces on it: weight force mAg pulling down, normal force N pushing up, and tension force T pulling down.
Block B has two forces on it: weight force mBg pulling down, and tension force T pulling up.
Sum of forces on A in the y direction:
∑F = ma
T + N − mAg = mAa
N = mAa + mAg − T
N = mA (a + g) − T
Sum of forces on B in the y direction:
∑F = ma
T − mBg = mBa
T = mBa + mBg
T = mB (a + g)
Plug in values:
T = (1.80 kg) (-1.60 m/s² + 10 m/s²)
T = 15.12 N
N = (3.60 kg) (-1.60 m/s² + 10 m/s²) − 15.12 N
N = 15.12 N
So the answers to (d) and (e) are both 15.12 N.
Answer:
it depends on the weight's ratio
(sorry)
Current= voltage divided by resistance
120/30=4
Answer:
α = 3.27 rad/s²
No
Explanation:
Given that
L= 3 m
Lets take mass of tree = m kg
The torque to mass m = m g .L
The mass moment of inertia of tree ,I
I = mL² kg.m²
We know that
τ = I α
m g .L = mL² .α
α= g/L
Now by putting the value
α = 3.27 rad/s²
No we can find the mass of the from the above information because angular acceleration is not depend on the mass of tree.
