Answer:
1470kgm²
Explanation:
The formula for expressing the moment of inertial is expressed as;
I = 1/3mr²
m is the mass of the body
r is the radius
Since there are three rotor blades, the moment of inertia will be;
I = 3(1/3mr²)
I = mr²
Given
m = 120kg
r = 3.50m
Required
Moment of inertia
Substitute the given values and get I
I = 120(3.50)²
I = 120(12.25)
I = 1470kgm²
Hence the moment of inertial of the three rotor blades about the axis of rotation is 1470kgm²
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There should be an image that should accompanied your question, I was able to chcek it from other sources. the acceleration of the block when the scale reads 32N is <span>4.0 m/s*s</span>
Answer: length of B =4.00
Explanation:
for the vectors A and B and the angle between them as x.
Magnitude of the sum of A and B is given as = √(A²+B²+2ABcosx
where
Magnitude of A = 3.00
Magnitude of the sum of A and B is 5.00
5.00=√(A²+B²+2ABcos90°
5.00= √3² +b² +0
5²= 3² +b²
25=9+b²
b²= 25-9
b² = 16
b= √16
b= 4
PART a)
here when stone is dropped there is only gravitational force on it
so its acceleration is only due to gravity
so we will have

Part b)
Now from kinematics equation we will have

now we have
y = 25 m
so from above equation


Part c)
If we throw the rock horizontally by speed 20 m/s
then in this case there is no change in the vertical velocity
so it will take same time to reach the water surface as it took initially
So t = 2.26 s
Part D)
Initial speed = 20 m/s
angle of projection = 65 degree
now we have




PART E)
when stone will reach to maximum height then we know that its final speed in y direction becomes zero
so here we can use kinematics in Y direction



so it will take 1.85 s to reach the top
Answer: For leverage and so the can be easily handled.
Explanation: