Answer:
The amplitude of the oscillation is 2.82 cm
Explanation:
Given;
mass of attached block, m = 4.1 kg
energy of the stretched spring, E = 3.8 J
period of oscillation, T = 0.13 s
First, determine the spring constant, k;

where;
T is the period oscillation
m is mass of the spring
k is the spring constant

Now, determine the amplitude of oscillation, A;

where;
E is the energy of the spring
k is the spring constant
A is the amplitude of the oscillation

Therefore, the amplitude of the oscillation is 2.82 cm
The hero attending a funeral is safe behavior while the hero driving fast is riskier behavior
<span>You want to focus on the car's acceleration.
(Hope this helps you! :) Have a nice day!)</span>
Answer:
a. -8 cm
Explanation:
= distance of the object = 4 cm
= distance of the image = ?
= focal length of the converging lens = 8 cm
using the lens equation


= - 8 cm
Answer:
Explanation:
a )
Each blade is in the form of rod with axis near one end of the rod
Moment of inertia of one blade
= 1/3 x m l²
where m is mass of the blade
l is length of each blade.
Total moment of moment of 3 blades
= 3 x
x m l²
ml²
2 )
Given
m = 5500 kg
l = 45 m
Putting these values we get
moment of inertia of one blade
= 1/3 x 5500 x 45 x 45
= 37.125 x 10⁵ kg.m²
Moment of inertia of 3 blades
= 3 x 37.125 x 10⁵ kg.m²
= 111 .375 x 10⁵ kg.m²
c )
Angular momentum
= I x ω
I is moment of inertia of turbine
ω is angular velocity
ω = 2π f
f is frequency of rotation of blade
d )
I = 111 .375 x 10⁵ kg.m² ( Calculated )
f = 11 rpm ( revolution per minute )
= 11 / 60 revolution per second
ω = 2π f
= 2π x 11 / 60 rad / s
Angular momentum
= I x ω
111 .375 x 10⁵ kg.m² x 2π x 11 / 60 rad / s
= 128.23 x 10⁵ kgm² s⁻¹ .