Answer:
Explanation:
Given that,
The compression in the spring, x = 0.0647 m
Speed of the object, v = 2.08 m/s
To find,
Angular frequency of the object.
Solution,
We know that the elation between the amplitude and the angular frequency in SHM is given by :
A is the amplitude
In case of spring the compression in the spring is equal to its amplitude
So, the angular frequency of the spring is 32.14 rad/s.
Answer:
1 greater distances fallen in successive seconds
Explanation:
When a body falls freely it is subjected to the action of the force of gravity, which gives an acceleration of 9.8 m / s2, consequently, we are in an accelerated movement
If we use the kinematic formula we can find the position of the body
Y = Vo t + ½ to t2
Where the initial velocity is zero or constant and the acceleration is the acceleration of gravity
Y = - ½ g t2 = - ½ 9.8 t2 = -4.9 t2
Let's look for the position for successive times
t (s) Y (m)
1 -4.9
2 -19.6
3 -43.2
The sign indicates that the positive sense is up
It can be clearly seen that the distance is greatly increased every second that passes
Answer:
The work done to get you safely away from the test is 2.47 X 10⁴ J.
Explanation:
Given;
length of the rope, L = 70 ft
mass per unit length of the rope, μ = 2 lb/ft
your mass, W = 120 lbs
mass of the 70 ft rope = 2 lb/ft x 70 ft
= 140 lbs.
Total mass to be pulled to the helicopter, M = 120 lbs + 140 lbs
= 260 lbs
The work done is calculated from work-energy theorem as follows;
W = Mgh
where;
g is acceleration due gravity = 32.17 ft/s²
h is height the total mass is raised = length of the rope = 70 ft
W = 260 Lb x 32.17 ft/s² x 70 ft
W = 585494 lb.ft²/s²
1 lb.ft²/s² = 0.0421 J
W = 585494 lb.ft²/s² = 2.47 X 10⁴ J.
Therefore, the work done to get you safely away from the test is 2.47 X 10⁴ J.
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