Since the angular acceleration α is constant,
0² - (30 rev/s)² = 2α (60 rev)
Solve for α :
α = -(30 rev/s)²/(120 rev) = -7.5 rev/s²
One revolution corresponds to an angular displacement of 2π radians, so
α = (-7.5 rev/s²) × (2π rad/rev) ≈ -47.12 rad/s²
<h2>
Answer:</h2>
400N/m
<h2>
Explanation:</h2>
When n identical springs of stiffness k, are attached in series, the reciprocal of their equivalent stiffness (1 / m) is given by the sum of the reciprocal of their individual stiffnesses. i.e
= ∑ⁿ₁ [
] -----------------------(i)
That is;
=
+
+
+ . . . +
-------------------(ii)
If they have the same value of stiffness say s, then equation (ii) becomes;
= n x
-----------------(iii)
Where;
n = number of springs
From the question,
There are 3 identical springs, each with stiffness of 1200N/m and they are attached in series. This implies that;
n = 3
s = 1200N/m
Now, to calculate the effective stiffness,m, (i.e the stiffness of a longer spring formed from the series combination of these springs), we substitute these values into equation (iii) above as follows;
= 3 x 
= 
= 
Cross multiply;
m = 400N/m
Therefore, the stiffness of the longer spring is 400N/m
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<h2>
Option A is the correct answer.</h2>
Explanation:
For a simple pendulum we have at any position
Total energy = Constant
Kinetic energy + Potential energy = Constant
At maximum displacement of pendulum, velocity is zero, hence kinetic energy is zero.
At maximum displacement the pendulum only have potential energy.
Given that maximum potential energy is 10 J.
That is at any position
Kinetic energy + Potential energy = 10 J
Now we need to find kinetic energy when potential energy is 5 J.
Kinetic energy + 5 J = 10 J
Kinetic energy = 5 J
Option A is the correct answer.