Answer:
2440.24 J
Explanation:
Moment of inertia, I1 = 5 kg m^2
frequency, f1 = 3 rps
ω1 = 2 x π x f1 = 2 x π x 3 = 6 π rad/s
Moment of inertia, I2 = 2 kg m^2
Let the new frequency is f2.
ω2 = 2 x π x f2
here no external torque is applied, so the angular momentum remains constant.
I1 x ω1 = I2 x ω2
5 x 6 π = 2 x 2 x π x f2
f2 = 7.5 rps
ω2 = 2 x π x 7.5 = 15 π
Initial kinetic energy, K1 = 1/2 x I1 x ω1^2 = 0.5 x 5 x (6 π)² = 887.36 J
Final kinetic energy, K2 = 1/2 x I2 x ω2^2 = 0.5 x 3 x (15 π)² = 3327.6 J
Work done, W = Change in kinetic energy = 3327.6 - 887.36 = 2440.24 J
A Venn diagram may also be called a primary diagram, set diagram or logic diagram. It is a diagram that shows all possible logical relations between a finite collection of different sets. ... In Venn diagrams, the curves are overlapped in every possible way.
Answer:
PHD
Explanation:
phd is the one that can help you
The value of 
is -3. 
<h2>Procedure - Differentiability</h2><h3 /><h3>Chain rule and derivatives</h3><h3 />
We derive an expression for by means of chain rule and differentiation rule for a product of functions:
![\frac{d}{dx}[f(x)\cdot g(y)] = [f'(x)\cdot \frac{dx}{dx}]\cdot g(y) + f(x) \cdot [g'(y)\cdot \frac{dy}{dx} ]](https://tex.z-dn.net/?f=%5Cfrac%7Bd%7D%7Bdx%7D%5Bf%28x%29%5Ccdot%20g%28y%29%5D%20%3D%20%5Bf%27%28x%29%5Ccdot%20%5Cfrac%7Bdx%7D%7Bdx%7D%5D%5Ccdot%20g%28y%29%20%2B%20f%28x%29%20%5Ccdot%20%5Bg%27%28y%29%5Ccdot%20%5Cfrac%7Bdy%7D%7Bdx%7D%20%5D)
![\frac{d}{dx}[f(x)\cdot g(y)] = f'(x)\cdot g(y) +f(x)\cdot g'(y)\cdot \frac{dy}{dx}](https://tex.z-dn.net/?f=%5Cfrac%7Bd%7D%7Bdx%7D%5Bf%28x%29%5Ccdot%20g%28y%29%5D%20%3D%20f%27%28x%29%5Ccdot%20g%28y%29%20%2Bf%28x%29%5Ccdot%20g%27%28y%29%5Ccdot%20%5Cfrac%7Bdy%7D%7Bdx%7D)
(1)
If we know that 
, 
,<em> </em>
,<em> </em>
and
, then we have the following expression:
The value of is -3.
To learn more on differentiability, we kindly invite to check this verified question: brainly.com/question/24062595
<h3>Remark</h3>
The statement is incomplete and full of mistakes. Complete and corrected form is presented below:
<em>The point (-2, 4) lies on the curve in the xy-plane given by the equation </em>
<em>, where </em>
<em> is a differentiable function of </em>
<em> and </em>
<em> is a differentiable function of </em>
<em>. Selected values of </em><em>, </em>
<em>, </em><em> and </em>
<em> are given below: </em><em>, </em><em>, </em><em>, </em><em>. </em>
<em />
<em>What is the value of </em><em> at the point </em>
<em>?</em>
