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Lunna [17]
3 years ago
9

A man (mass=68 kg) on a parachute is falling at terminal velocity (v=59 m/s)

Physics
1 answer:
raketka [301]3 years ago
3 0

Explanation:

It is given that,

Mass of the man, m = 68 kg

Terminal velocity of the man, v = 59 m/s

We need to find the rate at which the internal energy of the man and of the air around him increase. The gravitational potential energy of the man is given by :

E=mgh

Differentiating equation (1) wrt t as :

\dfrac{dE}{dt}=mg\dfrac{dh}{dt}

Since, v=\dfrac{dh}{dt}=59\ m/s

\dfrac{dE}{dt}=68\times 9.8\times 59

\dfrac{dE}{dt}=39317.6\ J/s

So, the internal energy of the man and the air around him is increasing at the rate of 39317.6 J/s. Hence, this is the required solution.

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A) Moment of inertia about an axis passing through the point where the two segments meet : $I_A=\frac{1}{12} M L^2$

B) Moment of inertia passing through the point where the midpoint of the line connects to its two ends: $I x=\frac{1}{3} M L^2$

What is Moment of inertia?

The term "moment of inertia" refers to a physical quantity that quantifies a body's resistance to having its speed of rotation along an axis changed by the application of a torque (turning force). The axis might be internal or exterior, fixed or not.

A) The moment of inertia about an axis passing through the point where the two segments meet is $I_A=\frac{1}{12} M L^2$given that the rod is bent at the center and distance from all the points to the axis remains the same, the moment of inertia about the center will remain the same.

B) Determine the moment of inertia about an axis passing through the point midpoint of the line which connects the two ends

First step: determine the distance between the ends ( d )

After applying Pythagoras theorem$\mathrm{d}=\frac{\sqrt{2}}{2} L$

Next step : determine distance between the two axis $(\mathrm{x})$

After applying Pythagoras theorem

\mathrm{x}=\frac{\sqrt{2}}{4} L$$

Final step : Calculate the value of $\mathrm{I}_{\mathrm{x}}$

applying Parallel Axis Theorem

$$I_x=I_8+M x^2$$

$$\begin{aligned}& =\frac{1}{12} M L^2+\frac{1}{4} M L^2 \\& \therefore \quad I x=\frac{1}{3} M L^2 \\&\end{aligned}$$

Hence we can conclude that Moment of inertia about an axis passing through the point where the two segments meet: $I_A=\frac{1}{12} M L^2$, Moment of inertia passing through the point where the midpoint of the line connects its two ends: $I x=\frac{1}{3} M L^2$

To learn more about moment of inertia visit:brainly.com/question/15246709

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