Question

Guys pls help me in this 2 Questions i want with suitable explanation​

Answer 1

Explanation:

$\huge\underbrace\mathfrak{Answer}$

The change in kinetic energy of the body from t = 2 s to t = 4 s is 56 Joules.

The power delivered by a body is given by :

P=3t^2P=3t²

t is the time

We need to find the change in kinetic energy of the body from t = 2 s to t = 4 s. Power of a body is given by the work done per unit time. It is given by :

P= W/T

In integral form,

$\frac{(</u></strong><strong><u>dW=</u></strong><strong><u>)(</u></strong><strong><u>P</u></strong><strong><u>)}{(</u></strong><strong><u>.</u></strong><strong><u>)(</u></strong><strong><u>d</u></strong><strong><u>t</u></strong><strong><u>)} </u></strong><strong><u>$

W = 56 Joules

W = 56 JoulesSo, the change in kinetic energy of the body from t = 2 s to t = 4 s is 56 Joules. Hence, this is the required solution.

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Answer 2

Topic :-

Work, Power and Energy

Given :-

Power delivered to a body varies as P = 3t².

To Find :-

n if 14n is Change in Kinetic Energy of body from t = 2 s to t = 4s.

Formula to be Used :-

$\boxed{\Delta K.E.=\displaystyle \int_{t_1}^{t_2}Pdt}$

$\boxed{\displaystyle \int x^ndx=\dfrac{x^{n+1}}{n+1}}$

Solution :-

P = 3t²

$t_1=2\:s$

$t_2=4\:s$

Put values in formula,

$\Delta K.E.=\displaystyle \int_{2}^{4}3t^2dt$

$\Delta K.E.=3\displaystyle \int_{2}^{4}t^2dt$

$\Delta K.E.=3\left [ \dfrac{t^3}{3} \right ]_{2}^{4}$

$\Delta K.E.=3\left [ \dfrac{4^3}{3}-\dfrac{2^3}{3}\right ]$

$\Delta K.E.=3\left [ \dfrac{64-8}{3} \right ]$

$\Delta K.E.=\not{3}\left [ \dfrac{56}{\not{3}} \right ]$

∆K.E. = 56 Joules

∆K.E. = 14n Joules

14n = 56

n = 4

Answer :-

So, value of n is 4.