Ask & Answer: India's Largest Education Community

Yes. Mechanical energy is the sum of both kinetic and potential energies. Kinetic energy cannot be a negative value since mass is always positive and velocity square leads to a positive value. However, potential energy can be negative depending on the reference point.

$\left|f\left(x\right)-f\left(y\right)\right|\le \left|{\left(x-y\right)}^2\right|$

$\left|\frac{f\left(x\right)-f\left(y\right)}{x-y}\right|\le x-y$

Taking the limit y x on both sides

$\underset{y\to x}{Lt}\left|\frac{f\left(x\right)-f\left(y\right)}{x-y}\right|\le \underset{y\to x}{Lt}\left(x-y\right)$

$\left|f\text{'}\left(x\right)\right|\le 0$

Hence, modulus cannot be zero. Hence f’ (x) = 0. Integrating, we get f (x) = c

at x = 0, f (0) = c = 1

$\therefore f\left(x\right)=1>0,\forall x\in R$

Hence, option (A) is correct option.

Yes, listed below are the top Law colleges in Udaipur along with their tuition fees:

Disclaimer: This information is sourced from official website and may vary.

About 4 top Law colleges are there in Udaipur. Students can secure admission on merit-based or through entrance exam like CLAT. Top LLB colleges in Udaipur include Mohanlal Sukhadia University, Bhupal Nobles' University, Pacific University, and Dr. Anushka Law College.

Of these, 2 colleges are privately owned and 1 college is owned by public/government organisations.

Not in all cases. This happens due to the interruption of non conservative forces like friction, air resistance, etc. which are responsible for decreasing the mechanical energy by converting it into other forms of energy such as heat, noise, etc.

No official update has been released by NCTE. But a post-graduation degree and prior experience might be needed.

x – y = 0 . (i)

x + 2y = 3. (ii)

2x + y = 6 . (iii)

Solving (i) & (ii), we get (1, 1)

Solving (ii) & (iii), we get (3, 0)

Solving (iii) & (i), we get (2, 2)     

$AB=\sqrt[]{5,}BC=\sqrt[]{5},CA=\sqrt[]{2}$

$\therefore AB=BC$ Isosceles triangle

Hence, option (B) is correct answer.

$\underset{h\to 0}{Lt}\frac{2\times 2\left[\frac{\sqrt[]{3}}{2}sin\left(\frac{\pi }{6}+h\right)-\frac{1}{2}cos\left(\frac{\pi }{6}+h\right)\right]}{2\sqrt[]{3}h\left[\frac{\sqrt[]{3}}{2}cosh-\frac{1}{2}sinh\right]}$

$=\underset{h\to 0}{Lt}4\frac{sin\left(\frac{\pi }{6}+h-\frac{\pi }{6}\right)}{2\sqrt[]{3}hsin\left(\frac{\pi }{3}-h\right)}=\frac{2}{\sqrt[]{3}}\times 1\times \frac{1}{\frac{\sqrt[]{3}}{2}}=\frac{4}{3}$

Conservation of Mechanical Energy is linked to newton's second law of motion (F = ma) in combination with work energy theorem. The work energy theorem states that total work done on an object equals to change in kinetic energy. This theorem is used to prove that the sum of kinetic energy and potential energy remains constant.

 ${\displaystyle \sum _{n=1}^{100}{\displaystyle \underset{n-1}{\overset{n}{\int }}{e}^{x-\left[x\right]}dx={\displaystyle \sum _{n=1}^{100}{\displaystyle \underset{n-1}{\overset{n}{\int }}{e}^{x-\left[x\right]}dx}}}}$

${\displaystyle \underset{0}{\overset{1}{\int }}{e}^{x-\left[x\right]}dx+{\displaystyle \underset{1}{\overset{2}{\int }}{e}^{x-\left[x\right]}dx}}+{\displaystyle \underset{2}{\overset{3}{\int }}{e}^{x-\left[x\right]}dx+.....+{\displaystyle \underset{99}{\overset{100}{\int }}{e}^{x-\left[x\right]}dx}}$

$=e-1+\frac{1}{e}\left(e^2-e\right)+\frac{1}{e^2}\left(e^3-e^2\right)+......+\frac{1}{e^{99}}\left(e^{100}-e^{99}\right)$

$=e-1+\left(e-1\right)+\left(e-1\right)+......+\left(e-1\right),100times$

$=100\left(e-1\right)$

2nd method

${\displaystyle \sum _{n=1}^{100}{\displaystyle \underset{n-1}{\overset{n}{\int }}{e}^{x-\left[x\right]}dx={\displaystyle \sum _{n=1}^{100}{\displaystyle \underset{n-1}{\overset{n}{\int }}{e}^{\left\{x\right\}}dx=}}{\displaystyle \sum _{n=1}^{100}\left({\displaystyle \underset{0}{\overset{1}{\int }}{e}^{x}dx}\right)}}}={\displaystyle \sum _{n=1}^{100}\left(e-1\right)}=100\left(e-1\right)$

Blue cupric metaborate is reduced to cuprous metaborate in a luminous flame.

  $2Cu{\left(B{O}_2\right)}_2+2NaB{O}_2+C\underset{}{\overset{Luminous}{\to }}2CuB{O}_2+N{a}_2{B}_4{O}_7+CO$            

Cupric metaborate is obtained by heating boric anhydride & copper sulphate in a non-luminous flame as

$CuS{O}_4+B_2{O}_3\underset{}{\overset{Non-LuminiousFlame}{\to }}\underset{Blue}{Cu{\left(B{O}_2\right)}_2}+S{O}_3.$            

The points on the curve are (0, 0), (2, 2) and $\left(3,\frac{21}{2}\right)$

$\therefore \frac{dy}{dx}=2x^3-15x^2+36x-19$

$at\left(0,0\right),\frac{dy}{dx}=-19,at\left(2,2\right),\frac{dy}{dx}=9at\left(3,\frac{21}{2}\right),\frac{dy}{dx}=8$

Hence maximum slope at (2, 2) is 9.

Al < Mg < Si < S < P

1^st I. E increase along period with exception on moving from group 2 to group 13 and group 15 to group 16

$\frac{GM}{{\left(3R/2\right)}^2}=\frac{GM}{R^3}\times r$

$\Rightarrow OA=\frac{4R}{9}=r$

$AB=R-\frac{4R}{9}=\frac{5R}{9}\Rightarrow OA:AB=4:5=x:y\Rightarrow x=4$