Tags Work, Energy and Power

Work, Energy and Power

Get insights from 175 questions on Work, Energy and Power, answered by students, alumni, and experts. You may also ask and answer any question you like about Work, Energy and Power

Follow Ask Question

175

Questions

Discussions

Active Users

Followers

New answer posted

2 months ago

Work, Energy and Power Class 11th

In the given figure, the block of mass m is dropped from the point 'A'. The expression for kinetic energy of block when it reaches point 'B' is

0 Follower 3 Views

Payal Gupta

Contributor-Level 10

According to conservation of energy, we can write

Gain in kinetic energy = Loss in potential energy

K_f – K_in = U_in - U_f

K - = mgy – mg (y – y₀) = mgy₀

0 0

New answer posted

2 months ago

Work, Energy and Power Class 11th

A car cover AB distance with first one-third at velocity $υ_{1} m s^{- 1},$ second one-third at $υ_{2} m s^{- 1}$ and last one-third at $υ_{3} m s^{- 1} .$ If then the average velocity of the car is__________ ms^-1.

0 Follower 3 Views

alok kumar singh

Contributor-Level 10

$t_{1} = \frac{L}{3 V_{1}}$

$t_{2} = \frac{L}{3 V_{2}}$

$t_{3} = \frac{L}{3 V_{3}}$

$V_{a m g} = \frac{L}{t_{1} + t_{2} + t_{3}} = \frac{L}{\frac{L}{3 V_{1}} + \frac{L}{3 V_{2}} + \frac{L}{3 V_{3}}}$

$V_{a m g} = \frac{3}{\frac{1}{v_{1}} + \frac{1}{v_{2}} + \frac{1}{v_{3}}}$

$= \frac{3}{\frac{1}{11} + \frac{1}{2 * 11} + \frac{1}{3 * 11}}$

= $\frac{3 * 11 * 6}{6 + 3 + 2}$

= 3 * b

= 18 m/sec

0 0

New answer posted

2 months ago

Work, Energy and Power Physics NCERT Exemplar Solutions Class 12th Chapter Six

Water falls from a 40m height dam at the rate of 9 * 10⁴ kg per hour. Fifty percentage of gravitational potential energy can be converted into electrical energy. Using this hydro electric energy number of 100 W lamps, that can be lit, is:

(Take g = 10 ms^-2)

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

Given : water flow rate = $(\overset{o}{m})$ = 9 * 10⁴kg/hr

= $\frac{9 * 1 0^{4} k g / s e c}{60 * 60}$

Potential energy available on water per unit time

$\Rightarrow \overset{o}{m} g h = \frac{9 * 1 0^{4}}{60 * 60} * 10 * 40$

$\overset{o}{m} g h = 1 0^{4} w a t t$

50% of it converts into electrical energy = 50% of $\overset{o}{m} g h = \frac{1 0^{4}}{2} w a t t$

Let $η$ be the number of bulbs n = 100 = $\frac{1 0^{4}}{2}$

n = 50

0 0

New answer posted

2 months ago

Work, Energy and Power Physics NCERT Exemplar Solutions Class 12th Chapter Six

A block of mass 2kg moving on a horizontal surface with speed of 4 ms^-¹ enters a rough surface ranging from x = 0.5m to x = 1.5m. The retarding force in this range of rough surface is related to distance by F = -kx where k = 12Nm^-¹. The speed of the block as it just crosses the rough surface will be:

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

by, WET $\int F d x = Δ k . 6$

$\int - k x d x = \frac{1}{2} m [v_{t}^{2} - v_{i}^{2}]$

$\Rightarrow \frac{- k}{2} {[x_{f}^{2} - x_{i}^{2}]}_{0.5}^{1.5} = \frac{1}{2} * 2 [v_{f}^{2} - 16]$ $[? v_{i} = 4 m / s e c]$

$\frac{- 12}{2} [1 . 5^{2} - 0 . 5^{2}] = [v_{f}^{2} - 16]$

$v_{f}^{2} = 16 - 12 = 4$

vf = 2m/sec

0 0

New answer posted

2 months ago

Work, Energy and Power physics ncert exemplar solutions class 12th chapter two

A system to 10 balls each of mass 2kg are connected via massless and unstretchable string. The system is allowed to slip over the edge of a smooth in figure. Tension on the string between the 4^th and 8^th ball is_________N when 6^th ball just leaves the table.

0 Follower 4 Views

Vishal Baghel

Contributor-Level 10

12 g – T = 12 a- (1)

T – T' = 2a- (2)

T' = 6a- (3)

From (1) & (2)

12g – 12a – T' = 2a- (4)

From (4) & (3)

12g – 12a – T' = 14

$12 g = \frac{10}{3} T' \Rightarrow T' = \frac{120 * 3}{10} = 36$

0 0

New answer posted

2 months ago

Work, Energy and Power Physics NCERT Exemplar Solutions Class 12th Chapter Six

A fighter jet is flying horizontally at a certain with a speed of 200ms^-¹. When it passes directly overhead ad anti-aircraft gun, a bullet is fired from the gun, at an angle q with the horizontal, to hit the jet. If the bullet speed is 400m/s, the value of q will be _____________.

0 Follower 2 Views

Payal Gupta

Contributor-Level 10

B fighter jet

A anti-Air craft gun

Draw velocity diagram of A w.r.t. B

If A hits B

Then Relative velocity perpendicular to the line joining A to B will be zero.

That means 400 cos = 200

$\begin{array}{l} c o s θ = \frac{1}{2} \\ θ = 60 ° \end{array}$

0 0

New question posted

2 months ago

Work, Energy and Power Physics NCERT Exemplar Solutions Class 12th Chapter Six

How many alpha and beta particles are emitted when Uranium ₉₂U²³⁸ decays to lead ₈₂Pb²⁰⁶?

0 Follower 2 Views

New answer posted

2 months ago

Work, Energy and Power physics ncert exemplar solutions class 12th chapter two

Arrange the four graphs in descending order of total work done; where W₁, W₂, W₃ and W₄ are the work done corresponding to figure a, b, c and d respectively.

(a) W₃ > W₂ > W₁ >W₄ (b) W₃ > W₂ > W₄ > W₁ (c) (d) W₂> W₃ > W₁ > W₄

0 Follower 5 Views

Vishal Baghel

Contributor-Level 10

Work done represents are under

F vs x diagram

w₃ > w₂ > w₁ > w₄

0 0

New answer posted

2 months ago

Work, Energy and Power Physics NCERT Exemplar Solutions Class 12th Chapter Six

A particle of mass 500 gm is moving in a straight line with velocity = The work done by the net force during its displacement from x = 0 to x = 4m is:

(Take b = 0.25m^-3/2s^-1).

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

$ω = Δ k . E = \frac{1}{2} m [v_{t}^{2} - v_{i}^{2}]$

$= \frac{1}{2} m {[b * 4^{5 / 2}]}^{2} - \frac{1}{2} m {[0]}^{2}$

$= \frac{1}{2} * 0.5 [0.2 5^{2} * 4^{5}] = 2^{4} = 16 J$

0 0

New answer posted

2 months ago

Work, Energy and Power Physics NCERT Exemplar Solutions Class 12th Chapter Six

A body mass 8kg and another of mass 2kg are moving with equal kinetic energy. Thee ratio of their respective momenta will be:

0 Follower 3 Views

alok kumar singh

Contributor-Level 10

$K E = \frac{ρ^{2}}{2 m}$

$\Rightarrow ρ^{2} α m$

$\Rightarrow {(\frac{P_{1}}{P_{2}})}^{2} = \frac{8}{2} = \frac{4}{1}$

$\Rightarrow \frac{P_{1}}{P_{2}} = \frac{2}{1}$

0 0

Get authentic answers from experts, students and alumni that you won't find anywhere else

On Shiksha, get access to

65k Colleges
1.2k Exams
688k Reviews
1800k Answers

Community GuidelinesUser Points System

QuestionsDiscussionsTags

Need guidance on career and education? Ask our experts

Your Question

Work, Energy and Power

Questions

Discussions

Active Users

Followers

All

Discussions

Unanswered Questions

Share Your College Life Experience