Laws of Motion

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New answer posted•a year ago

Laws of Motion Class 11th

A body of mass ‘m’ is launched up on a rough inclined plane making an angle of 30° with the horizontal. The coefficient of friction between the body and plane is $\frac{\sqrt[]{x}}{5}$  if the time of ascent is half of the time of descent. The value of x is__________.

10 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

For ascent

$t_{a} = \sqrt[]{\frac{2 l}{a_{a s c e n t}}} = \sqrt[]{\frac{2 l}{g (s i n θ + μ c o s θ)}}$

For descent

$t_{d} = \sqrt[]{\frac{2 l}{a_{d e s c e n t}}} = \sqrt[]{\frac{2 l}{g (s i n θ - μ c o s θ)}}$

According question, we can write

$\Rightarrow μ = \frac{3}{5} t a n θ = \frac{3}{5} t a n 30 ° = \frac{\sqrt[]{3}}{5}$

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New answer posted•a year ago

Laws of Motion Class 11th

The normal reaction ‘N’ for a vehicle of 800kg mass, negotiating a turn on a 30° banked road at maximum possible speed without skidding is_________× 10³ kg m/s².

[Given cos30° = 0.87,  $μ_{5} = 0.2]$

11 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

In the frame of vehicle, vehicle is in equilibrium under the influence of pseudo force F_P

$F_{P} = \frac{m v^{2}}{R}$

N = mg cos 30° + F_P sin 30°

$\Rightarrow N = m g c o s 30 ° + \frac{m v^{2}}{R} s i n 30 ° . (i)$

, and

$f_{S} = \frac{m v^{2}}{R} c o s 30 ° - m g s i n 30 °$

By doing (1) × cos 3

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New answer posted•a year ago

Laws of Motion Class 11th

The temperature of an ideal gas in 3-dimensions is 300K. The corresponding de-Broglie wavelength of the electron approximately at 300 K, is:

[m_e = mass of electron = 9 × 10^-³¹kg

h = Planck constant = 6.6 × 10^-³⁴ Js

k_B = Boltzmann constant = 1.38 × 10^-²³ JK^-¹]

4 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

De Broglie wavelength

$λ = \frac{h}{m V} = \frac{h}{\sqrt[]{2 m E}} E \to K . E .$

$E = \frac{3}{2}$ KT for gas

$S o λ = \frac{h}{\sqrt[]{3 m K T}} = \frac{6.6 \times 1 0^{- 34}}{\sqrt[]{3 \times 9 \times 1 0^{- 31} \times 1.38 \times 1 0^{- 23} \times 300}}$

$\begin{array}{l} λ = 6.26 \times 1 0^{- 9} m \\ λ = 6.26 n m \end{array}$

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New answer posted•a year ago

Laws of Motion Class 11th

A block of mass m slides on the wooden wedge, which in turn slides backward on the horizontal surface. The acceleration of the block with respect to the wedge is: Given m = 8 kg, M = 16kg. Assume all the surfaces shown in the figure to be frictionless.

19 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

Suppose acceleration of wedge is a and acceleration of block w. r.t. wedge is a₁ then N cos60° = Ma = 16 a -> N = 32 a

For block w.r.t. wedge

N + 8a sin 30° = 8g cos 30°

N = 8g cos 30° - 8a sin 30°

->32a = 8g cos 30° - 8a sin 30°

->a = $\frac{\sqrt[]{3}}{9} g$

Now for 8 kg,

$8 g s i n 30 ° + 8 a c o s 30 ° = m a_{1}$

$a_{1} = g \times \frac{1}{2} + \frac{\sqrt[]{3}}{9} g \times \frac{\sqrt[]{3}}{2}$

$= \frac{2}{3} g$

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New answer posted•a year ago

Laws of Motion Class 11th

The boxes of masses 2kg and 8kg are connected by a massless string passing over smooth pulleys. Calculate the time taken by box of mass 8kg to strike the ground starting from rest. (use g = 10 m/s²)

266 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

For 8 kg block,

8g – 2T = 8a . (i)

For 2kg block,

T – 2g = 2 × 2a . (ii)

Solving equations (i) and (ii),

$a = \frac{g}{4} = 2.5 m / s^{2}$

$t = \sqrt[]{\frac{2 s}{a}} = \sqrt[]{\frac{2 \times 0.2}{2.5}} = 0.4 s$

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New answer posted•a year ago

Laws of Motion Class 11th

The resultant of these forces $\vec{O P}, \vec{O Q}, \vec{Q R}, \vec{O S} a n d \vec{O T}$  is approximately. . . . . . .N

[Take $\sqrt[]{3} = 1.7, \sqrt[]{2} = 1.4 .$  Given $\hat{i} a n d \hat{j}$  unit vectors along x, y axis]

42 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

${\vec{F}}_{x} = [10 \times \frac{\sqrt[]{3}}{2} + 20 \times \frac{1}{2} + \frac{20}{\sqrt[]{2}} - \frac{15}{\sqrt[]{2}} - \frac{15 \sqrt[]{3}}{2}] = 9.25 \hat{i}$

${\vec{F}}_{y} = [15 \times \frac{1}{2} + 20 \times \frac{\sqrt[]{3}}{2} + 10 \times \frac{1}{2} - \frac{15}{\sqrt[]{2}} - \frac{20}{\sqrt[]{2}}] = 5 \hat{j}$