Tags Physics

Physics

Get insights from 5.6k questions on Physics, answered by students, alumni, and experts. You may also ask and answer any question you like about Physics

Follow Ask Question

5.6k

Questions

Discussions

Active Users

Followers

New answer posted

a year ago

Physics Vector Algebra

Three particles P, Q and R are moving along the vectors $\vec{A} = \hat{i} + \hat{j}, \vec{B} = \hat{j} + \hat{k}$ and $\vec{C} = - \hat{i} + \hat{j}$ respectively. They strike on point and start to move in different directions. Now particle P is moving normal to the plane which contains vector $\vec{A} a n d \vec{B} .$ Similarly particle Q is moving normal to the plane which contains vector $\vec{A} a n d \vec{C} .$ The angle between the direction of motion of P and Q $c o s^{- 1} (\frac{1}{\sqrt[]{x}}) .$ Then the value of x is_________.

0 Follower 5 Views

Vishal Baghel

Contributor-Level 10

Cross product will represent a vector perpendicular to both the vector.

Dir. Of Motion of P : $= (\hat{i} + \hat{j}) * (\hat{j} * \hat{k})$

= $\hat{k} - \hat{j} + \hat{i}$

Dir. Of Motion of Q = $(\hat{i} + \hat{j}) * (- \hat{i} + \hat{j})$

$= 2 \hat{k}$

$c o s θ = \frac{\vec{a} . \vec{b}}{a b} = \frac{2}{2 \sqrt[]{3}} = \frac{1}{\sqrt[]{3}} \Rightarrow θ = c o s^{- 1} (\frac{1}{\sqrt[]{3}})$

0 0

New answer posted

a year ago

Physics Class 12th

In a given circuit diagram, a 5 V zener diode along with a series resistance is connected across a 50 V power supply. The minimum value of the resistance required, if the maximum zener current is 90 mA be________ $Ω$ .

0 Follower 9 Views

Vishal Baghel

Contributor-Level 10

For 90 mA in zener diode, current in R should be greater than 90 mA.

$\frac{45}{R} = 90 * 1 0^{- 3} \Rightarrow R = \frac{45}{90 * 1 0^{- 3}} = 500 Ω$

0 0

New answer posted

a year ago

Physics Class 12th

In an electric circuit, a cell of certain emf provides a potential difference of 1.25 V across a load resistance of $5 Ω$ . However, it provides a potential difference of 1V across a load resistance of $2 Ω .$ The emf of the cell is given by $\frac{x}{10} V .$ Then the value of x is_________.

0 Follower 7 Views

Vishal Baghel

Contributor-Level 10

$i = \frac{ε}{R + r}$

$v_{R} = \frac{ε R}{R + r}$

$\frac{1.25 = \frac{ε (5)}{5 + r}}{1 = \frac{ε (2)}{2 + r}}$

r = 1

Using above equ : $ε = \frac{3}{2} = \frac{15}{10} V$

0 0

New answer posted

a year ago

Physics Mechanical Properties of Solids

The area of cross – section of a railway track is 0.01 m². The temperature variation is 10°C, Coefficient of linear expansion of material of track is 10^-⁵ / °C. The energy stored per meter in the track is_________J/m.

(Young's modulus of material of track is 10¹¹ Nm^-²)

0 Follower 12 Views

Vishal Baghel

Contributor-Level 10

Energy density (u) = $\frac{1}{2} * σ * ε = \frac{1}{2} * Y ε * ε = \frac{1}{2} ε^{2} Y$ $\frac{}{} \frac{}{} \frac{}{}^{}$

$\Rightarrow e n e r g y s t o r e d / m^{2} = \frac{1}{2} ε^{2} Y A$

Strain $(ε) = \frac{Δ l}{l_{0}} = α Δ T = 1 0^{- 5} * 10 = 1 0^{- 4}$ $\frac{}{_{}}^{}^{}$

$\Rightarrow e n e r g y s t o r e d / m^{2} = \frac{1}{2} * 1 0^{- 8} * 1 0^{11} * 1 0^{- 2} = 5$

0 0

New answer posted

a year ago

Physics Thermodynamics

In 5 minutes, a body cools from 75°C to 65°C at room temperature of 25°C. The temperature of body at the end of next 5 minutes is__________°C.

0 Follower 4 Views

Vishal Baghel

Contributor-Level 10

from newtan's law of cooling :

$\frac{d T}{d t} = - k (T - T_{0})$

Using average value :

$\frac{75 - 65}{5} = - k [\frac{(75 + 65) 2}{2} - 25]$

$\frac{(65 - T)}{5} = - k [\frac{(65 + T)}{2} - 25]$

$\Rightarrow \frac{10}{65 - T} = \frac{45}{\frac{65 + T}{2} - 25}$

$\Rightarrow T = 57 ° C$

0 0

New answer posted

a year ago

Physics Units and Measurement

Three students S₁, S₂ and S₂ perform an experiment for determining the acceleration due to gravity (g) using a simple pendulum. They use different lengths of pendulum and record time for different number of oscillations. The observations are as shown in the table.

Student No.	Length of Pendulum (cm)	No. of oscillations (n)	Total for n oscillations	Time period (s)
1	64.0	8	128.0	16.0
2	64.0	4	64.0	16.0
3	20.0	4	36.0	9.0

Least count of length = 0.1 cm

least count for time = 0.1 s)

If E₁, E₂ and E₃ are the percentage errors 'g' for students 1, 2 and 3 respectively, then the minimum percentage error is obtained by student no.___________.

0 Follower 14 Views

Vishal Baghel

Contributor-Level 10

$T = 2 π \sqrt[]{\frac{l}{g}}$

$g = 4 π^{2} \frac{l}{T^{2}}$

$\Rightarrow \frac{Δ g}{g} = \frac{Δ l}{l} + \frac{2 Δ T}{T}$

Student 1 will get least error because it has more number of oscillation

0 0

New answer posted

a year ago

Physics System of Particles and Rotational Motion

The centre of a wheel rolling on a plane surface moves with a speed $υ_{0} .$ A particle on the rim of the wheel at the same level as the centre will be moving at a speed $\sqrt[]{x} υ_{0} .$ Then the value of x is__________.

0 Follower 4 Views

Vishal Baghel

Contributor-Level 10

$V_{P} = \sqrt[]{V_{0}^{2} + V_{0}^{2}} = \sqrt[]{2} V_{0}$

0 0

New answer posted

a year ago

Physics Class 12th

A ray of light passing through a prism $(μ = \sqrt[]{3})$ surface minimum deviation. It is found that the angle of incidence is double the angle of refraction within the prism. Then, the angle of prism is________(in degrees).

0 Follower 2 Views

Vishal Baghel

Contributor-Level 10

At minimum deviation :

i = i'

r = r'

$r + r' = A \Rightarrow r = \frac{A}{2}$

Snell's law :

1 sin I = $\sqrt[]{3} s i n r = \sqrt[]{3} s i n \frac{i}{2}$ $\frac{}{}$

$\Rightarrow 2 c o s \frac{i}{2} = \sqrt[]{3}$

$\Rightarrow i = 60 °$

$\begin{array}{l} i = 2 r \Rightarrow r = 30 ° \\ \Rightarrow A = 60 ° \end{array}$

0 0

New answer posted

a year ago

Physics Class 12th

The total charge enclosed in an incremental volume of 2 * 10^-⁹ m³ located at the origin is _________nC, if electric flux density of its field is found as

D = e^-^x sin $y \hat{i} - e^{- x} c o s y \hat{j} + 2 z \hat{k} C / m^{2} .$

0 Follower 16 Views

Vishal Baghel

Contributor-Level 10

Using gauss law in differential form

$\vec{\nabla} . \vec{E} = \frac{ρ}{ε_{0}}$

$\vec{\nabla} . (ε_{0} \vec{E}) = ρ$

$\Rightarrow \vec{\nabla} . \vec{D} = ρ$

$\Rightarrow (\frac{\partial}{\partial x} \hat{i} + \frac{\partial}{\partial y} \hat{j} + \frac{\partial}{\partial z} \hat{k}) . \vec{D} = ρ$

$\begin{array}{l} \Rightarrow ρ = - e^{- x} s i n y + e^{- x} s i n y + 2 = 2 \\ \Rightarrow Q = 2 (V o l u m e) = 4 * 1 0^{- 9} c = 4 n c \end{array}$

0 0

New answer posted

a year ago

Physics Electric Charges and Fields

An electric dipole is placed on x – axis in proximity to a line charged of linear density 3.0 * 10^-⁶ C/m. Line charge is placed on z – axis and positive and negative charge of dipole is at a distance of 10 mm and 12 mm from the origin respectively. If total force of 4N is exerted on the dipole, find out the amount of positive charge of the dipole.

0 Follower 5 Views

Vishal Baghel

Contributor-Level 10

$F = [\frac{2 k λ}{r_{1}} - \frac{2 k λ}{r_{2}}] q = 2 k λ q [\frac{1}{r_{1}} - \frac{1}{r_{2}}]$

$2 * 9 * 1 0^{9} * 3 * 1 0^{- 6} [\frac{1000}{10} - \frac{1000}{12}] q$

$\begin{array}{l} \Rightarrow 4 = 9 * 1 0^{5} q \\ \Rightarrow q = 4.44 μ C \end{array}$

0 0

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

On Shiksha, get access to

67k Colleges
1.2k Exams
718k Reviews
1850k Answers

Community GuidelinesUser Points System

QuestionsDiscussionsTags

Need guidance on career and education? Ask our experts

Your Question

Physics

Questions

Discussions

Active Users

Followers

All

Discussions

Unanswered Questions

Share Your College Life Experience