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7 months ago

Physics Class 12th

Assuming the sun to be a spherical body (e = 1) of radius R at a temperature of T K, evaluate the total radiant power, incident on Earth having radius r₀, at a distance r from the Sun, where r₀ is the radius of the earth and $σ$ is Stefan's constant.

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alok kumar singh

Contributor-Level 10

\frac{e A_{s u n} σ T^{4}}{4 π r^{2}} * A_{p r o j E a r t h}

\frac{4 π R^{2}}{4 π r^{2}} σ T^{4} π r_{0}^{2}

π \frac{R^{2}}{r^{2}} r_{0}^{2} σ T^{4}

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7 months ago

Physics Physics Moving Charges and Magnetism

Magnetic field exists in the region between lines PQ and RS as shown in the figure. A particle having charge to mass ratio of a enters the magnetic field as shown in the figure with a speed v. Lines PQ and RS are parallel to each other. Find the value of d such that the charged particle exits the magnetic field region with velocity perpendicular to line RS.

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Vishal Baghel

Contributor-Level 10

d = Rsin 60°

⇒ d = (mv√3)/ (qB 2) = (√3v)/ (2aB)

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7 months ago

Physics Physics Waves

The absolute temperature of air in a region linearly increases from $T_{1}$ to $T_{2}$ in a space of width d. The time taken by the sound wave to travel in this region is (The velocity of sound at 273 k is v ):

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Raj Pandey

Contributor-Level 9

$V = \sqrt[]{\frac{γ R T}{M}},$ where, $T = T_{1} + \frac{T_{2} - T_{1}}{d} x$ $_{} \frac{_{}_{}}{}$

Also, $v = \sqrt[]{\frac{γ R 273}{M}} \Rightarrow \sqrt[]{\frac{γ R}{M}} = \frac{v}{\sqrt[]{273}}$ $\frac{}{} \frac{}{} \frac{}{}$

Also, $V = \frac{d x}{d t} = \sqrt[]{\frac{γ R T}{M}}$ $\Rightarrow \int_{0}^{d} ? \frac{d x}{\sqrt[]{T_{1} + \frac{T_{2} - T_{1}}{d} x}} = \sqrt[]{\frac{γ R}{M}} \int_{0}^{t} ? d t$ $\frac{}{} \frac{}{}$

$\Rightarrow t = \frac{\sqrt[]{273}}{v} \cdot \frac{2 d}{T_{2} - T_{1}} (\sqrt[]{T_{2}} - \sqrt[]{T_{1}}) ∴ t = \frac{2 d}{v} \frac{\sqrt[]{273}}{\sqrt[]{T_{1}} + \sqrt[]{T_{2}}}$

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7 months ago

Physics Class 11th

An inverted cone is rotating about its vertical axis. A particle is kept on the inner surface of cone and it is at rest relative to the cone at a height of 0.4 m above its vertex. The coefficient of friction between the surface of cone and the particle is 0.6 and the apex angle of cone is 90°. The maximum angular velocity of revolution of the cone can be: (take g = 10/s²)

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Vishal Baghel

Contributor-Level 10

N = (mg/√2) + (mω²h/√2)
(mω²h/√2) = (mg/√2) + f?
(mω²h/√2) = (mg/√2) + μ (mg/√2) + (mω²h/√2)
ω = √ (g/h) (1+μ)/ (1-μ) = 10rad/s

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7 months ago

Physics Physics Oscillations

A solid cylinder attached to horizontal massless spring can roll without slipping along horizontal surface. Find time period of oscillation.

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alok kumar singh

Contributor-Level 10

$τ_{A} = I α$

$K x R = \frac{3}{2} M R^{2} α$

$K R θ R = \frac{3}{2} M R^{2} α$

$α = \frac{2 K}{3 M}$

$∴ T = 2 π \sqrt[]{\frac{3 M}{2 K}}$

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7 months ago

Physics Motion in a Plane

A particle is moving in a circle of radius 1m with angular velocity (ω) given as a function of angular displacement ( θ ) by the relation ω = θ² + 2θ. The total acceleration of the particle when θ = 1rad is:

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Vishal Baghel

Contributor-Level 10

ω = θ² + 2θ
α = (ωdω)/dθ = (θ² + 2θ) (2θ + 2)
At θ = 1rad.
ω = 3rad/s and α = 12rad/s²
a? = αR = 12 m/s² a? = ω²R = 9 m/s² A? = √ (a? ² + a? ²) = 15 m/s²

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7 months ago

Physics Physics Atoms

Amount of radiation energy emitted by unit surface area per unit time by a body is defined as its emissive power. Dimension of Emissive power is:

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Raj Pandey

Contributor-Level 9

$E = \frac{Energy}{time * Area} \Rightarrow [E] = \frac{[Energy]}{[time] [Area]} = \frac{M L^{2} T^{- 2}}{L^{2} T}$

$[E] = M L^{0} T^{- 3}$

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7 months ago

Physics Class 12th

In the circuit shown swith S is connected to position 2 for a long time and then joined to position 1. The total heat produced in resistance R₁ is:

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alok kumar singh

Contributor-Level 10

$I = \frac{E}{R_{2}}$

$U_{I n d u c t o r} = H e a t l o s s i n R_{1} = \frac{1}{2} L {(\frac{E}{R_{2}})}^{2}$

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7 months ago

Physics Class 12th

At t < 0, the capacitor is charged and the switch is opened. At t = 0 the switch is closed. The shortest time T at which the charge on the capacitor will be zero is given by:

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alok kumar singh

Contributor-Level 10

The time period of LC oscillations,

$T = 2 π \sqrt[]{L C}$

The time at which charge on the capacitor will be zero is $\frac{T}{4} .$

So t $= \frac{π}{2} \sqrt[]{L C}$

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7 months ago

Physics Class 11th

How is the change of state related to latent heat?

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