Physics

Contributor-Level 9

$T ↓ (300 Kto 70 K)$

$\begin{array}{l} T ↓ R_{meta 1} ↓ R ↑ \\ (A l) (Si) \end{array}$ semi? conductor

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4 months ago

The diagram showing the variation of gravitational potential of earth with distance from the centre of earth is

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Contributor-Level 9

$V_{in} = \frac{- G M}{2 R} [3 - {(\frac{r}{R})}^{2}],$

$V_{s u r f a c e} = \frac{- G M}{R}, V_{out} = \frac{- G M}{r}$

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4 months ago

A 20 kg block B is suspended from a cord attached to a 40 kg cart A. Find the ratio of the acceleration of block in cases (i) and (ii) shown in the figure immediately after the system is released from rest. (Neglect friction)

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Contributor-Level 9

Case I:

$T - N = 40 a$

and $20 g - T = 20 a$

Also $N = 20 a$

After simplifying, we get

$a = \frac{g}{4}$

Acceleration of block $B, = \sqrt[]{2} a = \frac{g}{2 \sqrt[]{2}} .$

Case II:

$T = 40 a$

and $20 g - T = 20 a$

After simplifying above equation, we get

$a = g / 3$

Ratio $= \frac{g / 2 \sqrt[]{2}}{g / 3} = \frac{3}{2 \sqrt[]{2}} .$

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4 months ago

Physics Class 12th

Two balls of same mass and carrying equal charge are hung by threads of length $l$ from a fixed support. At electrostatic equilibrium, assuming that angles made by each thread is small, the separation, $x$ between the balls is proportional to:

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Physics Physics Moving Charges and Magnetism

Contributor-Level 9

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4 months ago

In gravity free space, a bead of charge $1 μ C$ and mass 3mg is threaded on a rough rod of friction coefficient $μ$ = 0.3. A magnetic field of magnitude 0.2 T exists perpendicular to the rod. The bead is projected along the rod with a speed 4m/s. How much distance (in m) will the bead cover before coming to rest?

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Contributor-Level 10

N = qvB

$- μ q v B = \frac{m d v}{d t} = m v \frac{d v}{d S}$

$μ q B S = m v$

$S = \frac{3 * 1 0^{- 6} * 4}{0.3 * 1 0^{- 6} * 0.2} = 200 m$

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4 months ago

The new white belt of a long horizontal conveyor is moving with a constant speed v = 3.0m/s. A small block of carbon is placed on the belt with zero initial velocity relative to the ground. The block will slip a bit before moving with the belt, leaving a black mark on the belt (figure). How long is that mark (in m) if the coefficient of kinetic friction between the carbon block and the belt is 0.20 and the coefficient of static friction is 0.30?

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Contributor-Level 10

$0^{2} = v^{2} - 2 μ_{k} g S_{r e l}$

$\frac{9}{2 * 2} = S_{r e l} = 2.25 m$

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4 months ago

A calorimeter consists of 400g of water at $24 ° C .$ A 500g piece of copper at $100 ° C$ is thrown into the water and the equilibrium temperature is found to be $36.5 ° C .$ What is the molar heat capacity of copper in cal/mol k, given that 1 mol of copper has a mass of 63.5g? Neglect the heat capacity of the container.

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Physics Physics Ray Optics and Optical Instruments

Contributor-Level 10

400 * 1 * 12.5 = 500 * 5 * (100 – 36.5)

$S = \frac{10}{63.5} c a l / g m ° C$

$1 g m = \frac{1}{63.5} m o l$

S = 10 cal/mol °C

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4 months ago

You are given a converging lens with equal radii of curvature and a diverging lens with the same radii of curvature as those of the converging lens. The lenses are made of material with n = 1.50, the radii of curvature are all 35cm. They are placed at opposite ends of a tube 15cm long, and the nearer lens which is converging is 10cm from an object. What is the location of the image that results from the two refractions? Write the final distance (in cm) from diverging lens.

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Contributor-Level 10

$\frac{1}{u} + \frac{1}{- 10} = \frac{1}{35}$

$\begin{array}{l} \frac{1}{u} = \frac{1}{35} - \frac{1}{10} = \frac{2 - 7}{70} = \frac{- 1}{14} \\ \frac{1}{u} = \frac{- 1}{35} - \frac{1}{29} = \frac{- 64}{35 * 29} \\ u = \frac{35 * 29}{64} = 15.85 t o 15.86 \end{array}$

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4 months ago

Physics Class 12th

An electron moves in the field produced by a charged sphere of radius 10cm along the radius connecting the points separated by 12 and 15 cm from the centre of the sphere. The velocity of the electron changes thereby from 2 * 10⁵ to 2 * 10⁶ m/s. Determine the surface charge (in nC/m²) density of the sphere. (Mass of electron = 9.1 * 10^-31kg)

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Contributor-Level 10

$\frac{k Q e}{0.12} + \frac{1}{2} m u^{2} = \frac{k Q e}{0.15} + \frac{1}{2} m u^{2}$

$\frac{k Q e * 0.03}{0.12 * 0.15} = \frac{1}{2} m * (v^{2} - u^{2})$

$Q = \frac{9.1 * 99 * 1 0^{- 11}}{12}$

$= \frac{75.075 * 1 0^{- 12}}{4 π * {(0.1)}^{2}} = 5.97 n C / m^{2}$

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4 months ago

A small elastic ball of mass m is placed at the apex of a $45 °$ inclined plane as shown in the figure below. The ball is allowed to slip without friction down the plane (along the dotted line), hit the ground (as shown) and bounce along it. If the height of the inclined plane is h and the coefficient of restitution between the ball and the ground is 0.5, then the distance AB, as marked on the figure, will be N * h, where N is

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