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5 months ago

Physics Physics System of Particles and Rotational Motion

A disc of mass 1kg and radius R is free to rotate about a horizontal axis passing through its centre and perpendicular to the plane of disc. A body of same mass as that of disc is fixed at the highest point of the disc. Now the system is released, when the body comes to the lowest position, its angular speed will be $\sqrt[]{\frac{x}{3 R} r a d s^{- 1}}$ where x = __________.

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Payal Gupta

Contributor-Level 10

Loss in P.E. = Gain in k.E

2 mg R = $\frac{1}{2} (\frac{1}{2} m R^{2} + m R^{2}) ω^{2}$

$ω = \sqrt[]{\frac{8 g}{3 R}} = 4 \sqrt[]{\frac{g}{2 * 3 R}}$

$x = \frac{g}{2} = 5$

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5 months ago

Physics Class 11th

If the initial velocity in horizontal direction of a projectile is unit vector $\hat{i}$ and the equation of trajectory is y = 5x(1 – x). The y component vector of the initial velocity is ________ $\hat{j}$ .

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Payal Gupta

Contributor-Level 10

y = x5 (1 – x) = x tan θ $(1 - \frac{x}{R})$

tan = 5, R = 1

$s i n θ = \frac{5}{\sqrt[]{26}}, c o s θ = \frac{1}{\sqrt[]{26}}$

$R = \frac{u^{2} s i n 2 θ}{g} = 1$

$\Rightarrow u^{2} = 26 \Rightarrow u = \sqrt[]{26} m / s$

y – component of initial velocity

= u sin θ

= $= \sqrt[]{26} * \frac{5}{\sqrt[]{26}}$

= 5 m/s

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5 months ago

Physics Units and Measurement

A screw gauge of pitch 0.5mm is used to measure the diameter of uniform wire of length 6.8cm, the main scale reading is 1.5mm and circular scale reading is 7. The calculated curved surface area wire to appropriate significant figures is:

[Screw gauge has 50 divisions on its circular scale]

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Payal Gupta

Contributor-Level 10

Least count = $\frac{0.5}{50} m m$ = 0.01 mm

Diameter, d = 1.5 + 7 * 0.01

= 1.57

$∴$ Surface Area = (2r) $l$

= 3.4 cm²

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5 months ago

Physics Class 11th

Two projectiles thrown at $30 ° a n d 45 °$ with the horizontal respectively, reach the maximum height in same time. The ratio of their initial velocities is

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Payal Gupta

Contributor-Level 10

H₁ = H₂

$\frac{u_{1}^{2} s i n^{2} θ_{1}}{2 g} = \frac{u_{2}^{2} s i n^{2} θ_{2}}{2 g}$

$\Rightarrow u_{1}^{2} {(s i n 30 °)}^{2} = {u_{2}}^{2} {(s i n 45 °)}^{2}$

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

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5 months ago

Physics Class 11th

A monkey of mass 50kg climbs on a rope which can withstand the tension (T) of 350 N. If monkey initially climbs down with an acceleration of 4 m/s² and then climbs up with an acceleration of 5 m/s². Choose the correct option (g = 10 m/s²).

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Payal Gupta

Contributor-Level 10

For climbing downward

50 g – T = 50a

T = 500 – 50 * 4

= 300 N < 350 N

for climbing upwards

T – 50 g = 50 a

T – 500 = 50 * 5

T = 750 N > 350 N

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5 months ago

Physics Work, Energy and Power

As per the given figure, two blocks each of mass 250g are connected to a spring of spring constant 2Nm^?1. If both are given velocity in opposite directions, then maximum elongation of the spring is

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Payal Gupta

Contributor-Level 10

Let m = 250 g = 0.25 kg

By Reduced mass method

$m_{r} = \frac{m_{1} m_{2}}{m_{1} + m_{2}} = \frac{m m}{m + m} = \frac{m}{2}$

By wet

$w_{S P} = Δ K . E .$

$- \frac{1}{2} k x^{2} = 0 - \frac{1}{2} (\frac{m}{2}) {(2 v)}^{2}$

$\frac{2}{2} x^{2} = 0.25 v^{2}$

$x^{2} = 0.25 v^{2}$

$x = \frac{v}{2}$

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5 months ago

Physics Physics Gravitation

The percentage decrease in the weight of a rocket, when taken to a height of 32 km above the surface of earth will, be: (Radius of earth will, be

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Payal Gupta

Contributor-Level 10

$g^{1} = g (1 - \frac{2 h}{R}) = g (1 - 2 * \frac{32}{6400})$

$g^{1} = \frac{99 g}{100} = 0.99 g$

% decrease is wt = $\frac{g - g^{1}}{g} * 100 = 1 %$

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5 months ago

Physics Class 11th

A water drop of radius 1cm is broken into 729 equal droplets. If surface tension of water is 75 dyne/cm, then the gain in surface energy upto first decimal place will be:

(Given p = 3.14)

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Payal Gupta

Contributor-Level 10

$\frac{4}{3} π R^{3} = 729 \frac{4}{3} π r^{3}$

R³= 729r³

$R = (729) \frac{1}{3} (r) (\frac{1}{3})$

R = 9r

$Δ u = T (4 π r^{2}) * 729 - T * 4 π R^{2}$

$= T * 4 π * 8 R^{2}$

$\begin{array}{l} = 75.39 * 1 0^{- 5} J \\ Δ u = 7.5 * 1 0^{- 4} J \end{array}$

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5 months ago

Physics Physics Thermodynamics

A monoatomic gas at pressure P and volume V is suddenly compressed to one eight of its original volume. The final pressure at constant entropy will be

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Payal Gupta

Contributor-Level 10

Constant Entropy means Adiabatic process

$P v^{y} = C$

$P_{1} {v_{1}}^{y} = P_{2} {v_{2}}^{y}$

$P_{2} = P_{1} {(\frac{v_{1}}{v_{2}})}^{y} = P {(\frac{v}{\frac{1}{8} v})}^{y}$

$P_{2} = P (8) \frac{5}{3}$

P2 = 32 P

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5 months ago

Physics Physics Thermodynamics

7 mol of a certain monoatomic ideal gas undergoes a temperature increase of 40K at constant pressure. The increase in the internal energy of the gas in this process is:

(Given R = 8.3 JK^-1 mol^-1)

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Payal Gupta

Contributor-Level 10

$Δ u = n C_{V} Δ T$

$= n \frac{3 R}{2} Δ T$

$= 7 * \frac{3}{2} * 8.3 * 40$

$= 3486 J$

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