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6 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Five

A ball of mass 0.15 kg hits the wall with its initial speed of 12 ms^-1and bounces back without changing its initial speed. If the force applied by the wall on the ball during the contact is 100 N, calculate the time during of the contact of ball with the wall.

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alok kumar singh

Contributor-Level 10

$F = \frac{Δ ρ}{Δ t} = \frac{1.8 - (- 1.8)}{Δ t} = 100$

$\Rightarrow Δ t = \frac{3.6}{100} = 0.036 s e c$

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6 months ago

Physics Motion in a Straight Line

A scooter accelerates from rest for time t₁ at constant rate a₁ and then retards at constant rate a₂ for time t₂ and comes to rest. The correct value of $\frac{t_{1}}{t_{2}}$ will be

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Payal Gupta

Contributor-Level 10

$t a n α = \frac{v_{m a x}}{t_{1}} = a_{1}$

$t a n β = \frac{v_{m a x}}{t_{2}} = a_{2}$

$\Rightarrow \frac{a_{1}}{a_{2}} = \frac{t_{2}}{t_{1}} \Rightarrow \frac{t_{1}}{t_{2}} = \frac{a_{2}}{a_{1}}$

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6 months ago

Physics physics ncert exemplar solutions class 12th chapter two

In a Vernier Calipers, 10 divisions of Vernier scale is equal to the 9 divisions of main scale. Then both jaws of Vernier calipers touch each other, the zero of the Vernier scale is shifted to the left of zero of the main scale and 4^th Vernier scale division exactly coincides with the main scale reading. One main scale division is equal to 1mm. While measuring diameter of a spherical body, the body is held between two jaws. It is now observed that zero of the Vernier scale lies between 30 and 31 divisions of main scale reading and 6^th Vernier scale division exactly coincides with the main scale reading. The diameter of the spherical bo

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alok kumar singh

Contributor-Level 10

1 MSD = 1mm

9 MSD = 10 VSD

1VSD = 0.9 MSD = 0.9 mm

Least count = 1 MSD – 1 VSD

= 0.1 mm = 0.01 cm

Zero Error = (10 – 4) * 0.1 = 0.6 mm

Reading = MSR + VSR – Zero error

= 3 cm + 6 * 0.01 – (0.06)

= 3.12 cm $?$ 3.10 cm

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6 months ago

Physics Oscillations

A particle executes S.H.M., the graph of velocity as a function of displacement is

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Payal Gupta

Contributor-Level 10

$v = ω \sqrt[]{A^{2} - x^{2}} \Rightarrow \frac{v^{2}}{A^{2} ω^{2}} + \frac{x^{2}}{A^{2}} = 1$ Path is ellipse.

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6 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Four

Two projectiles are thrown with same initial velocity making an angle of $45 °$ and $30 °$ with the horizontal respectively. The ration of their respectively ranges will be:

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alok kumar singh

Contributor-Level 10

$R_{1} = \frac{u^{2} s i n 90}{g}$

$R_{2} = \frac{u^{2} s i n 60}{g}$

$\Rightarrow \frac{R_{1}}{R_{2}} = \frac{2}{\sqrt[]{3}}$

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6 months ago

Physics Laws of Motion

Two masses A and B, each of mass M are fixed together by a massless spring. A force acts on the mass B as shown in figure. If the mass A starts moving away from mass B with acceleration 'a', then the acceleration of mass B will be

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Payal Gupta

Contributor-Level 10

Let the spring is in extension state and $a_{B} > a_{A}, {\vec{a}}_{A B} = - a_{A} \hat{i} - (- a_{B} \hat{i}) = (- a_{A} + a_{B}) \hat{i}$

Hence we can say that block moves away from block B in the frame of B

F – kx = Ma_B . (1)

kx = Ma . (2)

$\Rightarrow a_{B} = \frac{F}{M} - a$

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6 months ago

Physics Ray Optics and Optical Instruments

The incident ray, reflected ray and the outward drawn normal are denoted by the unit vectors $\vec{a}, \vec{b} a n d \vec{c}$ respectively. Then choose the correct relation for these vectors.

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Payal Gupta

Contributor-Level 10

$\hat{r} - \hat{i} = 2 c o s θ \hat{n} . . . . . . (1)$

$\hat{i} . \hat{n} = - c o s θ . . . . . . (2)$

$\Rightarrow \hat{r} = \hat{i} - 2 (\hat{i} . \hat{n}) \hat{n}$

$\Rightarrow \vec{b} = \vec{a} - 2 (\vec{a} . \vec{c}) \vec{c}$

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6 months ago

Physics Electric Charges and Fields

An inclined plane making an angle of 30Â° with the horizontal is placed in a uniform horizontal electric field $200 \frac{N}{C}$ as shown in the figure. A body of mass 1kg and charge 5 mC is allowed to slide down from rest at a height of 1m. If the coefficient of friction is 0.2, find the time taken by the body to reach the bottom.

$[g = 9.8 m / s^{2}; s i n 30 ° = \frac{1}{2}; c o s 30 ° = \frac{\sqrt[]{3}}{2}]$

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Payal Gupta

Contributor-Level 10

$N = m g c o s 30 ° + q E s i n 30 °$

$a = \frac{m g s i n θ - q E c o s θ - μ N}{m} = 2.30 m / s^{2}$

$S = u t + \frac{1}{2} a t^{2}$

$\Rightarrow t = \sqrt[]{\frac{2 l}{a}} = 1.31 s e c$

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6 months ago

Physics Electromagnetic Waves

A tuning fork A of unknown frequency produces 5 beats/s with a fork of known frequency 340 Hz. When fork A is filed, the beat frequency decreases to 2 beats/s. What is the frequency of fork A?

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Payal Gupta

Contributor-Level 10

Frequency increases on filing. So, initial frequency of A is 335 Hz.

f = 340 – 5 = 335 Hz

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6 months ago

Physics Alternating Current

Find the peak current and resonant frequency of the following circuit (as shown in figure).

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Payal Gupta

Contributor-Level 10

$Z = \sqrt[]{R^{2} + {(X_{L} - X_{C})}^{2}}$

$Z = \sqrt[]{{(120)}^{2} + {(10 - 100)}^{2}} = 150 Ω$

$ω = \frac{1}{\sqrt[]{L C}} = \frac{1}{\sqrt[]{1 0^{- 1} * 1 0^{- 4}}} = \sqrt[]{1 0^{5}}$

$? ω = 2 π f$

$\Rightarrow f = \frac{1 0^{3}}{2 π \sqrt[]{10}} = 50 H z$

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