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Physics Laws of Motion

The resultant of these forces $\vec{O P}, \vec{O Q}, \vec{Q R}, \vec{O S} a n d \vec{O T}$ is approximately. . . . . . .N

[Take $\sqrt[]{3} = 1.7, \sqrt[]{2} = 1.4 .$ Given $\hat{i} a n d \hat{j}$ unit vectors along x, y axis]

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Vishal Baghel

Contributor-Level 10

${\vec{F}}_{x} = [10 * \frac{\sqrt[]{3}}{2} + 20 * \frac{1}{2} + \frac{20}{\sqrt[]{2}} - \frac{15}{\sqrt[]{2}} - \frac{15 \sqrt[]{3}}{2}] = 9.25 \hat{i}$

${\vec{F}}_{y} = [15 * \frac{1}{2} + 20 * \frac{\sqrt[]{3}}{2} + 10 * \frac{1}{2} - \frac{15}{\sqrt[]{2}} - \frac{20}{\sqrt[]{2}}] = 5 \hat{j}$

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Physics Class 11th

A copper block of mass 5.0kg is heated to a temperature of $500 ° C$ and is placed on a large ice block. What is the maximum amount of ice that can melt?

[Specific heat of copper : 0.39 J g^-1 ⁰C^-1 and latent heat of fusion of water : 335 J g^-1]

0 Follower 3 Views

Raj Pandey

Contributor-Level 9

Heat Released by block = Heat gain by large Ice block

$\Rightarrow m c Δ T = M_{i c e} L$

5 * 0.39 * 500 = m_ice * 335

$m_{i c e} = \frac{5 * 0.39 * 500}{335}$

= 2.91 kg

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a year ago

Physics Electromagnetic Waves

If E and H represents the intensity of electric field and magnetizing field respectively, then the unit of E/H will be:

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Vishal Baghel

Contributor-Level 10

$\frac{E}{H} = \frac{V o l t / m e t r e}{A m p e r e / m e t r e} = \frac{V o l t}{A m p e r e} = o h m$

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a year ago

Physics Units and Measurement

Match List II with List II

List I List II

(A) R_H(Rydberg constant) (i) kg m^-¹s^-¹

(B) h (Planck

...more

Match List II with List II

List I List II

(A) R_H(Rydberg constant) (i) kg m^-¹s^-¹

(B) h (Planck's constant) (ii) kg m²s^-¹

(C) $μ_{B}$ (Magnetic field energy density) (iii) m^-¹

(D) $η$ (coefficient of viscocity (iv) kg m^-¹s^-²

less

0 Follower 5 Views

alok kumar singh

Contributor-Level 10

R_H : m^-1

h : kgm² s^-1

$μ_{B} : k g m^{- 1} s^{- 2}$

$η : k g m^{- 1} s^{- 1}$

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a year ago

Physics Oscillations

The variation of displacement with time of a particle executing free simple harmonic moton is shown in the figure. The potential energy U(x) versus time (t) plot of the particle is correctly shown in figure:

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Vishal Baghel

Contributor-Level 10

Potential Energy is maximum at extreme position

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a year ago

Physics Class 11th

A mass of 50 kg is placed at the centre of a uniform spherical shell of mass 100kg and radius 50m. If the gravitational potential at a point, 25m from the centre is V kg/m. The value of V is:

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alok kumar singh

Contributor-Level 10

$V_{A} = - \frac{G M_{1}}{r} - \frac{G M_{2}}{R} = - \frac{50 G}{25} - \frac{100 G}{50} = - 4 G$

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a year ago

Physics Physics Moving Charges and Magnetism

Two ions of masses 4 amu and 16 amu have charges +2e and +3e respectively. These ions pass through the region of constant perpendicular magnetic field. The kinetic energy of both ions is same. Then:

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Vishal Baghel

Contributor-Level 10

$R = \frac{\sqrt[]{2 E m}}{B . q}$

$\Rightarrow R * \frac{\sqrt[]{m}}{q}$

$\Rightarrow \frac{R_{1}}{R_{2}} = \frac{\sqrt[]{4}}{2} * \frac{3}{\sqrt[]{16}} = \frac{3}{4}$

$\Rightarrow R_{2} = \frac{4 R_{1}}{3}$

$s i n θ = \frac{d}{R} \Rightarrow θ α \frac{1}{2} \Rightarrow θ_{2} < θ_{1}$

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a year ago

Physics Physics Motion in Plane

A player kicks a football with an initial speed of 25 ms^-¹ at an angle of 45° from the ground. What are the maximum height and the time taken by the football to reach at the highest point during motion? (Take g = 10 ms^-²)

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alok kumar singh

Contributor-Level 10

$H_{m a x} = \frac{u^{2} s i n^{2} θ}{2 g} = \frac{{(25 * s i n 45 °)}^{2}}{2 * 10} = 15.625 m$

$T = \frac{u s i n θ}{g} = \frac{25 * s i n 45}{10} = 1.77 s$

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a year ago

Physics Class 12th

There are 10¹⁰ radioactive nuclei in a given radioactive element, its half-time is 1 minute. How many nuclei will remain after 30 seconds?

$(\sqrt[]{2} = 1.414)$

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Vishal Baghel

Contributor-Level 10

$\frac{N}{N_{0}} = {(\frac{1}{2})}^{t |_{t_{1} | 2}}$

$\Rightarrow N = 1 0^{10} * {(\frac{1}{2})}^{\frac{1}{2}} = \frac{1 0^{10}}{\sqrt[]{2}} ? 7 * 1 0^{9}$

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New answer posted

a year ago

Physics Class 11th

A solid metallic cube having total surface area 24 m² is uniformly heated. if its temperature is increased by 10⁰C, calculate the increase in volume of the cube.

(Given a = 5.0 * 10^-4 ⁰C^-1).

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Raj Pandey

Contributor-Level 9

Given

6a² = 24

a2 = 4

a = 2m

$Δ v = v (γ) Δ T$

$= 3 α v Δ T$

$= 3 * 5 * 1 0^{- 4} * (a^{3}) * 10$

$= 3 * 5 * 1 0^{- 4} * 8 * 10$

$\begin{array}{l} = 120 * 1 0^{3} \\ = 1.2 * 1 0^{5} c m^{3} \end{array}$

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