Physics

Contributor-Level 10

$r = \frac{m v}{q B} = \frac{\sqrt[]{2 m k . E}}{q B}$

$r \propto \frac{\sqrt[]{m}}{q}$

m_p = m

q_p = q

m_d = 2m

q_d = q

m_a = 4m

q_a = 2q

$\frac{r_{p}}{r_{d}} = \frac{\sqrt[]{m_{p}} * q_{d}}{q_{P} * \sqrt[]{m_{d}}}$

$\frac{r_{d}}{r_{α}} = \sqrt[]{2}$

$r_{p} : r_{d} : r_{α} = 1 : \sqrt[]{2} : 1$

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9 months ago

The soft-iron is a suitable material for making an electromagnet. This is because soft iron has

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Contributor-Level 10

High permeability and low retentivity

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9 months ago

What will be the most suitable combination of three resistors A = $2 Ω$ , B = $4 Ω$ , C = $6 Ω$ so that $(\frac{22}{3}) Ω$ is equivalent resistance of combination

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Physics Physics Oscillations

Contributor-Level 10

Given

$\begin{array}{l} A = 2 Ω \\ B = 4 Ω \\ C = 6 Ω \\ R_{e q} = \frac{2 * 4}{2 + 4} + 6 \end{array}$

$R_{e q} = \frac{22}{3} Ω$

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9 months ago

Two massless springs with spring constants 2k and 9k, carry 50g and 100g masses at their free ends. These two masses oscillate vertically such that their maximum velocities are equal. Then the ratio of their respective amplitudes will

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Contributor-Level 10

$ω_{1} = \sqrt[]{\frac{k}{m}} = \sqrt[]{\frac{2 k}{0.05}}$

$\begin{array}{l} \Rightarrow A_{1} ω_{1} = A_{2} ω_{2} \\ \frac{A_{1}}{A_{2}} = \frac{ω_{2}}{ω_{1}} = \sqrt[]{\frac{9 k}{0.1}} * \sqrt[]{\frac{0.05}{2 k}} \end{array}$

$\frac{A_{1}}{A_{2}} = \frac{3}{2}$

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9 months ago

Physics Class 11th

A carnot engine takes 5000 kcal of heat from a reservoir at $727 ° C$ and gives heat to a sink at $127 ° C .$ The work done by the engine is

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Physics System of Particles and Rotational Motion

Contributor-Level 10

T₁ = 727 + 273 = 1000k

T₂ = 127° + 273 = 400 k

Q₁ = 5 * 10³ k cal

$\Rightarrow \frac{600}{1000} = \frac{w}{5000}$

w = 12.6 * 10⁶ J

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9 months ago

A metre scale is balanced on a knife edge at its centre. When two coins, each of mass 10 g are put one on the top of the other at the 10.0 cm mark the scale is found to be balanced at 40.0 cm mark. The mass of the metre scale is found to be x * 10^-2 kg. The value of x is _________.

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Contributor-Level 10

$τ_{A} = 0$

2mg *30 = Mg *10

$\Rightarrow 2 * 0.01 g * 30 = M g * 10$

M = 6 * 10^-2kg

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9 months ago

In a potentiometer arrangement, a cell gives a balancing point at 75 cm length of wire. This cell is now replaced by another cell of unknown emf. If the ratio of the emef's of two cells respectively is 3:2, the difference in the balancing length of the potentiometer wire in above two cases will be __________ cm.

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Contributor-Level 10

$E α l$

$\Rightarrow \frac{E_{1}}{E_{2}} = \frac{l_{1}}{l_{2}}$

$\Rightarrow \frac{3}{2} = \frac{75}{l_{2}}$

$l_{2} = \frac{75 * 2}{3}$

$l_{2} = 50 c m$

$& l_{1} = 75 c m$

$Δ l = l_{1} - l_{2} = 25 c m$

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9 months ago

Physics Laws of Motion

A ball of mass 100 g is dropped from a height h = 10 cm on a platform fixed at the top of a vertical spring (as shown in figure). The ball stays on the platform and the platform is depressed by a distance $\frac{h}{2} .$ the spring constant is _________ Nm^-1. (Use g = 10 ms^-2)

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Contributor-Level 10

COME

$k \cdot E_{i} + P \cdot E_{i} = k \cdot E_{f} + P \cdot E_{f}$

$\Rightarrow 0 + m g (h + \frac{h}{2}) = 0 + \frac{1}{2} k {(\frac{h}{2})}^{2}$ $\Rightarrow \frac{3 h}{2} m g = \frac{1}{2} k \frac{h^{2}}{4}$

$\Rightarrow 3 m g h = \frac{k}{4} h^{2} \Rightarrow 12 \frac{m g}{h} = k \Rightarrow k = \frac{12 * 0.1 * 10}{0.1}$

k = 120Nm^-1

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9 months ago

Physics Class 11th

From the top of a tower, a ball is thrown vertically upward which reaches the ground in 6s. A second ball thrown vertically downward from the same position with the same speed reaches the ground in 1.5 s. A third ball released, from the rest from the same location, will reach the ground in __________ s.

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Contributor-Level 10

For first ball

s = ut + $\frac{1}{2} a t^{2}$

$\Rightarrow - h = u * 6 - \frac{1}{2} * 10 * 6 * 6$

$\Rightarrow - h = 6 u - 180$

h = 180 - 6u -(1)

for second boy

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

$h = u * 1.5 + \frac{1}{2} * 10 * 1.5 * 1.5$

h = 1.5u + 11.25 -(2)

from (1) & (2)

180 - 6u = 1.5u + 11.25

7.5u = 180 - 11.25

$u = \frac{168.75}{7.5} = 22.5$

h = 180 - 6u

= 180 - 6 * 22.5

45m

for third ball

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

$h = 0 * t + \frac{1}{2} * 10 t^{2}$

$\Rightarrow 45 = 5 t^{2}$

t² = 9

t = 3 sec

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9 months ago

Two identical charged particles each having a mass 10g and charge 2.0 * 10^-7 C are placed on a horizontal table with a separation of L between them such that they stay in limited equilibrium. If the coefficient of friction between each particle and the table is 0.25, find the value of L. [Use g = 10ms^-2]

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