Two positive ions, each carrying a charge q, are separated by a distance d. If F is the force of repulsion between the ions, the number of electrons missing from each ion will be ( e being the charge of an electron) ?

An electron is moving round the nucleus of a hydrogen atom in a circular orbit of radius r. The Coulomb force F between the two is ___ ?

 

When air is replaced by a dielectric medium of force constant K, the maximum force of attraction between two charges, separated by a distance ___ ?

Point charges +4q, −q and +4q are kept on the X-axis at points x = 0, x = a and x = 2a respectively. Then ___ ?

If the force on a rocket moving with a velocity of 300 m/sec is 345 N, then the rate of combustion of the fuel, is

 

A 3 kg ball strikes a heavy rigid wall with a speed of 10 m/s at an angle of 60◦. It gets reflected with the same speed and angle as shown here. If the ball is in contact with the wall for 0.20 s, what is the average force exerted on the ball by the wall?

A satellite in a force free space sweeps stationary interplanetary dust at a rate (dM/dt) = αv. The acceleration of satellite is

Physical independence of force is a consequence of

A particle of mass m is moving with a uniform velocity $v_1$. It is given an impulse such that its velocity becomes $v_2$. The impulse is equal to

A 600 kg rocket is set for a vertical firing. If the exhaust speed is 1000 $ms^{−1}$, the mass of the gas ejected per second to supply the thrust needed to overcome the weight of rocket is

A block B is pushed momentarily along a horizontal surface with an initial velocity V. If μ is the coefficient of sliding friction between B and the surface, block B will come to rest after a time

A 100 N force acts horizontally on a block of 10 kg placed on a horizontal rough surface of coefficient of friction μ = 0.5. If the acceleration due to gravity (g) is taken as 10 $ms^{−2}$, the acceleration of the block (in $ms^{−2}$ ) is

trial chatgpt codes

\[ \frac{F_{\text{electric}}}{F_{\text{gravity}}}\]

\[ \frac{F_{\text{electric}}}{F_{\text{gravity}}} = \frac{\frac{1}{4\pi \varepsilon_0} \cdot \frac{q_1 q_2}{r^2}}{G \cdot \frac{m_1 m_2}{r^2}} = \frac{(9 \times 10^9) \cdot \frac{(1.6 \times 10^{-19})(1.6 \times 10^{-19})}{r^2}}{(6.67 \times 10^{-11}) \cdot \frac{(1.67 \times 10^{-27})(9.11 \times 10^{-31})}{r^2}} = 2.4 \times 10^{39} \] \[ I = m_1 x^2 + m_2 (L - x)^2 \] \[ \frac{dI}{dx} = m_1 \cdot 2x + m_2 \cdot 2(L - x)(0 - 1) \] \[ \text{Minimum when } \frac{dI}{dx} = 0 \] \[ m_1 \cdot 2x = m_2 \cdot 2(L - x) \] \[ m_1 \cdot 2x = m_2 \cdot 2L - m_2 \cdot 2x \] \[ (m_1 + m_2) \cdot 2x = m_2 \cdot 2L \] \[ \Rightarrow x = \frac{m_2 L}{m_1 + m_2} \] 

\[ \frac{\Delta L}{L} = ? \] \[ \text{Given: } \text{Hypotenuse } = L + \Delta L, \quad \text{Opposite side } = x \] Using Pythagoras’ Theorem: \[ L^2 + x^2 = (L + \Delta L)^2 \] \[ L^2 + x^2 = \left( L \left(1 + \frac{\Delta L}{L} \right) \right)^2 \] \[ L^2 + x^2 = L^2 \left(1 + \frac{\Delta L}{L} \right)^2 \] Divide both sides by \( L^2 \): \[ 1 + \frac{x^2}{L^2} = \left(1 + \frac{\Delta L}{L} \right)^2 \] \[ \sqrt{1 + \frac{x^2}{L^2}} = 1 + \frac{\Delta L}{L} \] Using the approximation \( \sqrt{1 + x} \approx 1 + \frac{x}{2} \) for small \( x \), we get: \[ 1 + \frac{x^2}{2L^2} \approx 1 + \frac{\Delta L}{L} \] \[ \Rightarrow \frac{\Delta L}{L} \approx \frac{x^2}{2L^2} \] \[ \boxed{\frac{\Delta L}{L} = \frac{x^2}{2L^2}} \]

\[ \sum F = ma \]\[ 100 - \mu mg = ma \] \[ 100 - (0.5)(10)(10) = (10)a \] \[ 50 = 10\hspace{2mm}a \] \[ a = 5 \, \text{m/s}^2 \]