Physics Questions and Answers

In general, the energy released during nuclear fission is given by the Einstein's mass-energy equation, given by

\( E = \Delta MC^2 \)

Where \( \Delta m\) is the mass defect, and C is the speed of light.
Therefore
\( \Delta m = 0.01% \text{of} 1.0g = \frac{0.01}{100}
\times 1.0g \\

= 1.0 \times 10^{-4} \\
= 1.0 \times 10^{-7}kg\\

\text{Energy Released } = \Delta MC^2 \\

= 1.0 \times 10^{-7} \times (3.0 \times 10^8)^2\\
= 1.0 \times 10^{-2} \times 9.0 \times 10^{16} \\

= 9.0 \times 10^9 J \)

From the uncertainty principle the product of the position and momentum of a sub-atomic particle is equal or greater than planck's constant.

i.e \( \Delta x. \Delta p \geq h. \\
\text{Thus} 5 \times 10^{-10} \times \Delta p = 6.6 \times 10^{-34} \\

\Delta p = \frac{6.6 \times 10^{-34}}{5 \times 10^{-10}} = 1.32 \times 10^{-24}Ns\\

\text{Therefore Uncertainty in the momentum } = 1.32 \times 10^{-24}Ns \)

The speed of sound in a material is given as:

\( V = \sqrt{\frac{E}{P}}, Where\\

E = \text{Young's modulus} \\
P = \text{Density }\\

\text{Therefore } V = \sqrt{\frac{7.0 \times 10^{10}}{2.7 \times 10^3}} = 5091.75\\

= 5.1 \times 10^3MS^{-1} \)

From Newton's second law:

\( F = \frac{mV - mµ}{t} \)

Let the final velocity = V; and since the football is initially at rest, its initial vel µ = 0, M = 0.8kg

\( t = 0.8s \\

\implies 100 = \frac{0.8V - 0.8 \times 0}{0.8} \\

\text{Therefore } V = \frac{100 \times 0.8}{0.8} = 100ms^{-1} \)

Atomic emission and absorption of energy is usually in packets referred to as photon or quantum of energy of magnitude E =hf, where f is the frequency and h is the plancks constant

Energy emitted = hf

= 6.6 x 10^-34 x 8.0 x 10^-14
= 5.28 x 10^-19J