1. **State the problem:** We are given the expression $$\frac{90x^2 p}{35x^2 m_e} = 2.8228348433771 \times 10^{29} \times \frac{1}{m \cdot s}$$ where $m_e$ is the electron mass, $9.1093837015 \times 10^{-31}$ kg. We need to find the value of $p$. 2. **Simplify the expression:** Since $x^2$ appears in numerator and denominator, it cancels out: $$\frac{90 p}{35 m_e} = 2.8228348433771 \times 10^{29} \times \frac{1}{m \cdot s}$$ 3. **Rearrange to solve for $p$:** $$p = \frac{35 m_e}{90} \times 2.8228348433771 \times 10^{29} \times \frac{1}{m \cdot s}$$ 4. **Substitute the value of $m_e$:** $$p = \frac{35}{90} \times 9.1093837015 \times 10^{-31} \times 2.8228348433771 \times 10^{29} \times \frac{1}{m \cdot s}$$ 5. **Calculate the numerical value:** First calculate the fraction: $$\frac{35}{90} = 0.3888888889$$ Multiply constants: $$0.3888888889 \times 9.1093837015 \times 10^{-31} = 3.543 \times 10^{-31}$$ Then multiply by $2.8228348433771 \times 10^{29}$: $$3.543 \times 10^{-31} \times 2.8228348433771 \times 10^{29} = 3.543 \times 2.8228 \times 10^{-31+29} = 10.0 \times 10^{-2} = 0.1$$ 6. **Final answer:** $$p = 0.1 \times \frac{1}{m \cdot s} = 0.1 \ \text{per meter-second}$$ So, the value of $p$ is approximately $0.1 \ \frac{1}{m \cdot s}$.