answer:A possible reason for the reduced torque and the sound you're hearing could be related to the rear brakes. It's important to have them inspected for proper clearance, as they might be slightly locking up while riding. This not only affects the torque output but also reduces fuel efficiency. Additionally, check the condition of your air filter. A dirty or clogged air filter can restrict airflow, impacting engine performance, especially in hot weather. If it hasn't been changed recently, consider replacing it.

answer:Since the plutonium nucleus starts at rest and has no initial angular momentum, the total linear momentum before and after the decay must be conserved. According to the law of conservation of momentum: {eq}m_{Pu}vec v_{Pu} = m_Uvec v_U + m_{alpha}vec v_{alpha} {/eq} Given that the plutonium nucleus decays, {eq}vec v_{Pu} = 0 {/eq}. This simplifies the equation to: {eq}0 = m_Uvec v_U + m_{alpha}vec v_{alpha} {/eq} From the kinetic energy of the alpha particle, we have: {eq}K_{alpha} = frac{1}{2}m_{alpha}v_{alpha}^2 {/eq} Solving for {eq}v_{alpha} {/eq} gives: {eq}v_{alpha} = sqrt{frac{2K_{alpha}}{m_{alpha}}} {/eq} Now, using the conservation of momentum, we can find the velocity of the uranium nucleus: {eq}vec v_U = -frac{m_{alpha}}{m_U}vec v_{alpha} {/eq} Substituting the expression for {eq}v_{alpha} {/eq} into this equation: {eq}vec v_U = -frac{m_{alpha}}{m_U}sqrt{frac{2K_{alpha}}{m_{alpha}}} hat{v_{alpha}} {/eq} {eq}v_U = frac{m_{alpha}}{m_U}sqrt{frac{2K_{alpha}}{m_{alpha}}} {/eq} Hence, the speed of the uranium-235 nucleus is: {eq}v_U = frac{m_{alpha}}{m_U}sqrt{frac{2K_{alpha}}{m_{alpha}}} {/eq} The direction of the uranium nucleus is opposite to that of the alpha particle ({eq}hat{v_{alpha}} {/eq}).

answer:Step 1: Calculate the line charge density: {eq}lambda=dfrac{q}{dy}=dfrac{20~nC}{10~m}=dfrac{2~nC}{m}{/eq} Step 2: Use the equation for the electric field due to a line charge: {eq}E=dfrac{klambda}{r}{/eq} where {eq}r=5~m{/eq}. Step 3: Substitute the values and solve for the magnitude of the electric field: {eq}E=dfrac{left(9times 10^9 dfrac{Ncdot m^2}{C^2}right)left(2times10^{-9}dfrac{C}{m}right)}{5~m} E=3.6~N/C {/eq}

answer:The expression frac{ln(x^{2})}{ln(x)} simplifies as follows: frac{ln(x^{2})}{ln(x)} = frac{2ln(x)}{ln(x)} Since ln(x^2) is the logarithm of x^2 with base e, it can be written as 2ln(x) (as x^2 = e^{2ln(x)}). When we divide this by ln(x), the ln(x) terms cancel out, resulting in: 2 Alternatively, we can use the change of base formula to express the original expression in terms of the logarithm base x: frac{ln left(x^2right)}{ln x}=log_x!!left(x^2right) By the properties of logarithms, log_x(x^2) equals the exponent, which is 2: 2 This equality holds for all x > 0 except when x = 1, because the logarithm of 1 is 0, making the original expression undefined.