Question

1. a body acted upon by a variable force, f, produces a corresponding velocity each time. assume that the power of the body is constant, f and v are related by the expression a. ( f propto v^{-1} ) b. ( f propto v^{1/2} ) c. ( f propto v^2 ) d. ( f propto v^{-2} ) 28. the distance between two successive troughs of a wave is 16.5 cm and the velocity of the wave is ( 247.5 , \text{ms}^{-1} ). calculate the frequency of the wave. a. ( 15.00 , \text{hz} ) b. ( 40.84 , \text{hz} ) c. ( 4083.75 , \text{hz} ) d. ( 1500.00 , \text{hz} )

Explanation:

Question 27

Step1: Recall the power formula

Power \( P \) is given by the product of force \( F \) and velocity \( V \), i.e., \( P = F \times V \).

Step2: Analyze the relationship when \( P \) is constant

If power \( P \) is constant, we can rewrite the formula as \( F=\frac{P}{V} \). Since \( P \) is a constant, we can say that \( F \) is inversely proportional to \( V \), or \( F\propto V^{-1} \).

Step1: Recall the wave formula

The velocity \( v \) of a wave is related to its frequency \( f \) and wavelength \( \lambda \) by the formula \( v = f\lambda \).

Step2: Determine the wavelength

The distance between two successive troughs is the wavelength \( \lambda \). Given \( \lambda = 16.5\space cm = 0.165\space m \) (converting to meters) and \( v = 247.5\space ms^{-1} \).

Step3: Solve for frequency \( f \)

From \( v = f\lambda \), we can solve for \( f \) as \( f=\frac{v}{\lambda} \). Substituting the values, we get \( f = \frac{247.5}{0.165} \).

Step4: Calculate the value

\( \frac{247.5}{0.165}=1500 \space Hz \).

Answer:

A. \( \boldsymbol{F \propto V^{-1}} \)

Question 28