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Assume that a skater of mass m, moving at a constant speed Vo, suddenly stands stiffly with skates pointed directly forward and allows herself to coast to a stop. Neglecting air resistance, how far will she travel (on two blades) before she stops? Give the answer X as a function of (Vo, m, L, h, H. M.

Assume that a skater of mass m, moving at a constant speed Vo, suddenly stands stiffly with skates pointed directly forward and allows herself to coast to a stop. Neglecting air resistance, how far will she travel (on two blades) before she stops? Give the answer X as a function of (Vo, m, L, h, H. M.

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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### Multiple Choice Question Choose the correct formula for \( X \): 1. ☐ \(X = \frac{V_0 rmh_u}{2LW}\) 2. ☐ \(X = \frac{V_0^2 rmh}{2\mu LW}\) 3. ☐ \(X = \frac{V_0 h_u L}{2mW}\) 4. ☐ \(X = \frac{V_0 rmh}{2\mu LW}\) Each option consists of variables and constants arranged in a specific formula format. Here's a breakdown of the components involved: - \( V_0 \): Initial velocity or some constant value - \( rmh_u \): Product of variables \( r \), \( m \), and \( h_u \) - \( LW \): Product of variables \( L \) and \( W \) - \( \mu \): Coefficient of friction or some other multiplier - \( h \): Variable potentially representing height or some other measure - \( L \): Length or some other measure - \( W \): Width or some other measure Make sure to carefully analyze each formula to choose the correct one for \( X \).
Transcribed Image Text:### Multiple Choice Question Choose the correct formula for \( X \): 1. ☐ \(X = \frac{V_0 rmh_u}{2LW}\) 2. ☐ \(X = \frac{V_0^2 rmh}{2\mu LW}\) 3. ☐ \(X = \frac{V_0 h_u L}{2mW}\) 4. ☐ \(X = \frac{V_0 rmh}{2\mu LW}\) Each option consists of variables and constants arranged in a specific formula format. Here's a breakdown of the components involved: - \( V_0 \): Initial velocity or some constant value - \( rmh_u \): Product of variables \( r \), \( m \), and \( h_u \) - \( LW \): Product of variables \( L \) and \( W \) - \( \mu \): Coefficient of friction or some other multiplier - \( h \): Variable potentially representing height or some other measure - \( L \): Length or some other measure - \( W \): Width or some other measure Make sure to carefully analyze each formula to choose the correct one for \( X \).
### Understanding Ice Skating Physics **Key Concept:** When a person ice-skates, the ice surface actually melts beneath the blades, creating a thin film of water between the blade and the ice. **Problem Statement:** Assume that a skater of mass \( m \), moving at a constant speed \( V_0 \), suddenly stands stiffly with skates pointed directly forward and allows themselves to coast to a stop. Neglecting air resistance, how far will they travel (on two blades) before they stop? **Solution Approach:** To find the distance \( X \) that the skater travels before stopping, we need to consider the forces acting on the skater and how they influence the motion. **Given Variables:** - \( m \): Mass of the skater - \( V_0 \): Initial speed of the skater - \( \mu \): Coefficient of friction between the skates and the ice - \( g \): Acceleration due to gravity (standard value \( 9.8 \, \text{m/s}^2 \)) **Simplified Solution Derivation:** To find the distance \( X \), apply the principles of kinematics and frictional force: 1. The frictional force that the ice exerts on the skater is given by: \[ F_{\text{friction}} = \mu \cdot m \cdot g \] 2. Since this is the force causing deceleration (negative acceleration), we can write the equation for the deceleration as: \[ a = \frac{F_{\text{friction}}}{m} = \mu \cdot g \] 3. Using the kinematic equation for motion under constant acceleration: \[ V_f^2 = V_0^2 + 2aX \] where \( V_f \) (final velocity) is zero since the skater stops. 4. Substituting \( a = \mu \cdot g \) and rearranging to solve for \( X \): \[ 0 = V_0^2 + 2(\mu \cdot g)(X) \] \[ X = \frac{V_0^2}{2 \mu g} \] Thus, the distance \( X \) in terms of the given variables is: \[ X = \frac{V_0^2}{2
Transcribed Image Text:### Understanding Ice Skating Physics **Key Concept:** When a person ice-skates, the ice surface actually melts beneath the blades, creating a thin film of water between the blade and the ice. **Problem Statement:** Assume that a skater of mass \( m \), moving at a constant speed \( V_0 \), suddenly stands stiffly with skates pointed directly forward and allows themselves to coast to a stop. Neglecting air resistance, how far will they travel (on two blades) before they stop? **Solution Approach:** To find the distance \( X \) that the skater travels before stopping, we need to consider the forces acting on the skater and how they influence the motion. **Given Variables:** - \( m \): Mass of the skater - \( V_0 \): Initial speed of the skater - \( \mu \): Coefficient of friction between the skates and the ice - \( g \): Acceleration due to gravity (standard value \( 9.8 \, \text{m/s}^2 \)) **Simplified Solution Derivation:** To find the distance \( X \), apply the principles of kinematics and frictional force: 1. The frictional force that the ice exerts on the skater is given by: \[ F_{\text{friction}} = \mu \cdot m \cdot g \] 2. Since this is the force causing deceleration (negative acceleration), we can write the equation for the deceleration as: \[ a = \frac{F_{\text{friction}}}{m} = \mu \cdot g \] 3. Using the kinematic equation for motion under constant acceleration: \[ V_f^2 = V_0^2 + 2aX \] where \( V_f \) (final velocity) is zero since the skater stops. 4. Substituting \( a = \mu \cdot g \) and rearranging to solve for \( X \): \[ 0 = V_0^2 + 2(\mu \cdot g)(X) \] \[ X = \frac{V_0^2}{2 \mu g} \] Thus, the distance \( X \) in terms of the given variables is: \[ X = \frac{V_0^2}{2
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