A fisherman reels in his fishing line onto a spool with a diameter of 0.100 meters. If he rotates the spool 3.5 times, how much fishing line does he reel in, in meters?

Bill cut 4 1/12 yards of fishing line from a new reel. How much is this in inches? Write your answer as a whole number or a mixed number in simplest form. Include the correct unit in your answer.

Miguel cut 413 yards of fishing line from a new reel. How much is this in inches?Write your answer as a whole number or a mixed number in simplest form. Include the correct unit in your answer.

A kite 100 ft100 ft above the ground moves horizontally at a speed of 3 ft/s3 ft/s . At what rate is the angle (in radians) between the string and the horizontal decreasing when 200 ft200 ft of string have been let out?

To solve this problem, we need to determine the time required for the reel to reach an angular velocity of \(5 \, \text{rad/s}\) when subjected to a force \(F = 400 \, \text{N}\). The reel has a mass of \(50 \, \text{kg}\), a radius of gyration \(k_G = 2 \, \text{m}\), and the coefficient of kinetic friction between the reel and the ground is \(\mu_k = 0.2\). ### Step-by-Step Solution: 1. **Determine the Moment of Inertia:** The moment of inertia \(I\) of the reel about its center of mass \(G\) is given by: \[ I = m k_G^2 \] where \(m = 50 \, \text{kg}\) and \(k_G = 2 \, \text{m}\). \[ I = 50 \times 2^2 = 50 \times 4 = 200 \, \text{kg} \cdot \text{m}^2 \] 2. **Calculate the Torque:** The torque \(\tau\) applied by the force \(F\) is: \[ \tau = F \times r \] where \(r = 3 \, \text{m}\) (the radius at which the force is applied). \[ \tau = 400 \times 3 = 1200 \, \text{N} \cdot \text{m} \] 3. **Determine the Frictional Force:** The normal force \(N\) is equal to the weight of the reel: \[ N = mg = 50 \times 9.8 = 490 \, \text{N} \] The frictional force \(f_k\) is: \[ f_k = \mu_k N = 0.2 \times 490 = 98 \, \text{N} \] 4. **Calculate the Frictional Torque:** The frictional torque \(\tau_f\) is: \[ \tau_f = f_k \times r_f \] where \(r_f = 2 \, \text{m}\) (the radius at which the friction acts). \[ \tau_f = 98 \times 2 = 196 \, \text{N} \cdot \text{m} \] 5. **Net Torque:** The net torque \(\tau_{net}\) is: \[ \tau_{net} = \tau - \tau_f = 1200 - 196 = 1004 \, \text{N} \cdot \text{m} \] 6. **Angular Acceleration:** The angular acceleration \(\alpha\) is given by: \[ \alpha = \frac{\tau_{net}}{I} = \frac{1004}{200} = 5.02 \, \text{rad/s}^2 \] 7. **Time to Reach Angular Velocity:** Using the kinematic equation for angular motion: \[ \omega = \omega_0 + \alpha t \] where \(\omega = 5 \, \text{rad/s}\), \(\omega_0 = 0 \, \text{rad/s}\) (initial angular velocity), and \(\alpha = 5.02 \, \text{rad/s}^2\). Solving for \(t\): \[ 5 = 0 + 5.02 t \] \[ t = \frac{5}{5.02} \approx 0.996 \, \text{s} \] So, the time required for the reel to obtain an angular velocity of \(5 \, \text{rad/s}\) is approximately \(0.996 \, \text{s}\).

To solve this problem, we need to determine the time required for the reel to obtain an angular velocity of \(5 \, \text{rad/s}\) when subjected to a force \(F = 400 \, \text{N}\). We will use the principles of rotational dynamics and kinematics. ### Given: - Mass of the reel, \( m = 50 \, \text{kg} \) - Radius of gyration, \( k_G = 2 \, \text{m} \) - Force applied, \( F = 400 \, \text{N} \) - Desired angular velocity, \( \omega = 5 \, \text{rad/s} \) - Coefficient of kinetic friction, \( \mu_k = 0.2 \) - Radius of the reel, \( R = 3 \, \text{m} \) ### Steps to solve: 1. **Determine the moment of inertia \( I \) of the reel:** The moment of inertia \( I \) about the center of mass \( G \) is given by: \[ I = m k_G^2 = 50 \, \text{kg} \times (2 \, \text{m})^2 = 200 \, \text{kg} \cdot \text{m}^2 \] 2. **Determine the torque \( \tau \) applied by the force \( F \):** The torque \( \tau \) applied by the force \( F \) is: \[ \tau = F \times R = 400 \, \text{N} \times 3 \, \text{m} = 1200 \, \text{N} \cdot \text{m} \] 3. **Determine the frictional force \( f_k \):** The normal force \( N \) is equal to the weight of the reel: \[ N = m g = 50 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 490.5 \, \text{N} \] The kinetic frictional force \( f_k \) is: \[ f_k = \mu_k N = 0.2 \times 490.5 \, \text{N} = 98.1 \, \text{N} \] 4. **Determine the net torque \( \tau_{\text{net}} \):** The net torque \( \tau_{\text{net}} \) is the applied torque minus the torque due to friction. The frictional force acts at the radius \( R = 3 \, \text{m} \): \[ \tau_{\text{friction}} = f_k \times R = 98.1 \, \text{N} \times 3 \, \text{m} = 294.3 \, \text{N} \cdot \text{m} \] Therefore, the net torque is: \[ \tau_{\text{net}} = \tau - \tau_{\text{friction}} = 1200 \, \text{N} \cdot \text{m} - 294.3 \, \text{N} \cdot \text{m} = 905.7 \, \text{N} \cdot \text{m} \] 5. **Determine the angular acceleration \( \alpha \):** Using the relation between torque and angular acceleration: \[ \tau_{\text{net}} = I \alpha \] \[ \alpha = \frac{\tau_{\text{net}}}{I} = \frac{905.7 \, \text{N} \cdot \text{m}}{200 \, \text{kg} \cdot \text{m}^2} = 4.5285 \, \text{rad/s}^2 \] 6. **Determine the time \( t \) required to reach the angular velocity \( \omega \):** Using the kinematic equation for rotational motion: \[ \omega = \omega_0 + \alpha t \] where \(\omega_0 = 0 \, \text{rad/s}\) (initial angular velocity). Solving for \( t \): \[ t = \frac{\omega}{\alpha} = \frac{5 \, \text{rad/s}}{4.5285 \, \text{rad/s}^2} \approx 1.104 \, \text{s} \]