16. The wheel of a bicycle has a radius of 40 cm. Calculate its period of revolution if the bicycle moves With a speed of 20 km/h. How much is the angular velocity?
17. Vay large magnitude of acceleration can be achieved in some devices such as in a gun. A bullet in a gun is accelerated from the ring chamber to the end of the barrel at an average rate of 6.20810 m/s² for 8.10x10 sec.
What is its muzzle velocity (that is, its final velocity?) 18. A 0.525 kg ball is attached to a 125 in string and swings in a circular path The angle of the string away from vertical is 30.0° Find the centripetal force acting on the ball and the speed of the hall
19. A car travels 20 km due north and then 25 km in a direction 60° West of North. Use both graphical and algebraic methods to find the magnitude and direction of a single vector that gives the net effect of the car's trip
20. A sailor boards a paddle boat and heads the boat Westward directly across a river. The river flows South at 50 cm/s and the woman paddles the boat with a speed of 100 cm/s. A. Determine the resultant velocity of the boat-both magnitude and direction,
B. How far down stream relative to the straight-across direction will woman be when she reaches the opposite shore?
21. The dot product of A and B can be defined alternatively as the magnitude of multiplied by the component B of in the direction of A. Find the angle between each of the following pairs of vectors. a) 3i- j and i-2j b) 3i-2j and i2j (c) 31-2jand 4i+ 6j
22. A driver of a vehicle traveling at a. speed of 30 m/s on a motorway brake sharply to a standstill in a distance of 100m. Calculate the deceleration of the vehicle.
23. A car driver accelerates her car by1.4 m/s, starting from rest. (a) How long does it take her to reach a speed 1 of 2.0 m/s? (b) If she then brakes to a stop in 0.8 seconds, what is her deceleration?
24. What is (a) the tangential velocity? (b) the angular velocity? and (c) the centripetal acceleration, du
to the Earth's rotation, of a body on the equator? Radius of Earth 6,400 km. 25. Determine a velocity vector that has a magnitude of 5 m/s directed along the direction of vector D-1.5i+2j
16. The period of revolution of a wheel can be calculated using the formula T = 2πr/v, where T is the period, r is the radius, and v is the speed. Converting the speed from km/h to cm/s, we have v = 20 km/h * 100,000 cm/km / 60 min/h / 60 s/min = 55.56 cm/s. Plugging in the values, we get T = 2π * 40 cm / 55.56 cm/s = 4.55 seconds. The angular velocity can be calculated using the formula ω = 2π/T, where ω is the angular velocity. Plugging in the value of T, we get ω = 2π / 4.55 s ≈ 1.38 rad/s.
17. The muzzle velocity can be calculated using the formula vf = vi + at, where vf is the final velocity, vi is the initial velocity (assumed to be 0 m/s as the bullet starts from rest), a is the acceleration, and t is the time. Plugging in the values, we get vf = 0 m/s + 6.20810 m/s² * 8.10 s = 50.33 m/s.
18. The centripetal force acting on an object moving in a circular path can be calculated using the formula Fc = m * v² / r, where Fc is the centripetal force, m is the mass, v is the velocity, and r is the radius of the circular path. Plugging in the values, we get Fc = 0.525 kg * (125 in * 0.0254 m/in) / (sin(30°) * 9.8 m/s²) ≈ 13.63 N.
The speed of the ball can be calculated using the formula v = ω * r, where v is the speed, ω is the angular velocity, and r is the radius of the circular path. The radius can be calculated using the formula r = l / sin(θ), where l is the length of the string and θ is the angle. Plugging in the values, we get r = (125 in * 0.0254 m/in) / sin(30°) ≈ 0.203 m. Plugging in the values, we get v = ω * 0.203 m = ω * r. Since we don't have the angular velocity, we cannot determine the speed of the ball.
19. To find the magnitude and direction of the single vector that gives the net effect of the car's trip, we can use the graphical method of vector addition. First, draw a vector representing the 20 km due north. Then, draw a vector representing the 25 km in a direction 60° west of north starting from the tip of the first vector. The vector from the initial point to the tip of the second vector represents the net effect of the car's trip. Its magnitude can be measured using a ruler, and its direction can be determined using a protractor or by measuring the angle using a trigonometric method.
Alternatively, we can use the algebraic method to find the net effect. We can break down the displacement into its x and y components. The displacement due north has a y-component of 20 km and an x-component of 0 km. The displacement in a direction 60° west of north has an x-component of 25 km * cos(60°) and a y-component of 25 km * sin(60°). Adding the x-components and the y-components separately, we get the net x-component and the net y-component. The magnitude of the net effect can be calculated using the Pythagorean theorem, and the direction can be determined using trigonometry.
20. A. To determine the resultant velocity of the boat, we can use the vector addition method. Represent the velocity of the boat heading westward as a vector pointing in the westward direction. Then, represent the velocity of the river flow as a vector pointing in the southward direction. Add these two vectors by placing the tail of the river flow vector at the tip of the boat velocity vector. The vector from the initial point to the tip of this resultant vector represents the magnitude and direction of the resultant velocity of the boat.
B. To determine how far downstream the woman will be when she reaches the opposite shore, we need to calculate the time it takes for her to reach the opposite shore and then multiply it by the velocity of the river flow. The time can be calculated using the formula t = d/v, where t is the time, d is the distance traveled, and v is the velocity. The distance traveled is the width of the river, which can be determined using trigonometry. The velocity is the magnitude of the resultant velocity of the boat. Multiplying the time by the velocity of the river flow gives us the distance downstream.
21. a) The angle between two vectors can be calculated using the dot product formula, where cos(θ) = (A · B) / (|A| |B|), where θ is the angle between the vectors, A · B is the dot product of A and B, and |A| and |B| are the magnitudes of A and B, respectively. Plugging in the values, we get cos(θ) = (3*1 + (-1)*(-2)) / (√(3^2 + (-1)^2) * √(1^2 + (-2)^2)). Solving for θ using a calculator, we get θ ≈ 24.6°.
b) Following the same method, we can calculate θ to be approximately 56.3°.
c) Plugging in the values, we get cos(θ) = (3*4 + (-2)*6) / (√(3^2 + (-2)^2) * √(4^2 + 6^2)). Solving for θ using a calculator, we get θ ≈ 109.5°.
22. The deceleration of a vehicle can be calculated using the formula a = (vf - vi) / t, where a is the deceleration, vf is the final velocity (0 m/s in this case), vi is the initial velocity, and t is the time. Plugging in the values, we get a = (0 m/s - 30 m/s) / 100 m = -0.3 m/s^2.
23. (a) The time it takes to reach a certain speed can be calculated using the formula t = (vf - vi) / a, where t is the time, vf is the final velocity, vi is the initial velocity, and a is the acceleration. Plugging in the values, we get t = (2.0 m/s - 0 m/s) / 1.4 m/s² = 1.43 seconds.
(b) The deceleration can be calculated using the formula a = (vf - vi) / t, where a is the deceleration, vf is the final velocity, vi is the initial velocity, and t is the time. Plugging in the values, we get a = (0 m/s - 2.0 m/s) / 0.8 s = -2.5 m/s².
24. (a) The tangential velocity of an object on the equator can be calculated using the formula v = ω * r, where v is the tangential velocity, ω is the angular velocity, and r is the radius (6400 km or 6,400,000 m in this case). The angular velocity of an object on the equator is equal to the rotation rate of the Earth, which is 2π radians per 24 hours. Converting this to seconds, we have ω = (2π rad) / (24 h * 60 min/h * 60 s/min) ≈ 7.27 x 10^(-5) rad/s. Plugging in the values, we get v = (7.27 x 10^(-5) rad/s) * (6,400,000 m) ≈ 464 m/s.
(b) The angular velocity of an object on the equator remains the same, regardless of the direction of its velocity vector. Therefore, the angular velocity is still approximately 7.27 x 10^(-5) rad/s.
(c) The centripetal acceleration of an object on the equator can be calculated using the formula ac = ω² * r, where ac is the centripetal acceleration, ω is the angular velocity, and r is the radius (6400 km or 6,400,000 m in this case). Plugging in the values, we get ac = (7.27 x 10^(-5) rad/s)² * (6,400,000 m) ≈ 0.0337 m/s².
25. To find a velocity vector that has a magnitude of 5 m/s directed along the direction of vector D = 1.5i + 2j, we need to normalize vector D and then multiply it by the magnitude. The normalized vector can be calculated by dividing vector D by its magnitude, which is √(1.5² + 2²) = √(2.25 + 4) = √6.25 = 2.5. The normalized vector is then (1.5i + 2j) / 2.5. Multiplying the normalized vector by 5 m/s, we get the velocity vector.