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 i-2j (c) 3i-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 formula for the period of revolution of an object moving in a circular path is given by T = 2πr/v, where T is the period of revolution, r is the radius, and v is the speed. Plugging in the values, we have T = 2π(40 cm)/(20 km/h).
Converting the units, 1 km = 1000 m and 1 h = 3600 s, we have T = 2π(0.4 m)/(20 m/s) = 0.04π s.
Therefore, the period of revolution is approximately 0.1257 s.
To calculate the angular velocity, we can use the formula ω = v/r, where ω is the angular velocity, v is the linear velocity, and r is the radius. Plugging in the values, we have ω = (20 km/h)/(0.4 m) = 50 rad/s.
Therefore, the angular velocity is 50 rad/s.
17. The formula for calculating final velocity is given by vf = vi + at, where vf is the final velocity, vi is the initial velocity, a is the acceleration, and t is the time. Plugging in the values, we have vf = 0 + (6.20810 m/s²)(8.10x10 s) = 50.252 m/s.
Therefore, the muzzle velocity of the bullet is 50.252 m/s.
18. The centripetal force acting on an object moving in a circular path is given by F = mv²/r, where F is the centripetal force, m is the mass, v is the speed, and r is the radius. Plugging in the values, we have F = (0.525 kg)(v²)/(125 in) = 0.0042v² N.
To find the speed of the ball, we can use the formula v = ωr, where v is the speed, ω is the angular velocity, and r is the radius. Plugging in the values, we have v = (ω)(125 in) = (2π/T)(125 in).
Therefore, the centripetal force acting on the ball is 0.0042v² N and the speed of the ball is (2πr)/T.
19. The net effect of the car's trip can be found by adding the two displacement vectors together. Using the graphical method of vector addition, we can draw the two vectors to scale and then draw a line connecting the initial point of the first vector to the final point of the second vector. The magnitude and direction of this line will give the magnitude and direction of the net effect vector.
Using the algebraic method of vector addition, we can break the two displacement vectors into their x and y components and then add the x components together and the y components together to find the x and y components of the net effect vector. The magnitude and direction of the net effect vector can then be found using the Pythagorean theorem and trigonometric functions.
20.
A. To find the resultant velocity of the boat, we can use the equation v_resultant = √(v_boat^2 + v_river^2), where v_resultant is the resultant velocity, v_boat is the velocity of the boat, and v_river is the velocity of the river. Plugging in the values, we have v_resultant = √((100 cm/s)^2 + (50 cm/s)^2) = √(10000 cm^2/s^2 + 2500 cm^2/s^2) = √12500 cm^2/s^2 = 111.8 cm/s. Therefore, the magnitude of the resultant velocity of the boat is approximately 111.8 cm/s.
The direction of the resultant velocity can be found using trigonometric functions. The angle θ can be found using the equation θ = arctan(v_river/v_boat), where θ is the direction of the resultant velocity, v_river is the velocity of the river, and v_boat is the velocity of the boat. Plugging in the values, we have θ = arctan((50 cm/s)/(100 cm/s)) = arctan(0.5) = 26.6°.
Therefore, the direction of the resultant velocity of the boat is approximately 26.6° West of North.
B. To find how far downstream the woman will be when she reaches the opposite shore, we can use the equation d = v_river * t, where d is the distance downstream, v_river is the velocity of the river, and t is the time taken to cross the river. Plugging in the values, we have d = (50 cm/s) * (100 cm/100 cm/s) = 50 cm.
Therefore, the woman will be 50 cm downstream relative to the straight-across direction when she reaches the opposite shore.
21.
a) The angle between vectors A = 3i - j and B = i - 2j can be found using the dot product formula A · B = |A||B|cos(θ), where A · B is the dot product of A and B, |A| and |B| are the magnitudes of A and B, and θ is the angle between A and B. Plugging in the values, we have (3i - j) · (i - 2j) = √(3^2 + (-1)^2)√(1^2 + (-2)^2)cos(θ). Simplifying, we have (3i - j) · (i - 2j) = 3(1) + (-1)(-2) = 5 = √10√5cos(θ). Solving for θ, we have cos(θ) = 5/(√10√5) = 1/√2 = 1/√2. Taking the inverse cosine of both sides, we have θ ≈ 45°.
Therefore, the angle between vectors A = 3i - j and B = i - 2j is approximately 45°.
b) The angle between vectors A = 3i - 2j and B = i - 2j can be found using the same dot product formula. Plugging in the values, we have (3i - 2j) · (i - 2j) = √(3^2 + (-2)^2)√(1^2 + (-2)^2)cos(θ). Simplifying, we have (3i - 2j) · (i - 2j) = 3(1) + (-2)(-2) = 7 = √13√5cos(θ). Solving for θ, we have cos(θ) = 7/(√13√5) = 7/√65. Taking the inverse cosine of both sides, we have θ ≈ 21.8°.
Therefore, the angle between vectors A = 3i - 2j and B = i - 2j is approximately 21.8°.
c) The angle between vectors A = 3i - 2j and B = 4i + 6j can be found using the same dot product formula. Plugging in the values, we have (3i - 2j) · (4i + 6j) = √(3^2 + (-2)^2)√(4^2 + 6^2)cos(θ). Simplifying, we have (3i - 2j) · (4i + 6j) = 3(4) + (-2)(6) = 0 = √13√52cos(θ). Since the dot product is 0, we know that the vectors are perpendicular and the angle between them is 90°.
Therefore, the angle between vectors A = 3i - 2j and B = 4i + 6j is 90°.
22. The deceleration of the vehicle can be calculated using the formula a = Δv/Δt, where a is the deceleration, Δv is the change in velocity, and Δt is the time taken for the change in velocity. Plugging in the values, we have a = (0 m/s - 30 m/s)/(100 m) = -0.3 m/s².
Therefore, the deceleration of the vehicle is -0.3 m/s².
23.
a) The time taken to reach a speed of 2.0 m/s can be calculated using the formula t = Δv/a, where t is the time, Δv is the change in velocity, and a is the acceleration. Plugging in the values, we have t = (2.0 m/s - 0 m/s)/(1.4 m/s²) = 1.43 s.
Therefore, it takes 1.43 seconds for the car driver to reach a speed of 2.0 m/s.
b) The deceleration can be calculated using the formula a = Δv/Δt, where a is the deceleration, Δv is the change in velocity, and Δt is the time taken for the change in velocity. Plugging in the values, we have a = (0 m/s - 2.0 m/s)/(0.8 s) = -2.5 m/s².
Therefore, the deceleration of the car driver is -2.5 m/s².
24.
a) The tangential velocity of a body on the equator can be calculated using the formula v = ωr, where v is the tangential velocity, ω is the angular velocity, and r is the radius. Plugging in the values, we have v = (2π rad/day)(6400000 m) = 40000π m/day.
To convert to m/s, we divide by 24 hours in a day and 3600 seconds in an hour, so v = (40000π m/day)/(24 hours/day)(3600 s/hour) ≈ 183.95 m/s.
Therefore, the tangential velocity of a body on the equator is approximately 183.95 m/s.
b) The angular velocity of a body on the equator is equal to the rotational speed of the Earth, which is equal to 2π radians per day. Therefore, the angular velocity is 2π rad/day.
c) The centripetal acceleration of a body on the equator can be calculated using the formula a = ω²r, where a is the centripetal acceleration, ω is the angular velocity, and r is the radius. Plugging in the values, we have a = (2π rad/day)²(6400000 m) = 4π²(6400000 m)/(day²).
To convert to m/s², we divide by (24 hours/day)² and (3600 seconds/hour)², so a = (4π²(6400000 m)/(day²))/((24 hours/day)²)(3600 s/hour)² ≈ 0.0337 m/s².
Therefore, the centripetal acceleration of a body on the equator is approximately 0.0337 m/s².
25. To determine a velocity vector with a magnitude of 5 m/s directed along the direction of vector D = -1.5i + 2j, we can normalize vector D by dividing it by its magnitude and then multiply the resulting unit vector by the desired magnitude.
The magnitude of vector D is |D| = √((-1.5)^2 + 2^2) = √(2.25 + 4) = √6.25 = 2.5.
Therefore, the normalized unit vector in the direction of vector D is D_normalized = D/|D| = (-1.5i + 2j)/2.5.
To find the desired velocity vector, we can multiply the normalized unit vector by the desired magnitude, which is 5 m/s. Therefore, the velocity vector is 5(D/|D|) = 5((-1.5i + 2j)/2.5).