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. Very 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 ball.
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, due 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 can be calculated using the formula:
T = 2πr/v
where T is the period, r is the radius, and v is the linear velocity.
Converting the speed from km/h to cm/s:
20 km/h = 20,000 cm/h = 20,000/60 cm/s = 333.33 cm/s
Plugging in the values:
T = 2π(40 cm)/(333.33 cm/s)
T ≈ 2π(40 cm)/(333.33 cm/s)
T ≈ 0.239 seconds
The angular velocity can be calculated using the formula:
ω = v/r
Plugging in the values:
ω = 333.33 cm/s / 40 cm
ω ≈ 8.333 rad/s
17. The final velocity can be calculated using the formula:
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:
vf = 0 + (6.20810 m/s²)(8.10x10 s)
vf = 50.322 m/s
18. The centripetal force can be calculated using the formula:
Fc = m * ac
where Fc is the centripetal force, m is the mass, and ac is the centripetal acceleration.
The centripetal acceleration can be calculated using the formula:
ac = (v^2) / r
where v is the speed and r is the radius.
Plugging in the values:
ac = (100 m/s)^2 / 125 m
ac = 80 m/s²
The centripetal force can be calculated as:
Fc = (0.525 kg)(80 m/s²)
Fc = 42 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.
Plugging in the values:
v = ω * 125 m
v ≈ (8.333 rad/s)(125 m)
v ≈ 1041.62 m/s
19. The net effect of the car's trip can be found by adding the two displacement vectors together.
Using graphical methods, we can draw the two vectors on a coordinate plane. The first vector points directly north for 20 km. The second vector points in a direction 60° west of north for 25 km.
By adding these vectors head to tail, we can draw the resultant vector. The magnitude and direction of the resultant vector can be determined using the Pythagorean theorem and trigonometry.
Using algebraic methods, we can represent the first vector as (0, 20 km) and the second vector as (-25 cos(60°), -25 sin(60°)) km. We can then add the x-components and the y-components separately to find the resultant vector. The magnitude and direction of the resultant vector can again be determined using the Pythagorean theorem and trigonometry.
20. A. The resultant velocity of the boat can be found by adding the velocities of the boat and the river.
Using vector addition, we can add the velocity of the boat (100 cm/s) and the velocity of the river (-50 cm/s).
Resultant velocity = 100 cm/s + (-50 cm/s) = 50 cm/s
Therefore, the magnitude of the resultant velocity is 50 cm/s and the direction is westward.
B. To find how far downstream the woman will be when she reaches the opposite shore, we can calculate the displacement caused by the river during the crossing.
The time it takes for the woman to cross the river can be found using the equation:
t = d/v
where t is the time, d is the distance, and v is the velocity.
The time to cross the river is:
t = 100 cm / 100 cm/s
t = 1 second
During this time, the river will have carried the woman downstream by:
d = v * t
d = -50 cm/s * 1 s
d = -50 cm
Therefore, the woman will be 50 cm downstream relative to the straight-across direction when she reaches the opposite shore.
21. The angle between two vectors A and B can be found using the dot product:
Let vector A = 3i - j and vector B = i - 2j
a) Angle between A and B:
A · B = |A| |B| cosθ
(3i - j) · (i - 2j) = √((3)^2 + (-1)^2) √((1)^2 + (-2)^2) cosθ
(3*i*i + 3*(-1)(-2) + (-1)*i + (-1)*2j) · (i + (-2j)) = (3)(1)(1) + (-1)(-2)(-2) + (-1)(1) + (-1)(-2) cosθ
(3i^2 + 6 + (-i) + (-2j)) · (i + (-2j)) = 3 + 4 - 1 + 2 cosθ
(3(-1) + 6 + (-i) + (-2j)) · (i + (-2j)) = 6 + 2 cosθ
(-3 + 6 + (-i) + (-2j)) · (i + (-2j)) = 6 + 2 cosθ
(3i -2j) · (i - 2j) = 6 + 2 cosθ
(3*i*i + (-2)(-2) -2i - (-2)(2)j) · (i - 2j) = 6 + 2 cosθ
(3(i^2) + 4 + (-2)i - 4j) · (i - 2j) = 6 + 2 cosθ
(3(-1) + 4 + (-2)i - 4j) · (i - 2j) = 6 + 2 cosθ
(-3 + 4 + (-2)i - 4j) · (i - 2j) = 6 + 2 cosθ
(1 + (-2)i - 4j) · (i - 2j) = 6 + 2 cosθ
(i + (-2)j) · (i - 2j) = 6 + 2 cosθ
(1i + (-2)j) · (1i - 2j) = 6 + 2 cosθ
1(1) + (-2)(-2) = 6 + 2 cosθ
5 = 6 + 2 cosθ
-1 = 2 cosθ
cosθ = -1/2
θ = arccos(-1/2)
θ ≈ 120°
So, the angle between vector A and vector B is approximately 120°.
b) Angle between A and B:
Let vector A = 3i - 2j and vector B = i - 2j
A · B = |A| |B| cosθ
(3i - 2j) · (i - 2j) = √((3)^2 + (-2)^2) √((1)^2 + (-2)^2) cosθ
(3*i*i + 3*(-2)(2) + (-2)*i + (-2)*(-2)j) · (i + (-2j)) = (3)(1)(1) + (-2)(-2)(-2) + (-2)(1) + (-2)(-2) cosθ
(3(i^2) + 12 + (-2)i + 4j) · (i + (-2j)) = 3 + 8 - 2 - 8 cosθ
(3(-1) + 12 + (-2)i + 4j) · (i + (-2j)) = 3 + 8 - 2 - 8 cosθ
(-3 + 12 + (-2)i + 4j) · (i + (-2j)) = 3 + 8 - 2 - 8 cosθ
(9 + (-2)i + 4j) · (i + (-2j)) = 3 + 8 - 2 - 8 cosθ
(9i + (-2)j) · (i - 2j) = 3 + 8 - 2 - 8 cosθ
(9i + (-2)j) · (i - 2j) = 7 - 8 cosθ
(1i + (-2)j) · (1i - 2j) = 7 - 8 cosθ
1(1) + (-2)(-2) = 7 - 8 cosθ
5 = 7 - 8 cosθ
-2 = -8 cosθ
cosθ = -2/-8
cosθ = 1/4
θ = arccos(1/4)
θ ≈ 75.5°
So, the angle between vector A and vector B is approximately 75.5°.
c) Angle between A and B:
Let vector A = 3i - 2j and vector B = 4i + 6j
A · B = |A| |B| cosθ
(3i - 2j) · (4i + 6j) = √((3)^2 + (-2)^2) √((4)^2 + (6)^2) cosθ
(3*i*i + 3*(-2)(6) + (-2)*i + (-2)*6j) · (4i + 6j) = (3)(1)(4) + (-2)(6)(6) + (-2)(4) + (-2)(6) cosθ
(3(i^2) - 36 + (-2)i + (-12)j) · (4i + 6j) = 12 - 72 - 8 - 12 cosθ
(9 - 36 + (-2)i + (-12)j) · (4i + 6j) = 12 - 72 - 8 - 12 cosθ
(-27 - 2i - 12j) · (4i + 6j) = 12 - 72 - 8 - 12 cosθ
(-2i - 12j) · (4i + 6j) = -68 - 12 cosθ
(1i + (-6)j) · (4i + 6j) = -68 - 12 cosθ
(4 - 36) = -68 - 12 cosθ
-32 = -68 - 12 cosθ
12 cosθ = 68 - 32
12 cosθ = 36
cosθ = 36/12
cosθ = 3
Since the magnitude of cosine cannot exceed 1, there is no solution for θ in the range of real numbers for this case.
22. The deceleration of the vehicle can be calculated using the equation:
a = (v_f - v_i) / t
where a is the deceleration, v_f is the final velocity, v_i is the initial velocity, and t is the time.
Plugging in the values:
a = (0 m/s - 30 m/s) / 100 m
a = -30 m/s / 100 m
a = -0.3 m/s²
Therefore, the deceleration of the vehicle is -0.3 m/s².
23. (a) The time it takes to reach a speed of 2.0 m/s can be calculated using the equation:
t = (v_f - v_i) / a
where t is the time, v_f is the final velocity, v_i is the initial velocity, and a is the acceleration.
Plugging in the values:
t = (2.0 m/s - 0 m/s) / 1.4 m/s²
t = 2.0 s / 1.4
t ≈ 1.43 seconds
(b) The deceleration can be calculated using the equation:
a = (v_f - v_i) / t
where a is the deceleration, v_f is the final velocity, v_i is the initial velocity, and t is the time.
Plugging in the values:
a = (0 m/s - 2.0 m/s) / 0.8 s
a = -2.0 m/s / 0.8 s
a = -2.5 m/s²
Therefore, the deceleration is -2.5 m/s².
24. (a) The tangential velocity of a body on the Earth's 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.
Converting the radius of the Earth from km to m:
6,400 km = 6,400,000 m
Plugging in the values:
v = ω * 6,400,000 m
v ≈ (2π / 24 hours) * 6,400,000 m
v ≈ (2π / (24 * 60 * 60) s) * 6,400,000 m
v ≈ (2π / (86,400 s)) * 6,400,000 m
v ≈ 464.47 m/s
Therefore, the tangential velocity of a body on the equator is approximately 464.47 m/s.
(b) The angular velocity can be calculated using the formula:
ω = v / r
where ω is the angular velocity, v is the tangential velocity, and r is the radius.
Plugging in the values:
ω = 464.47 m/s / 6,400,000 m
ω ≈ 0.000072 degrees per second
Therefore, the angular velocity is approximately 0.000072 degrees per second.
(c) The centripetal acceleration can be calculated using the formula:
ac = (v^2) / r
where ac is the centripetal acceleration, v is the tangential velocity, and r is the radius.
Plugging in the values:
ac = (464.47 m/s)^2 / 6,400,000 m
ac ≈ 3.37 x 10^-3 m/s²
Therefore, the centripetal acceleration, due to the Earth's rotation, is approximately 3.37 x 10^-3 m/s².
25. The speed of the velocity vector can be determined using the magnitude of the vector:
|V| = √(Dx^2 + Dy^2)
where |V| is the magnitude of the velocity vector and Dx and Dy are the components of the vector.
Plugging in the values:
|V| = √((-1.5)^2 + (2)^2)
|V| = √(2.25 + 4)
|V| = √(6.25)
|V| ≈ 2.5 m/s
Therefore, the magnitude of the velocity vector is approximately 2.5 m/s.