physics testphysics exam ALL multiple CHOICE ONLY 20 QUESTION, i have uploaded the test review with the answers to review
Fall 2020 Final Exam Review Answers
Chapter 2
9. In order for the bear to catch the tourist over the distance d, the bear must reach the car at the same time as the tourist. During the time t that it takes for the tourist to reach the car, the bear must travel a total distance of d + 26 m. From Equation 2.1,
(1) and (2)
Equations (1) and (2) can be solved simultaneously to find d.
SOLUTION Solving Equation (1) for t and substituting into Equation (2), we find
Solving for d yields:
29. The average acceleration of the plane can be found by solving Equation 2.9 for a. Taking the direction of motion as positive, we have
The minus sign indicates that the direction of the acceleration is opposite to the direction of motion, and the plane is slowing down.
33.
Solving Equation 2.9 for x shows that the second part of the stopping distance is
Here, the acceleration is assigned a negative value, because we have assumed that the car is traveling in the positive direction, and it is decelerating. Since it is decelerating, its acceleration points opposite to its velocity. The stopping distance, then, is
40. As the plane decelerates through the intersection, it covers a total distance equal to the length of the plane plus the width of the intersection, so x = 59.7 m + 25.0 m = 84.7 m
The speed of the plane as it enters the intersection can be found from Equation 2.9. Solving Equation 2.9 for v0 gives
The time required to traverse the intersection can then be found from Equation 2.4. Solving Equation 2.4 for t gives
Chapter 3.
14. From Equation 3.5b we have that
The motion in the horizontal direction occurs at a constant velocity of and the displacement in the horizontal direction is . Thus, it follows that
Using this value for the time in the expression for y shows that
Since the displacement of the ball is 2.40 m downward, the ball is above the court when it leaves the racket.
27. In order to reach the highest possible fire, the displacement of the hose from the building is x, where, according to Equation 3.5a (with ax = 0 m/s2),
with t equal to the time required for the water the reach its maximum vertical displacement. The time t can be found by considering the vertical motion. From Equation 3.3b,
When the water has reached its maximum vertical displacement, vy = 0 m/s. Taking up and to the right as the positive directions, we find that
And
Therefore, we have
70. With upward taken as the positive direction, the time tH for the greyhound to reach its maximum vertical displacement H is given by Equation 3.3b as
The range of the leap is, therefore,
b. Ignoring air resistance, we know that the time for the projectile to rise from ground level to its maximum height is equal to the time for the projectile to fall back to ground level. Therefore, the greyhound is in the air for .
73. Equation 3.5b relates the time t to the three known variables. The terms in this equation can be rearranged to as to place it in a standard quadratic form:. The solution of this quadratic equation is
The first solution (t = 0.217 s) corresponds to the situation where the ball is moving upward and has a displacement of y = +5.50 m. The second solution represents the later time when the ball is moving downward and its displacement is also y = +5.50 m (see the drawing). This is the solution we seek, so t =.
Chapter 4.
15.
Substituting this result into Newton’s second law we obtain
The answer is negative, indicating that the force is directed away from the cushion.
55. The drawing shows the two forces, T and , that act on the tooth. To obtain the net force, we will add the two forces using the method of components (see Section 1.8).
SOLUTION The table lists the two vectors and their x and y components:
Vector
x component
y component
T
+T cos 16.0°
–T sin 16.0°
T¢
–T¢ cos 16.0°
–T¢ sin 16.0°
T + T¢
+T cos 16.0° – T¢ cos 16.0°
–T sin 16.0° – T¢ sin 16.0°
Since we are given that T = T¢ = 21.0 N, the sum of the x components of the forces is
SFx = +T cos 16.0° – T¢ cos 16.0° = +(21 N) cos 16.0° – (21 N) cos 16.0° = 0 N
The sum of the y components is
SFy = –T sin16.0° – T¢ sin 16.0° = -(21 N) sin 16.0° – (21 N) sin 16.0° = -11.6 N
The magnitude F of the net force exerted on the tooth is
83. The system is shown in the drawing. We will let , and . Then, will move upward, and will move downward. There are two forces that act on each object; they are the tension T in the cord and the weight mg of the object. The forces are shown in the free-body diagrams at the far right.
We will take upward as the positive direction. If the acceleration of is a, then the acceleration of must be –a. From Newton’s second law, we have for that
(1)
and for
(2)
a. Eliminating T between these two equations, we obtain
b. Eliminating a between Equations (1) and (2), we find
91. Newton’s second law for block 1 (10.0 kg) is T = m1a (1)
Block 2 (3.00 kg) has two ropes attached each carrying a tension T. Also, block 2 only travels half the distance that block 1 travels in the same amount of time so its acceleration is only half of block 1’s acceleration. Newton’s second law for block 2 is then (2)
Solving Equation (1) for a, substituting into Equation (2), and rearranging gives
b. Using this result in Equation (1) yields
Chapter 6.
24. The initial speed v0 of the skier can be obtained by applying the work-energy theorem: (Equation 6.3). This theorem indicates that the initial kinetic energy of the skier is related to the skier’s final kinetic energy and the work W done on the skier by the kinetic frictional force according to
Solving for the skier’s initial speed gives (1)
The work done by the kinetic frictional force is given by (Equation 6.1), where fk is the magnitude of the kinetic frictional force and s is the magnitude of the skier’s displacement. Because the kinetic frictional force points opposite to the displacement of the skier, q = 180°. According to Equation 4.8, the kinetic frictional force has a magnitude of , where is the coefficient of kinetic friction and is the magnitude of the normal force. Thus, the work can be expressed as
Substituting this expression for W into Equation (1), we have that (2)
Since the skier is sliding on level ground, the magnitude of the normal force is equal to the weight mg of the skier (see Example 10 in Chapter 4), so FN = mg. Substituting this relation into Equation (2) gives
Since the skier comes to a halt, vf = 0 m/s. Therefore, the initial speed is
26. we know that W = (fk cos θ) s. Substituting this result into the work-energy theorem gives (1)
In Equation (1), fk = 205 N (see the REASONING), the angle θ between the frictional force and the displacement is θ = 180º (see the REASONING), the final speed is vf = 0 m/s (the snowmobile coasts to a halt), and the initial speed is given as v0 = 5.50 m/s. Solving Equation (1) for s gives
b. As explained in the REASONING, we can use (Equation 2.7) to determine the time t during which the snowmobile coasts to a halt. Solving this equation for t gives
39. a. Applying the conservation of mechanical energy in the form of Equation 6.9b, we have
Solving for the final speed gives
b. The calculation and answer are the same as in part (a).
c. The calculation and answer are the same as in part (a).
54. The work-energy theorem states that
(6.8)
Solving for the final speed gives
Chapter 7:
19. The total linear momentum of the system before the lumberjack begins to move is zero, since all parts of the system are at rest. Momentum conservation requires that the total momentum remains zero during the motion of the lumberjack.
Here the subscripts “1” and “2” refer to the first log and lumberjack, respectively. Let the direction of motion of the lumberjack be the positive direction. Then, solving for v1f gives
The minus sign indicates that the first log recoils as the lumberjack jumps off.
b. Now the system is composed of the lumberjack, just before he lands on the second log, and the second log. Gravity acts on the system, but for the short time under consideration while the lumberjack lands, the effects of gravity in changing the linear momentum of the system are negligible. Therefore, to a very good approximation, we can say that the linear momentum of the system is very nearly conserved. In this case, the initial momentum is not zero as it was in part (a); rather the initial momentum of the system is the momentum of the lumberjack just before he lands on the second log. Therefore,
In this expression, the subscripts “1” and “2” now represent the second log and lumberjack, respectively. Since the second log is initially at rest, . Furthermore, since the lumberjack and the second log move with a common velocity, . The statement of momentum conservation then becomes
Solving for vf, we have
The positive sign indicates that the system moves in the same direction as the original direction of the lumberjack’s motion.
28. After Miranda (m2 = 58 kg) jumps onto the inner tube, she and Ashley (m1 = 71 kg) both have the same final velocity: vf = vf1 = vf2. Making this substitution in Equation 7.7b, and solving for their common final velocity, we obtain
Their common velocity after Miranda hops on is, therefore,
The common speed is the magnitude of this value or
61. Applying the momentum-conservation principle separately in terms of the x and y components of the total momentum, we have
Both the x and y components of the total momentum of the three-bullet system are zero after the collision, since the bullets form a stationary lump. The x component of the initial momentum of bullet 1 is positive and the y component is negative, because this bullet is fired to the right and downward in the drawing. The x and y components of the initial momentum of bullet 3 are negative, because this bullet is fired to the left and downward in the drawing. Either of the two equations presented above can be solved for the unknown mass m3. From the equation for the x component of the total momentum, we find that
Chapter 9:
4. The mass of the first child is m1 = 44.0 kg. This child is a distance d1 = 1.30 m from the tree trunk. The mass of the second child, hanging d2 = 2.10 m from the tree trunk, is m2 = 35.0 kg. Both children produce positive (counterclockwise) torques. The net torque exerted on the branch by the two children is then
Substituting the given values, we obtain
28. Although the crate is in translational motion, it undergoes no angular acceleration. Therefore, the net torque acting on the crate must be zero: (Equation 9.2). The four forces acting on the crate appear in the free-body diagram: its weight W, the kinetic friction force fk, the normal force FN, and the tension T in the strap. We will take the edge of the crate sliding along the floor as the rotation axis for applying Equation 9.2. Both the friction force and the normal force act at this point. These two forces, therefore, generate no torque about the axis, so the clockwise torque of the weight W must balance the counterclockwise torque of the tension T in the strap:
(1)
We will apply trigonometry to determine the lever arms and for the weight and the tension, respectively, and then calculate the magnitude T of the tension in the strap.
SOLUTION The lever arm of the crate’s weight is shown in the diagram at the right, and is given by , where d is the distance between the axis of rotation (lower edge of the crate) and the crate’s center of gravity, and q is the angle between that line and the bottom of the crate. The right triangle in the free-body diagram of the crate (drawing on the left) shows how we can use trigonometry to determine both the length d and the angle q. The length d is the hypotenuse of that right triangle, and the other sides are the half-height (H/2 = 0.20 m) and half-length (L/2 = 0.45 m) of the crate, so by the Pythagorean theorem (Equation 1.7) we find that
We can find the angle q from the inverse tangent function:
The lever arm of the tension force is illustrated in the drawing at the right, where we see that . Therefore, from Equation (1), the magnitude of the tension in the strap is
35. Using Equation 9.2, the net torque that acts on each wheel is given by , where I is the moment of inertia and a is the angular acceleration. Solving Equation 8.7 for the angular acceleration and substituting the result into Equation 9.2 gives
Table 9.1 indicates that the moment of inertia of a hoop is , while the moment of inertia of a disk is . The net external torques acting on the hoop and the disk are:
54. When applying energy conservation, we will assume that the zero level for measuring the height h is located at the bottom of the ramp. As applied to the marble, energy conservation gives
(1)
In Equation (1) we have used the fact that the marble starts at rest at the top of the ramp. Since the marble rolls without slipping, we know that (Equation 8.12), where r is the radius of the marble. Referring to Table 9.1, we also know that the marble’s moment of inertia is . Substituting these two expressions into Equation (1) gives
As applied to the cube, energy conservation gives
where we have used the fact that the cube starts at rest at the top of the ramp and does not rotate. The desired ratio of the center-of-mass speeds is
61. The person and the carousel have the same initial angular velocity w0, and the same final angular velocity wf. Therefore, the conservation of angular momentum principle can be expressed as
(1)
where IC is the moment of inertia of the carousel (without the person). Substituting and (Equation 9.6) for the person’s initial and final moments of inertia into Equation (1), we obtain
Solving for the mass m of the person yields
Chapter 10:
14. In phase 1 of the block’s motion (uniform acceleration) we find that the net force on the block is F1 – fk = ma where the force of friction is fk = µkmg. Therefore F1 = m(a + µkg), which is just the force exerted by the spring on the block, i.e., F1 = kx1. So we have
kx1 = m(a + µkg) (1)
We can find the acceleration using
In phase 2 of the block’s motion (constant speed) a = 0 m/s2, so the force exerted by the spring is
(2)
so that
Using this expression for mk in Equation (1) we obtain
a. Solving for k gives
b. Substituting this value for k into Equation (2) , we have
29. The expression for the total mechanical energy for an object on a spring is given by Equation 10.14, so that we have
The block does not rotate, so the angular speeds wf and w0 are zero. Since the block comes to a momentary halt on the spring and is dropped from rest, the translational speeds vf and v0 are also zero. Because the spring is initially unstrained, the initial displacement y0 of the spring is likewise zero. Thus, the above expression can be simplified as follows:
The block was dropped at a height of h0 – hf above the compressed spring. Solving the simplified energy-conservation expression for this quantity gives
42. The bullet (mass m) moves with speed v, strikes the block (mass M) in an inelastic collision and the two move together with a final speed V. We first need to employ the conservation of linear momentum to the collision to obtain an expression for the final speed:
The block/bullet system now compresses the spring by an amount x. During the compression the total mechanical energy is conserved so that
Substituting the expression for V into this equation, we obtain
Solving this expression for v gives
82. upward-directed restoring force from the spring. The magnitude of the weight is mg, and the magnitude of the restoring force is given by Equation 10.2 without the minus sign as kx. Thus, we have
b. The conservation of mechanical energy states that the final total mechanical energy Ef is equal to the initial total mechanical energy E0. The expression for the total mechanical energy for an object on a spring is given by Equation 10.14, so that we have
The block does not rotate, so the angular speeds wf and w0 are zero. Since the block comes to a momentary halt on the spring and is released from rest, the translational speeds vf and v0 are also zero. Because the spring is initially unstrained, the initial displacement x0 of the spring is likewise zero. Thus, the above expression can be simplified as follows:
The term h0 – hf is the amount by which the spring has compressed, or h0 – hf = xf. Making this substitution into the simplified energy-conservation equation gives
Solving for xf, we find
Chapter 11.
28. If the pressure is P1 at a certain point in a static fluid (density = ρ), the pressure P2 in the fluid at a distance h beneath this point is (Equation 11.4). As discussed in Section 11.4, this expression becomes when applied to the mercury barometer shown in Figure 11.11 of the text. Using Equation 11.4 with and , we have
In this result, hair is the height of a column of air that equals the height of the building. We can also express using the mercury barometer, which indicates that corresponds to a column of mercury that has a height of . Thus, we have
Equating the two expressions for , we obtain
Taking the density of mercury from Table 11.1 in the text and solving for hair gives
29. P = Patm + rwghw + rmghm
We want P = 2Patm, and we know h = hw + hm = 1.00 m. Using the above and rearranging gives
48. Solving Equation 4.2b for the acceleration ay gives
(1)
The weight W of an object is equal to its mass m times the magnitude g of the acceleration due to gravity, or W = mg (Equation 4.5). The mass, in turn, is equal to the product of an object’s density r and its volume V, so m = r V (Equation 11.1). Combining these two relations, the weight can be expressed as W = r V g.
According to Archimedes’ principle, the magnitude FB of the buoyant force is equal to the weight of the cool air that the balloon displaces, so FB = mcool airg = (rcool airV)g. Since we are neglecting the weight of the balloon fabric and the basket, the weight of the balloon is just that of the hot air inside the balloon. Thus, m = mhot air = rhot airV andW = mhot airg = (rhot airV)g .
Substituting the expressions FB = (rcool airV)g, m = rhot airV, and W = (rhot airV)g into Equation (1) gives
67. The pressure P, the fluid speed v, and the elevation y at any two points in an ideal fluid of density r are related by Bernoulli’s equation: (Equation 11.11), where 1 and 2 denote, respectively, the first and second floors. With the given data and a density of r = 1.00 ´ 103 kg/m3 for water (see Table 11.1), we can solve Bernoulli’s equation for the desired pressure P2.
Solving Bernoulli’s equation for P2 and taking the elevation at the first floor to be , we have
Chapter 12:
43. Since there is no heat lost or gained by the system, the heat lost by the water in cooling down must be equal to the heat gained by the thermometer in warming up. The heat Q lost or gained by a substance is given by Equation 12.4 as Q = cmDT, where c is the specific heat capacity, m is the mass, and DT is the change in temperature. Thus, we have that
We can use this equation to find the temperature of the water before the insertion of the thermometer.
SOLUTION Solving the equation above for , and using the value of from Table 12.2, we have
The temperature of the water before the insertion of the thermometer was
49. Let the system be comprised only of the metal forging and the oil. Then, according to the principle of energy conservation, the heat lost by the forging equals the heat gained by the oil, or . According to Equation 12.4, the heat lost by the forging is , where is the final temperature of the system at thermal equilibrium. Similarly, the heat gained by the oil is given by .
Solving for , we have
or
61. According to Equation 12.4, the heat lost by the mercury is . The heat required to vaporize the water is, from Equation 12.5, . Thus, the total amount of heat gained by the water is .
where and . The specific heats of mercury and water are given in Table 12.2, and the latent heat of vaporization of water is given in Table 12.3. Solving for the mass of the water that vaporizes gives
Chapter 14.
24. a. The number n of moles of chlorine originally in the tank is equal to the mass of chlorine gas (11.0 g) divided by the mass per mole of chlorine gas (70.9 g/mol), so n = (11.0 g)/(70.9 g/mol). The temperature of the gas is 82 °C, which, when converted to the Kelvin temperature scale, is (82 + 273 )K. The volume of the tank is
(14.1)
b. The mass mCl of chlorine gas that has leaked out of the tank is mCl = 11.0 g – mR, where mR is the mass of the gas remaining in the tank. The remaining mass, in turn, is equal to the number nR of moles of gas remaining in the tank times the mass per mole (70.9 g/mol) of chlorine gas. Thus,
The number of remaining moles of gas can be found from the ideal gas law, , so that
27. If the pressure at the surface is P1 and the pressure at a depth h is P2, we have that P2 = P1 + rgh. We also know that P1V1 = P2V2. Then, Therefore,
Equation 6.9b gives the principle of conservation of mechanical energy:
In this expression, we know that and that (since the atom comes to a halt at the top of its trajectory). Furthermore, we can take the height at the earth’s surface to be h0 = 0 m. Taking this information into account, we can write the energy-conservation equation as follows:
Using M to denote the molecular mass (in kilograms per mole) and recognizing that , where NA is Avogadro’s number and is the number of xenon atoms per mole, we have
Recognizing that kNA = R and that M = 131.29 g/mol = 131.29 × 10-3 kg/mol, we find
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