chaney (glc568) Work, Energy, Impulse, and Momentum murthy (21118) This print-out should have 49 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. 001 (part 1 of 2) 10.0 points Consider
chaney (glc568) Impulse, Momentum, Angular Motion murthy (21118) This print-out should have 49 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. 001 10.0 points Consider the elastic head
Homework #8 Solutions Chapter 28
28.37. Model: Energy is conserved. The charged spheres are point charges.
Visualize: Label the spheres 1, 2, 3, and 4 in a clockwise manner, with the sphere in the upper left-hand corner
being sphere 1.
Solve: The total po
Homework #11 Solutions Chapter 32
32.11. Model: The magnetic field is that of a current loop.
Solve: (a) From Equation 32.7, the magnetic field strength at the center of a loop is
2RBloop center 2(0.50 102 m)(2.5 103 T)
Bloop center = 0 I =
= 20 A
Homework #13 Solutions Chapter 33
To develop a motional emf the magnetic field needs to be perpendicular to both the velocity and the current, so
lets say its direction is into the page.
Solve: This is a straightforward use of Equation 33
Homework #12 Solutions Chapter 32
32.13. Model: Assume the wires are infinitely long.
The field vectors are tangent to circles around the currents. The net magnetic field is the vectorial sum of the
fields Btop and Bbottom . Points a and c are
Homework #9 Solutions Chapter 31
31.3. Model: Assume that the connecting wires are ideal.
Visualize: Please refer to Figure EX31.3.
Solve: The current in the 2 resistor is I1 = 4 V/2 = 2 A to the right. The current in the 5 resistor is
I 2 = (15 V)/ (5 )
Homework #14 Solutions Chapter 33
33.14. Model: Assume the field is uniform across the loop.
Visualize: Please refer to Figure EX33.14. There is a current in the loop so there must be an emf that is due to a
changing flux. With the loop fixed the area is
Homework #3 Solutions Chapter 25
25.1. Model: Use the charge model.
Solve: (a) In the process of charging by rubbing, electrons are removed from one material and transferred into the
other because they are relatively free to move. Protons, on the other ha
Homework #5 Solutions Chapter 25
25.23. Model: The electric field is that of a negative charge on the plastic bead. Model the bead as a point
charge (which is exact for r > rbead ).
Solve: The electric field is
8.0 109 C
E = K 2 r = (9.0 109 N m 2 /C2
Homework #10 Solutions Chapter 31
31.23. Model: The connecting wires are ideal with zero resistance.
For the first step, the two resistors in the middle of the circuit are in parallel, so their equivalent resistance is
Req 1 100 100
Homework #2 Solutions Chapter 21
21.7. Model: Reflections at both ends of the string cause the formation of a standing wave.
Solve: Figure EX21.7 indicates 5/2 wavelengths on the 2.0-m-long string. Thus, the wavelength of the standing
wave is = 2 (2.0 m)
Homework #4 Solutions Chapter 25
25.33. Model: The charges are point charges.
Visualize: Please refer to Figure P25.33.
Solve: The electric force on charge q1 is the vector sum of the forces F2 on 1 and F3 on 1, where q1 is the 1.0
nC charge, q2 is th
Homework #6 Solutions Chapter 26
26.3. Model: The electric field is that due to superposition of the fields of the two 3.0 nC charges located on the yaxis.
Visualize: We denote the top 3.0 nC charge by q1 and the bottom 3.0 nC charge by q2 . The electric
Homework #7 Solutions Chapter 28
28.7. Model: The charges are point charges.
Solve: For a system of point charges, the potential energy is the sum of the potential energies due to all distinct
pairs of charges:
Kqi q j
U elec =
= U12 + U13 + U 23
i, j ri
Homework Solutions Chapter 3
3.5. Visualize: The position vector r whose magnitude r is 10 m has an x-component of 8 m. It makes an
angle with the +x-axis in the first quadrant.
Solve: Using trigonometry, rx = r cos , or 8 m = (10 m)cos . This gives = 36.
chaney (glc568) Torque, Angular Momentum, and Rotational Equilibrium murthy (21118) 1 This print-out should have 51 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. 001 (part 1 of 2) 10
chaney (glc568) Rotations and Gravity murthy (21118) This print-out should have 47 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. 001 (part 1 of 3) 10.0 points A student builds and ca
chaney (glc568) Gravitational PE, Torque and Angular Momentum murthy (21118) This print-out should have 37 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. 001 10.0 points Energy is req
Homework 3 Solutions
2.15. Model: We are using the particle model for the skater and the kinematics model of motion under
Solve: Since we dont know the time of acceleration we will use
vf2 = vi2 + 2a(xf xi )
Homework 2 Solutions
2.23. Solve: x = (2t 2 t + 1) m
(a) The position t = 2 s is x2 s = [2(2)2 2 + 1] m = 7 m.
(b) The velocity is the derivative v = dx/dt and the velocity at t = 2 s is calculated as follows:
v = (4t 1) m/s v2 s = [4(2) 1] m/s
Homework 4 Solutions
2.43. Model: The car is a particle and constant-acceleration kinematic equations hold.
Solve: (a) This is a two-part problem. During the reaction time,
x1 = x0 + v0 (t1 t0 ) + 1/2a0 (t1 t0 )2
= 0 m + (20 m/s)(0.50
Homework Solutions Chapter 1
Solve: The player starts with an initial velocity but as he slides he moves slower and slower until coming to
Model: Represent the (child + sled) system as a particle.
Visualize: The dots in the figure are equ
Homework #9 Solutions Chapter 6
6.11. Model: We assume that the box is a point particle that is acted on only by the tension in the rope and the
pull of gravity. Both the forces act along the same vertical line.
Solve: (a) Since the box is at r
Homework #10 Solutions Chapter 7
7.5. Visualize: Please refer to Figure EX7.5.
Solve: (a) Gravity acts on both blocks. Block A is in contact with the floor and experiences a normal force and
friction. The string tension is the same on both blocks since th
Homework 8 Solutions Chapter 6
6.3. Model: We assume the speaker is a particle in static equilibrium under the influence of three forces: gravity
and the tensions in the two cables.
Solve: From the lengths of the cables and the distance below t
Homework 7 Solutions Chapter 5
5.9. Visualize: Please refer to Figure EX5.9.
Solve: Newtons second law is F = ma. Applying this to curves 1 at the point F = 3 rubber bands and to curve 2
at the point F = 5 rubber bands gives
3F = m1 (5a1 ) ! 3 5m1
Homework Solutions Chapter 4
4.7. Model: Model the rocket as a particle and assume constant acceleration in both directions (vertical and
horizontal) so use the kinematic equations in direction.
a (t)2 in each direction. t is the same in each
Fluids: Pressure, Buoyancy, Archimedes, Bernoulli | PHYS 1500
October 29, 2015
1 = 0 + ; pressure at depth h below
Part I: Pressure
1. What is the pressure in water (density 1000 kg per cubic meter) at depth 2 m below the surf