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Physics Problems with Solutions

Forces in physics, tutorials and problems with solutions.

Free tutorials on forces with questions and problems with detailed solutions and examples. The concepts of forces, friction forces, action and reaction forces, free body diagrams, tension of string, inclined planes, etc. are discussed and through examples, questions with solutions and clear and self explanatory diagrams. Questions to practice for the SAT Physics test on forces are also included with their detailed solutions. The discussions of applications of forces engineering system are also included.

Forces: Tutorials with Examples and Detailed Solutions

• Forces in Physics .
• Components of a Force in a System of Coordinates .
• Free Body Diagrams, Tutorials with Examples and Explanations .
• Forces of Friction .
• Addition of Forces .
• What is the Tension of a String or rope? .
• Newton's Laws in Physics .
• Hooke's Law, Examples with solutions .

Problems on Forces with Detailed Solutions

• Inclined Planes Problems in Physics with Solutions .
• Tension, String, Forces Problems with Solutions .

SAT Questions on Forces with Solutions

• Physics Practice Questions with Solutions on Forces and Newton's Laws .

Formulas and Constants

• Physics Formulas Reference
• SI Prefixes Used with Units in Physics, Chemistry and Engineering
• Constants in Physics, Chemistry and Engineering

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Check Your Understanding

Answer: d = 1720 m

Answer: a = 8.10 m/s/s

Answers: d = 33.1 m and v f = 25.5 m/s

Answers: a = 11.2 m/s/s and d = 79.8 m

Answer: t = 1.29 s

Answers: a = 243 m/s/s

Answer: a = 0.712 m/s/s

Answer: d = 704 m

Answer: d = 28.6 m

Answer: v i = 7.17 m/s

Answer: v i = 5.03 m/s and hang time = 1.03 s (except for in sports commericals)

Answer: a = 1.62*10 5 m/s/s

Answer: d = 48.0 m

Answer: t = 8.69 s

Answer: a = -1.08*10^6 m/s/s

Answer: d = -57.0 m (57.0 meters deep) 

Answer: v i = 47.6 m/s

Answer: a = 2.86 m/s/s and t = 30. 8 s

Answer: a = 15.8 m/s/s

Answer: v i = 94.4 mi/hr

Solutions to Above Problems

d = (0 m/s)*(32.8 s)+ 0.5*(3.20 m/s 2 )*(32.8 s) 2

Return to Problem 1

110 m = (0 m/s)*(5.21 s)+ 0.5*(a)*(5.21 s) 2

110 m = (13.57 s 2 )*a

a = (110 m)/(13.57 s 2 )

a = 8.10 m/ s 2

Return to Problem 2

d = (0 m/s)*(2.60 s)+ 0.5*(-9.8 m/s 2 )*(2.60 s) 2

d = -33.1 m (- indicates direction)

v f = v i + a*t

v f = 0 + (-9.8 m/s 2 )*(2.60 s)

v f = -25.5 m/s (- indicates direction)

Return to Problem 3

a = (46.1 m/s - 18.5 m/s)/(2.47 s)

a = 11.2 m/s 2

d = v i *t + 0.5*a*t 2

d = (18.5 m/s)*(2.47 s)+ 0.5*(11.2 m/s 2 )*(2.47 s) 2

d = 45.7 m + 34.1 m

(Note: the d can also be calculated using the equation v f 2 = v i 2 + 2*a*d)

Return to Problem 4

-1.40 m = (0 m/s)*(t)+ 0.5*(-1.67 m/s 2 )*(t) 2

-1.40 m = 0+ (-0.835 m/s 2 )*(t) 2

(-1.40 m)/(-0.835 m/s 2 ) = t 2

1.68 s 2 = t 2

Return to Problem 5

a = (444 m/s - 0 m/s)/(1.83 s)

a = 243 m/s 2

d = (0 m/s)*(1.83 s)+ 0.5*(243 m/s 2 )*(1.83 s) 2

d = 0 m + 406 m

Return to Problem 6

(7.10 m/s) 2 = (0 m/s) 2 + 2*(a)*(35.4 m)

50.4 m 2 /s 2 = (0 m/s) 2 + (70.8 m)*a

(50.4 m 2 /s 2 )/(70.8 m) = a

a = 0.712 m/s 2

Return to Problem 7

(65 m/s) 2 = (0 m/s) 2 + 2*(3 m/s 2 )*d

4225 m 2 /s 2 = (0 m/s) 2 + (6 m/s 2 )*d

(4225 m 2 /s 2 )/(6 m/s 2 ) = d

Return to Problem 8

d = (22.4 m/s + 0 m/s)/2 *2.55 s

d = (11.2 m/s)*2.55 s

Return to Problem 9

(0 m/s) 2 = v i 2 + 2*(-9.8 m/s 2 )*(2.62 m)

0 m 2 /s 2 = v i 2 - 51.35 m 2 /s 2

51.35 m 2 /s 2 = v i 2

v i = 7.17 m/s

Return to Problem 10

(0 m/s) 2 = v i 2 + 2*(-9.8 m/s 2 )*(1.29 m)

0 m 2 /s 2 = v i 2 - 25.28 m 2 /s 2

25.28 m 2 /s 2 = v i 2

v i = 5.03 m/s

To find hang time, find the time to the peak and then double it.

0 m/s = 5.03 m/s + (-9.8 m/s 2 )*t up

-5.03 m/s = (-9.8 m/s 2 )*t up

(-5.03 m/s)/(-9.8 m/s 2 ) = t up

t up = 0.513 s

hang time = 1.03 s

Return to Problem 11

(521 m/s) 2 = (0 m/s) 2 + 2*(a)*(0.840 m)

271441 m 2 /s 2 = (0 m/s) 2 + (1.68 m)*a

(271441 m 2 /s 2 )/(1.68 m) = a

a = 1.62*10 5 m /s 2

Return to Problem 12

• (NOTE: the time required to move to the peak of the trajectory is one-half the total hang time - 3.125 s.)

First use:  v f  = v i  + a*t

0 m/s = v i  + (-9.8  m/s 2 )*(3.13 s)

0 m/s = v i  - 30.7 m/s

v i  = 30.7 m/s  (30.674 m/s)

Now use:  v f 2  = v i 2  + 2*a*d

(0 m/s) 2  = (30.7 m/s) 2  + 2*(-9.8  m/s 2 )*(d)

0 m 2 /s 2  = (940 m 2 /s 2 ) + (-19.6  m/s 2 )*d

-940  m 2 /s 2  = (-19.6  m/s 2 )*d

(-940  m 2 /s 2 )/(-19.6  m/s 2 ) = d

Return to Problem 13

-370 m = (0 m/s)*(t)+ 0.5*(-9.8 m/s 2 )*(t) 2

-370 m = 0+ (-4.9 m/s 2 )*(t) 2

(-370 m)/(-4.9 m/s 2 ) = t 2

75.5 s 2 = t 2

Return to Problem 14

(0 m/s) 2 = (367 m/s) 2 + 2*(a)*(0.0621 m)

0 m 2 /s 2 = (134689 m 2 /s 2 ) + (0.1242 m)*a

-134689 m 2 /s 2 = (0.1242 m)*a

(-134689 m 2 /s 2 )/(0.1242 m) = a

a = -1.08*10 6 m /s 2

(The - sign indicates that the bullet slowed down.)

Return to Problem 15

d = (0 m/s)*(3.41 s)+ 0.5*(-9.8 m/s 2 )*(3.41 s) 2

d = 0 m+ 0.5*(-9.8 m/s 2 )*(11.63 s 2 )

d = -57.0 m

(NOTE: the - sign indicates direction)

Return to Problem 16

(0 m/s) 2 = v i 2 + 2*(- 3.90 m/s 2 )*(290 m)

0 m 2 /s 2 = v i 2 - 2262 m 2 /s 2

2262 m 2 /s 2 = v i 2

v i = 47.6 m /s

Return to Problem 17

( 88.3 m/s) 2 = (0 m/s) 2 + 2*(a)*(1365 m)

7797 m 2 /s 2 = (0 m 2 /s 2 ) + (2730 m)*a

7797 m 2 /s 2 = (2730 m)*a

(7797 m 2 /s 2 )/(2730 m) = a

a = 2.86 m/s 2

88.3 m/s = 0 m/s + (2.86 m/s 2 )*t

(88.3 m/s)/(2.86 m/s 2 ) = t

t = 30. 8 s

Return to Problem 18

( 112 m/s) 2 = (0 m/s) 2 + 2*(a)*(398 m)

12544 m 2 /s 2 = 0 m 2 /s 2 + (796 m)*a

12544 m 2 /s 2 = (796 m)*a

(12544 m 2 /s 2 )/(796 m) = a

a = 15.8 m/s 2

Return to Problem 19

v f 2 = v i 2 + 2*a*d

(0 m/s) 2 = v i 2 + 2*(-9.8 m/s 2 )*(91.5 m)

0 m 2 /s 2 = v i 2 - 1793 m 2 /s 2

1793 m 2 /s 2 = v i 2

v i = 42.3 m/s

Now convert from m/s to mi/hr:

v i = 42.3 m/s * (2.23 mi/hr)/(1 m/s)

v i = 94.4 mi/hr

Return to Problem 20

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Schaum's 3,000 Solved Problems in Physics (Schaum's Outlines) 1st Edition

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More than 40 million students have trusted Schaum’s Outlines for their expert knowledge and helpful solved problems. Written by renowned experts in their respective fields, Schaum’s Outlines cover everything from math to science, nursing to language. The main feature for all these books is the solved problems. Step-by-step, authors walk readers through coming up with solutions to exercises in their topic of choice.

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• ISBN-10 0071763465
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Multiply $-2$ times $9.81$

Multiply $-19.62$ times $3.2$

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Solving Fermi-Hubbard-type models by tensor representations of backflow corrections

Yu-tong zhou, zheng-wei zhou, and xiao liang, phys. rev. b 109 , 245107 – published 4 june 2024.

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The quantum many-body problem is an important topic in condensed matter physics. To efficiently solve the problem, several methods have been developed to improve the representation ability of wave functions. For the Fermi-Hubbard model under periodic boundary conditions, current state-of-the-art methods are neural network backflows and the hidden fermion Slater determinant. The backflow correction is an efficient way to improve the Slater determinant of free particles. In this work we propose a tensor representation of the backflow-corrected wave function; we show that for the spinless t − V model, the energy precision is competitive or even lower than current state-of-the-art fermionic tensor network methods. For models with spin, we further improve the representation ability by considering backflows on fictitious particles with different spins, thus naturally introducing nonzero backflow corrections when the orbital and the particle have opposite spins. We benchmark our method on molecules in the STO-3G basis and the Fermi-Hubbard model with periodic and cylindrical boundary conditions. We show that the tensor representation of backflow corrections achieves competitive or even lower-energy results than current state-of-the-art neural network methods.

• Received 12 March 2024
• Revised 12 May 2024
• Accepted 22 May 2024

DOI: https://doi.org/10.1103/PhysRevB.109.245107

©2024 American Physical Society

Physics Subject Headings (PhySH)

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• 1 CAS Key Laboratory of Quantum Information, University of Science and Technology of China , Hefei 230026, China
• 2 Synergetic Innovation Center of Quantum Information and Quantum Physics, University of Science and Technology of China , Hefei 230026, China
• 3 Hefei National Laboratory, University of Science and Technology of China , Hefei 230088, China
• 4 Pittsburgh Supercomputing Center , Carnegie Mellon University , Pittsburgh, Pennsylvania 15213, USA
• 5 Department of Physics, Carnegie Mellon University , Pittsburgh, Pennsylvania 15213, USA
• * [email protected]
• † [email protected]

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Vol. 109, Iss. 24 — 15 June 2024

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The matrix in the Slater determinant for N particles, with N ↑ and N ↓ being the particle numbers for spin up and spin down, respectively. The spin on the horizontal (vertical) axis denotes the spin of the particles (orbitals).

The energy convergence of the first-order gradient descent for the spinless t − V model with V / t = 1 on the 10 × 10 lattice under the OBC. The MC sample number is 128 000, and the parameter updating step size δ = 5 × 10 − 4 . The converged energy per site is − 0.4617 , while the reference energy obtained by fPEPS is − 0.4620 .

Comparison of energies (per site) of two kinds of tensor representations of backflow corrections for the Fermi-Hubbard model with n = 0.875 and U = 8 on the 4 × L lattice with the PBC. “Original BW” denotes the coefficient c i j , and the HF orbitals are represented by two separate tensors. “BW” denotes the backflow-corrected wave function is represented by a single tensor. Each energy is evaluated by the p = 0 wave function.

Energy comparisons for the Fermi-Hubbard model on 4 × L lattices under the PBC and U = 8 . The filling is n = 0.875 . Red upward-pointing triangles denote NN backflow, green downward-pointing triangles denote HDFS. BW1 energy results of p = 0 ( p = 1 ) are denoted by left-pointing (right-pointing) triangles. BW2 energy results of p = 0 ( p = 1 ) are denoted by squares (circles).

Comparison of spin correlation functions 〈 S ( 0 , 0 ) S ( x , y ) 〉 for BW1 (solid line) and BW2 (dashed line) for different lattice sizes; the filling is n = 1 , and U = 8 . The relative distance is defined as d = x 2 + y 2 ; all results are evaluated with p = 1 wave functions.

The spin density of the Fermi-Hubbard model with filling n = 0.875 and U = 8 under the PBC, evaluated using the p = 1 wave function. On the 4 × 8 lattice, the spin density obtained with (a) BW1 and (b) BW2. The spin density on the 4 × 24 lattice obtained with (c) BW1 and (d) BW2.

The spin density and the hole density achieved with BW2 on rectangular lattices under (a)–(d) the CBC and (e) and (f) the PBC, with a pinning field applied on both short edges. (a) and (b) depict the 4 × 16 lattice, with filling n = 0.875 . (c) and (d) depict the 4 × 20 lattice, with filling n = 0.9 . (e) and (f) depict the 8 × 16 lattice, with filling n = 0.875 .

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