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Welcome to Physics Homework - Questions & Answers

This site started back in 2008 and features, among others, a collection of physics homework exercises and tutorials for all levels of physics students (high school and college).

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Analog and Digital Communications (Physics 227/623)

2. Problem 2.11. Use the function and sample rate of problem 2.10. Let Δ = 0.5V (see p. 109 in text). Sketch the resulting staircase approximation (superimposed on s(t)) and… Click here to read more

Physics 227/623, Assignment #3 (Analog and Digital Communications)

1. Problem 2.1. Do parts (a) and (c), skipping (b). Sketch the waveform for the first 1 msec, rather than 3 msec. Use 3 bits plus a sign bit, i.e.,… Click here to read more

Coffee Drop Experiment

Coffee drop experiment For the coffee filter the graph of position shows that it does not bounce out. Although we have done a fit with a quadratic equation, the coefficient… Click here to read more

Ball Drop Experiment

Ball drop experiment Data was recorded for position and velocity as function of time. For position the floor is considered to be at a distance of 2 m. We have… Click here to read more

Effective Charge Laboratory

The definition of the effective charge In this paper we will determine the effective charge of an electron. The effective charge is by definition the ratio (e/m). Background Information and… Click here to read more

Spherical shell potential (Griffiths)

The charged sphere A spherical shell has spherical symmetry and because of this we write its potential as $V(R,\theta)=V_0(\theta)$ ($\theta$ is the “polar” angle of the spherical coordinate system). a)… Click here to read more

Icy slope and sled

The icy slope A frictionless icy slope of angle $\theta$ having a length $L$, followed by a horizontal portion, then by a bump ($R$) make a frictionless path as in… Click here to read more

Moment of Inertia

Circle moment of inertia Find the principal moments of inertia of a circle of radius $R$ and mass $m$. Use the integration over the circle length. (Hint: first define the… Click here to read more

Wien Bridge

The Bridge In the figure is presented a Wien Bridge. When its two arms are at equilibrium ($R1*R3=R2*R4$) the voltage on the Wien bridge diagonal cancels. You apply to this… Click here to read more

Analog-Digital Converter

The Analog-Digital Converter The input range of a 10 bit ADC varies from +0.1..+4.9V. a) What input step size has this analog-digital converter? b) For a full scale sine wave… Click here to read more

Pendulum Lab. Interference

Pendulum Laboratory The equations of pendulum are $T= C * \sqrt{L/g}$    (1) $T= 2\pi * \sqrt{L/g} * [1+(1/2)^2*\sin^2(x/2)+(3/4)^2*\sin^4(x/2)]$   (2) 1. For which angle is the period given by pendulum equation… Click here to read more

Supernova Expansion

The supernova expansion In a supernova explosion, initially the supernova expands freely. Then the supernova enters a new stage when in the existent gas around the star, shock waves push… Click here to read more

Acoustic Impedance

Pipe Acoustic Impedance Let us suppose that we want to send an acoustic wave from left towards right through a series of connected pipes having different areas. What is the… Click here to read more

Falling Electron (on Nucleus)

The falling electron Using the classic EM framework, a falling electron $e^-$ (a negative charge) emits energy/time as $W= (2/3)*(a^2e^2/c^3)$ (erg/s) where $a$ is the acceleration of the falling electron… Click here to read more

Motorola M68000 Assembler

MOVE and CLR 8. If the value stored in the data register of Motorola M68000 D0 = \$ABCD1234, what is the data found in register D0 at the end of… Click here to read more

Velocity versus Time

Velocity as function of time You are given the dependence of velocity as function of time from the figure. Please find the position as function of time from the given… Click here to read more

Planet Distance

Question 1 A classic radiotelecope detects a planet’s hydrogen emitted waves of $\lambda=21 cm$. If the signal emitted by a planet has a wavelength that is 0.28 cm longer than… Click here to read more

Right shift and left shift

In general dividing by 2 is equivalent to right shifting with 1. Also left shifting with 1 is equivalent to multiplication with 2. (a)   Is a right shift equivalent to… Click here to read more

Geometric Optics for EM wave

EM wave propagation An EM wave goes through the athmosphere of Earth (considered inhomogeneous) having an index of refraction of $n(z)$. This index of refraction depends on altitude $z$ as:… Click here to read more

Hermitian or not?

Hermitian definition Please find if $(\overrightarrow p*\overrightarrow r)\overrightarrow L$ is hermitian or not. Please state the definition of a hermitian matrix? Which are the matrixes or $\hat r$, $\hat p$… Click here to read more

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College Physics

College Physics for AP ® Courses

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College Physics Answers was created by Shaun Dychko, a teacher with more than 12 years experience teaching high school physics and mathematics, mostly at Point Grey Secondary in Vancouver, Canada. I have taught AP Physics, and all levels of high school mathematics. I studied physics at the University of British Columbia, where I obtained both my B.Sc., and B.Ed.

I sincerely hope these videos will be helpful with your, or your son/daughter's studies. I put a great deal of effort into making them as clear as possible, and finding a balance between being concise while also being thorough enough to raise flags on common mistakes, and making a bit of time for occasional intriguing digressions.

I am also the creator of Giancoli Answers which provides solutions to problems in the textbook Physics: Principles with Applications , 7th and 6th Editions, by Douglas Giancoli.

Problems & Exercises

(c) + 7 m + 7 m

a. 8  m + 2  m + 3  m = 13  m ¯ 8  m + 2  m + 3  m = 13  m ¯

b. 9  m ¯ 9  m ¯

c. Δ x = 11  m − 2  m = 9  m ¯ Δ x = 11  m − 2  m = 9  m ¯

(a) 3 . 0 × 10 4 m/s 3 . 0 × 10 4 m/s

2 × 10 7 years 2 × 10 7 years

34 . 689 m/s = 124 . 88 km/h 34 . 689 m/s = 124 . 88 km/h

(a) 40 . 0 km/h 40 . 0 km/h

(b) 34.3 km/h, 25º S of E . 25º S of E .

(c) average speed = 3.20 km/h, v - = 0. average speed = 3.20 km/h, v - = 0.

(a) 6 . 61 × 10 15 rev/s 6 . 61 × 10 15 rev/s

4 . 29 m/s 2 4 . 29 m/s 2

(a) 1 . 43 s 1 . 43 s

(b) − 2 . 50 m/s 2 − 2 . 50 m/s 2

(a) 10 . 8 m/s 10 . 8 m/s

38.9 m/s (about 87 miles per hour)

(a) 16 . 5 s 16 . 5 s

(b) 13 . 5 s 13 . 5 s

(c) − 2 . 68 m/s 2 − 2 . 68 m/s 2

(a) 20 . 0 m 20 . 0 m

(b) − 1 . 00 m/s − 1 . 00 m/s

(c) This result does not really make sense. If the runner starts at 9.00 m/s and decelerates at 2 . 00 m/s 2 2 . 00 m/s 2 , then she will have stopped after 4.50 s. If she continues to decelerate, she will be running backwards.

0 . 799 m 0 . 799 m

(a) 28 . 0 m/s 28 . 0 m/s

(b) 50 . 9 s 50 . 9 s

(c) 7.68 km to accelerate and 713 m to decelerate

(a) 51 . 4 m 51 . 4 m

(b) 17 . 1 s 17 . 1 s

(a) − 80 . 4 m/s 2 − 80 . 4 m/s 2

(b) 9 . 33 × 10 − 2 s 9 . 33 × 10 − 2 s

(a) 7 . 7 m/s 7 . 7 m/s

(b) − 15 × 10 2 m/s 2 − 15 × 10 2 m/s 2 . This is about 3 times the deceleration of the pilots, who were falling from thousands of meters high!

(a) 32 . 6 m/s 2 32 . 6 m/s 2

(b) 162 m/s 162 m/s

(c) v > v max v > v max , because the assumption of constant acceleration is not valid for a dragster. A dragster changes gears, and would have a greater acceleration in first gear than second gear than third gear, etc. The acceleration would be greatest at the beginning, so it would not be accelerating at 32 . 6 m/s 2 32 . 6 m/s 2 during the last few meters, but substantially less, and the final velocity would be less than 162 m/s.

(a) v = 12 . 2 m/s v = 12 . 2 m/s ; a = 4 . 07 m/s 2 a = 4 . 07 m/s 2

(b) v = 11 . 2 m/s v = 11 . 2 m/s

(a) y 1 = 6 . 28 m y 1 = 6 . 28 m ; v 1 = 10 . 1 m/s v 1 = 10 . 1 m/s

(b) y 2 = 10 . 1 m y 2 = 10 . 1 m ; v 2 = 5 . 20 m/s v 2 = 5 . 20 m/s

(c) y 3 = 11 . 5 m y 3 = 11 . 5 m ; v 3 = 0 .300 m/s v 3 = 0 .300 m/s

(d) y 4 = 10 .4 m y 4 = 10 .4 m ; v 4 = − 4 .60 m/s v 4 = − 4 .60 m/s

v 0 = 4 . 95 m/s v 0 = 4 . 95 m/s

(a) a = − 9 . 80 m/s 2 a = − 9 . 80 m/s 2 ; v 0 = 13 . 0 m/s v 0 = 13 . 0 m/s ; y 0 = 0 m y 0 = 0 m

(b) v = 0 m/s v = 0 m/s . Unknown is distance y y to top of trajectory, where velocity is zero. Use equation v 2 = v 0 2 + 2 a y − y 0 v 2 = v 0 2 + 2 a y − y 0 because it contains all known values except for y y , so we can solve for y y . Solving for y y gives

Dolphins measure about 2 meters long and can jump several times their length out of the water, so this is a reasonable result.

(c) 2 . 65 s 2 . 65 s

(b) 0.717 s

(a) -70.0 m/s (downward)

(a) 19 . 6 m 19 . 6 m

(b) 18 . 5 m 18 . 5 m

(b) 262 m, -29.2 m/s

(a) 115 m/s 115 m/s

(b) 5 . 0 m/s 2 5 . 0 m/s 2

(c) 3 m/s 2 3 m/s 2

(a) Car A is traveling faster at the checkpoint because it must go past the speed of car B to reach the same distance.

(b) i. Yes, the equation is consistent with the answer because the speed of car A is only a constant away from the correct answer. ii. Yes, the equation makes sense because V = 2 V 0 V = 2 V 0 .

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Physics 2210 - Fall 2015

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