# Giancoli Solutions on Video

Learn physics easily with guided practice.

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## Features

• 1,930 video solutions for all regular problems in Giancoli's 7th Edition and 1,681 solutions for most regular problems in the 6th Edition.
• Final answer provided in text form for quick reference above each video, and formatted nicely as an equation, like $E=mc^2$. This is useful if you are in the library or have a slow internet connection.

• Pen colors make the step-by-step solutions clear. Red is used to illustrate algebra steps, and to substitute numeric values in the final step of a solution. When a solution switches to a new train of thought a different pen color emphasizes the switch, so that solutions are very methodical and organized.
• Solutions are classroom tested, and created by an experienced physics teacher.
• Videos are delivered with a high performance content delivery network. No waiting for videos to load or buffer.
• Pause, rewind, repeat, and never miss what is being said.

## Sample solution

Giancoli 7th Edition, Chapter 5, Problem 12

(4:49)

### Giancoli 7th Edition, Chapter 24, Problem 53

Hello,
there is a second question in this problem: "What if the glass is to appear dark?"
I am curious to see that solution.
thanks,
Spanda

### Giancoli 7th Edition, Chapter 8, Problem 70

Hi Mr. Dychko,
For solving for the inertia of the horizontal rod, why do we use the I = (1/12 M L^2) equation and not the I = (1/3 M L^2) equation?
Thanks!

ur the best

### Giancoli 7th Edition, Chapter 8, Problem 1

GOOD JOB MR DITCHKO!!!!

### Giancoli 7th Edition, Chapter 6, Problem 42

Hi tmesser, good question. The tricky thing here is that acceleration is not constant. All of the kinematics formulas, like $d = \dfrac{v_f^2-v_i^2}{2a}$, assume acceleration is constant, so it can't be used in this question. Only conservation of energy can be used, plus the Hook's law for a spring, and $F_{net}=ma$, all three of which don't assume constant acceleration. The acceleration isn't constant since the force applied by the spring on the car changes depending on how much the spring is compressed. In the beginning the spring pushes only lightly, so the car's deceleration is small, but then as the spring is compressed it pushes harder on the car thereby increasing the deceleration.

Hope that helps,
Mr. Dychko