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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 ...
Three years in the making, this enormous collection demonstrates best practices for solving any type of physics problem. Each video is concise, but without skipping steps, to help get you on your way as quickly as possible. I hope you will find the help you need. Download the textbook for free from OpenStax. Best wishes with your studies!
Exercise 22.48. (a) A 200-turn circular loop of radius 50.0 cm is vertical, with its axis on an east-west line. A current of 100 A circulates clockwise in the loop when viewed from the east. The Earth’s field here is due north, parallel to the ground, with a strength of 3.00 \times 10^ {-5} \textrm { T} 3.00 ×10−5 T .
While pecking on a tree, the woodpecker's head comes to a stop from an initial velocity of 0.600 m/s in a distance of only 2.00 mm. (a) Find the acceleration in \textrm {m/s}^2 m/s2 and in multiples of g g (g = 9.80 \textrm { m/s}^2 g = 9.80 m/s2). (b) Calculate the stopping time.
Question. A 60.0-kg skier with an initial speed of 12.0 m/s coasts up a 2.50-m-high rise as shown in Figure 7.39. Find her final speed at the top, given that the coefficient of friction between her skis and the snow is 0.0800. (Hint: Find the distance traveled up the incline assuming a straight-line path as shown in the figure.)
19. OpenStax College Physics. Chapter 4: Dynamics: Force and Newton's Laws of Motion. Problem 19. Question. (a) Calculate the tension in a vertical strand of spider web if a spider of mass 8.00 \times 10^ {-5} \textrm { kg} 8.00 ×10−5 kg hangs motionless on it. (b) Calculate the tension in a horizontal strand of spider web if the same spider ...
This is College Physics Answers with Shaun Dychko. What fraction of the Earth's diameter is the greatest ocean depth? We need to look up these numbers in our data table in chapter 1 and the diameter of Earth is 10 to the 7 meters and the deepest ocean is 10 to the 4 meters so we have 10 to 4 divided by 10 to the 7 giving us 10 to the minus 3.
Then, to get the length of the velocity of the ship with respect to the Earth, we take the square root of the sum of the squares of each of its components, so it’s the square root of 1.1491 meters per second squared plus 7.9642 meters per second squared, that gives 8.05 meters per second. It’s direction, we’ll take the inverse tangent of ...
So the final answer for part (a) is that the field at position a is 1.25 times 10 to the 6 newtons per coulomb, 30.0 degrees above the x-axis to the left and you have to say to the left because otherwise this would also be 30.0 degrees above the x-axis but that's to the right so we have to distinguish between this direction versus this one by ...
So the range formula, equation 71 in chapter three says that, the range will be; the initial speed squared, times sine of two times the launch angle, divided by acceleration due to gravity. Now the sine function has a range between one and negative one. This is a graph of the sine function. This is y equals sine theta.