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**Correct answer: 1.43 sec for A, 1.65 sec for B**

Estimate the time it takes for a free fall drop from 10 meters height. Also, estimate the time a 10 m platform diver would be in the air if he takes off straight up with a vertical speed of 2 m/s (and clears the platform of course!). Please provide a brief description of how you arrived at your estimates. (A: 18 correct B 9 correct) 10m / 9.8m/s= 1.02s (UNITS!); 10m / 2m/s= 5s (RTFQ!!) It would take about .98 seconds for the ball to fall to the ground. The speed of the fall is 9.8 m/s. NO!!!!!! v2=v1-g.t; 2m/s=0-[9.8,/s^2].t; t= 0.2 s (wrong eqn) the answer to the question for just a free fall is 2 seconds. something that is free falling, falls at an acceleration of m/s2 velocity is m/s NO, ONLY IF v=const. Correct answer: 1.43 sec for A, sec for B

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**Estimate the time it takes for a free fall drop from 10 meters height**

Estimate the time it takes for a free fall drop from 10 meters height. Also, estimate the time a 10 m platform diver would be in the air if he takes off straight up with a vertical speed of 2 m/s (and clears the platform of course!). Please provide a brief description of how you arrived at your estimates. (A: 19 correct B 10 correct) I found this throughusing the formula y-y0= v0t-0.5at^2. because this is a free fall a=-g which is -9.8m/s^2 y=-10m because the fall is downwards and we assume to start at zero for simplicity and the initial velocity was 0, therefore, the formula becomes -10m-0= 0t- .5(9.8m/s^2)t^2 t^2 = 10/4.9 which leads t=2.04s, and taking into account sig figs becomes 2 s. (oops SQRT missed; again carrying units throughout the calculation would have helped). This example captured lots of silly mistakes, incorret units, using the wrong equation, not paying attention to just what the problem stated, and even not noticing that there was a second part to the question. Keep this in mind as you move forward in this class!!

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**DON'T confuse accel and velocity! **

What is the magnitude and direction (if any) of the acceleration of the ball at the instant it reaches the highest point in its trajectory? What is its acceleration just before it hits the ground upon its return journey? (Correct: 18 Wrong:13 No Ans: 16) At the peak of the trajectory flight, the ball has no magnitude or direction, for that instant; however it then immediately accelerates downwards at 9.8 m/s2 I don't know the streets names, but I guess, the magnitude of the acceleration of the ball at the highest point in trajectory, it will be 9.81m/s^2(same as gravitational force) and the direction is upward or no direction,because the direction was headed to upward during the time it goes up,but at the top,because it stops for a moment, maybe there's no direction. The magnitude of the acceleration of the ball the instant it reaches the highest point in its trajectory is 9.8 m/s2. The acceleration just before the ball hits the floor is 0 m/s2. DON'T confuse accel and velocity! The direction is DOWN (not just “negative”, the sign means nothing if you don't specify the coordinate system!).

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Examples– Chpt. 2

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Examples– Chpt. 2

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Examples– Chpt. 2

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Examples– Chpt. 3

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When you drive somewhere in your car, is the change in the reading on your odometer equal to, greater than, or less than the magnitude of the displacement you experienced on your journey? (30 more or correct, 4 incorrect, 14 no resp.). Equal to. The change in the odometer is the distance you've traveled. The magnitude of a displacement vector describes exactly that. When you drive somewhere in your car, the change in the reading on your odometer is equal to the magnitude of the displament you experienced on your journey. Displacement is the change in position and odometers measures the magnitude of that change. (True, but be careful!) The change in the reading on the odometer will be less than the magnitude of the displacement. The change in the reading would give the acceleration at that time which is much smaller than the position of the car. The reading on the odometer is greater than the magnitude of the displacement in most cases (unless you drove in a straight line for the entire drive).

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Please give a concise, simple description of the physical significance of the dot product of two vectors. By and large the class knew this involved a scalar result from multiplying two vectors, but many of the answers were fuzzy. If two vectors have the same starting point, we can use the dot product to find the cosine of the angle between those vectors (NOTE starting point of a vector is not defined in absolute terms, you put it where it needs to be for the manipulation of interest!!!) The dot product is the product of the magnitude of one vector along with the scalar component of another.(close to the book version). A dot product is the product of two quantities, the magnitude of one vector and the scalar component of the second vector along that of the first. Each vector has a component along the direction of the other vector. I would say the product of the magnitude of one with the SIGNED component of the other along the first, or the projection of one on the other times the magnitude of the latter. DVB

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Examples– Chpt. 3 (d) What is the angle between the two vectors?

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Examples– Chpt. 4

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P221 Lecture 6 Does the answer to either a or b depend on the angle with respect to the ground at which the ball is thrown? ( A: Correct: 26 Wrong: 7 Confused: 3 No Answer: 18 B: Correct: 30 Wrong: 0 Confused: 0 No Answer: 15 C: Correct: 25 Wrong: 6 Confused: 1 No Answer: 15) C) If the ball was thrown in any direction other than straight up, there would be two accelerations, one in the down direction and the other in the lateral direction the ball was thrown. Yes, the answer to b does depend on the angle with respect to the ground at which the ball is thrown. This is due to the fact that the velocities of the ball across the path vary. Thus, one must consider the components to determine the actual acceleration just before it hits the ground. C) Both a and b depend on the angle with respect to the ground the ball was thrown because it will change the direction and the magnitude of the acceleration. … emm NO !!!!

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1/29/03 Physics 103, Spring 2004, U. Wisconsin 1 Physics 103: Lecture 3 Position & Velocity with constant Acceleration l Today’s lecture will be on kinematic.

1/29/03 Physics 103, Spring 2004, U. Wisconsin 1 Physics 103: Lecture 3 Position & Velocity with constant Acceleration l Today’s lecture will be on kinematic.

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