a-carpenter-on-the-roof-of-a-building-accidentally-drops-her-hammer-as-the-hammer-falls-it-passes-tw

Question

A carpenter on the roof of a building accidentally drops her hammer. As the hammer falls it passes two windows of equal height, as shown in FlGURE $2-39$ . (a) Is the increase in speed of the hammer as it drops past window 1 greater than, less than, or equal to the increase in speed as it drops past window 2$?$ (b) Choose the best explanation from among the following:
\begin{equation}\begin{array}{l}{	ext { I. The greater speed at window } 2 	ext { results in a greater increase in }} \ {	ext { speed. }} \ {	ext { II. Constant acceleration means the hammer speeds up the }} \ {	ext { same amount for each window. }} \ {	ext { II. The hammer spends more time dropping past window 1. }}\end{array}\end{equation}

Nathan Silvano · Accepted Answer

Okay, so in this problem we have a homer falling down and he passes for two windows. That&#x27;s car, window one and window, too. The windows had the same height, and the problem wants us to tell if the increase in speed and window one is going to be equal greater than lesser than the increase in speed and window to this is the first item and the second itin. We have three statements and we need to tell which one of them er has the best explanation for this problem. Okay, so we know that even so, we have a constant acceleration, their hammer and window one. It&#x27;s going to pass slowly, then in window, too, because we know in the window, too. He has a bigger speed. So he going to pass for the window much faster than in the window one. So if we consider just that time the change in velocity in window to me is going to be bigger, because what&#x27;s the question? We used to describe velocity as a function of time? We just use V equals be zero. Tell us, Jeanty. So the miners v zero is the change in velocity we want to discover doubt v So this is going to be equal g t. So, since we already agreed at the time in Window one is much bigger then the time and the window, too. We know precisely that the difference in velocity off the difference in speed off Window one is going to be bigger than the increase in speed off the window. Truth, this is the answer. So let&#x27;s look to this statement now. The first a statement is the greatest Peter Window tune results in a greater increase of speed About this. We already proved that is wrong because the speed is going to increase based on the time and not their celebration, because acceleration is constant. So Delta V. One it&#x27;s going to have g do you want and out of E two going to have Jew T too. So what&#x27;s what matters? Here is the time and not ah, the gravity. So the second statement is constant acceleration, Mr Hammond, Speed is up at the same amount for each window. Well, obviously is. It is wrong. And a turd statement the hammer. Spain&#x27;s more time dropping fast window one. And that&#x27;s the answer. Is this statement? Three. This is the answer for

Nathan Silvano · Answer

Okay, so in this question, we have the graph off the movement of there Homer on a free falling in the shaded areas are the time that we see the hammer passing through the wind for the window one and Windell two. And the problem wants us to know if the era, the shaded area off window one is going to be equal last 10 or greater than the shaded area off window, too. So since this is a grass off the lost city versus time, we know the shaded area under the curve is just there displacement off the hammer. So if we know the shaded area is the displacement and the shaded area represents the time frame that we can see the hammer through the window. The windows has the same height, so these shaded areas must be equal because the window has the same height independently if the shape off the shaded area. So your first answer is equal, and the second item we need to how each of the statement our true Well, since we know the shaded areas are equal, the only statement there is possible is the 2nd 1 And this is the true answers for the question. Thanks for watching

Suzanne W. · Answer

at the edge of a roof. You throw ball one upwards with the speed of the, not a moment later you throw a ball to downwards with the same initial speed the balls land at the same time. Which of the following statements is true for the instant, just before the balls hit the ground. So is it a the speed of the ball Of Ball? one is greater than the speed of ball too? Be the speed of ball, one is equal to the speed of ball too, Or C at the speed of ball, one is less than the speed of ball too. Okay, so when If you throw ball one up, so you&#x27;re on the roof, If you throw a ball one up, going to have a velocity when it reaches that same position on the roof, it&#x27;s going to have the velocity negative phenomena. So since it also has a velocity of negative Vienna and when you throw a ball to it&#x27;s going to have a velocity of negative be not, that means that it is B the speed will be equal

Kristela Garcia · Answer

in this problem. We&#x27;re trying to determine which of these two situations inserts a greater force on the nail. So if you look at the first situation, you have a hammer coming down. We can call this an initial velocity of the O or V one as it&#x27;s given the problem and the second situation, it comes back with the same velocity. So you have a V one coming down and then everyone coming back up, and then this the fact that it&#x27;s in the opposite direction is going to be really important. So this is an impulse problem, and we know that impulse is forest times are changing time, which is equal to my change in momentum. And in this case, my mom stays the same. So it&#x27;s really just a change in velocity. So we have our force. Is our mass times change in velocity over time. Now, in this first situation, I&#x27;m just gonna go ahead and copy my equation down. In this first situation, you have a final velocity of zero. And so our force is going to be proportional to our mass. Times are in the shit velocity divided by time, and it is telling us that our time intervals are the same for both. But if you look into a second situation, we have a certain velocity going down and we have the exact same velocity coming back up. But it&#x27;s actually in the opposite direction, so we&#x27;re gonna call that negative. So if we put that information in, we get a negative V one minus. The one gives us a negative to be one. So that gives us n times negative to view one, which tells us that everything being the same, the masses being the same, the time intervals being the same, that our second situation is going to yield a force that is two times greater than the first.

Matt Dransfield · Answer

Okay. So in this problem, it asks you to explain what happens if you dropped two balls of equal diameter from the same height at the same time. Ball one. In this case, the black well is made of metal and has a greater Maston ball to made of wood. In my diagram, I&#x27;ve illustrated that is red. The upward force, due to air resistance is the same for both balls illustrated here as to black vector arrows acting on both balls equally. So the question is a is the drop time of ball one greater, less than or equal to the drop time of ball, too. And then part B s says choose the best explanation from among the following choice. One. The acceleration of gravity is the same for all objects. Regardless of mass choice to the more massive ball is harder to accelerate. Trace three. Air resistance has less effect on a more massive ball. So to approach this problem, we need to complete the vector diagrams. Ball one. The black ball has a greater mass. Newton&#x27;s equation for us equals mass times. Acceleration would dictate them. That force pulling down on this metal ball would be greater than the mat force pulling down on the lighter wooden ball, the less massive wooden ball. So here I&#x27;ve illustrated the force vector downward as larger for the metal ball ball, one in the force factor for a ball to the wooden ball. So the drop time of this metal ball would be less than the drop time of ball, too, because there&#x27;s a greater downward force acting on that blow. Then part two of the best explanation for this while statement one is true in a vacuum. It is not true here cause there&#x27;s a resistance acting on the ball, So the best choice is about choice. Three. Air resistance has less effect on the massive ball because there&#x27;s a greater downward.

Justin Hameline · Answer

problem. 69 of Chapter two asks us to determine how fast you have to throw a ball downwards in order for a ball at half the height to be dropped so that they both hit the ground at the same time. In order to do this, we know two things. We know that one of the heights is H, which will a velocity V not attached to it and then the other height is h over to and we know their final position will be y f equals zero. Thus, we can use the equations h equals beena times t, which is its initial velocity times time plus 1/2 A T squared, which is the acceleration due to gravity downwards. Therefore, it accounts for both the velocity and the acceleration due to gravity. Whereas the second ball, which is being dropped from height or two, is equal to just the 1/2 a T squared because its initial velocity from here, by dividing the equations we can solve for the night. Thus we dio h divided by each over two is equal to V zero t plus 1/2 80 squared, divided by 1/2 a T squared looking at this equation, we can cancel out some terms. Firstly, we know that the age on the top of the H on the bottom will cancel out. And on the right side, we know that we can take out a T from every term. Thus we end up with the equation. 1/1 over jail is equal to V zero plus 1/2 A T, divided by 1/2 80 one divided by 1/2 is also equal to two itself. We multiplied the left side by 1/2 A T squared, which gives us 80 is equal to VI zero plus 1/2 80 by subtracting 1/2 a t from each side. We end up with the equation. The initial is equal to 1/2 a t and that is the magnitude of the initial. In the scheme of things, a ball being thrown downward will have a negative velocity. But since we&#x27;re dealing with magnitudes, the velocity will be a positive value. From here, we need to determine the amount of time it takes for the second ball being dropped from halfway up to reach the round. Thus, we will use this equation in salt for tea To begin, we multiply the left side by two and divide by a causing us tohave H over A is equal to t squared. From here, we square both sides and find that T is equal to the square root of H over a this week and then plug in to the velocity equation that we found V zero equals 1/2 a T and we get V zero is equal to 1/2 a times route h over a. Now we can move the A inside the square root in order to get 1/2 square rate h A. Thus the magnitude of our velocity at the initial throwing point will be 1/2 square root h a.

Problem

FIGURE 2-40 shows a v-versus-t plot for the hamme…

Problem 74 Medium Difficulty

Answer

(a) Greater
(b) Statement III

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(a) Greater(b) Statement III

Physics

Christina K.

Andy C.

Liev B.

Zachary M.

Math Review - Intro

Algebra - Example 1

James S. WalkerPhysics

Christina K.

Andy C.

Liev B.

Zachary M.

(a) Greater
(b) Statement III

James S. Walker

Physics