![]() ![]() Determine McMahon's free cash flow for both years. reported the following on the company's statement of cash flows in Year 2 and Year 1: Year 2 Year 1 Net cash flow from operating activities $294,000 $280,000 Net cash flow used for investing activities (224,000) (252,000) Net cash flow used for financing activities (63,000) (42,000) Seventy percent of the net cash flow used for investing activities was used to replace existing capacity. ![]() The group demonstrates understanding of the information when they identify what location as an unlikely site for childhood cancer? 1) BrainĤ) Bladder How does environment shape humans McMahon Inc. They are reviewing how childhood cancers differ from adult cancers. Thus the fraction part of the total initial kinetic energy lost during the collision is 0.961.Ī group of nursing students are studying information about childhood cancers in preparation for a class examination. The kinetic energy is the half of the mass time square of its velocity. Now the kinetic energy lost is the ratio of difference of initial kinetic energy and final kinetic energy to the initial kinetic energy. Negative sine indicates the direction after the collision is changed. Put the values to find the value of velocity after the collision as, Here, is the mass of body one, body 2 respectively and are the velocities of body one, body two before the collision respectively. As the momentum is product of mass times velocity. The speed of the second piece is 0.75 m/s.īy the conservation of momentum, we know that the momentum of two body before the collision is equal to the momentum after the collision. The mass of the second piece is 600 grams. The kinetic energy of a body is half of the product of mass times square of its velocity. Kinetic energy is the energy of of the body, which it posses due to force of motion. The fraction part of the total initial kinetic energy lost during the collision is 0.961. ![]() Two pieces of clay are moves directly, and collide the momentum before and after the collision remain same. ![]() It is necessary to pass from kilometers to astronomical unit (AU), where 1 AU is equivalent to 150.000.000 Km ( )ġ AU is defined as the distance between the earth and the sun.īut 1 year is equivalent to 1 AU according with Kepler’s third law, since 1 year is the orbital period of the earth.Ĭircular velocity for the particle in the Encke Division:įor a better representation of the velocity, kilometers and days are changed to meters and seconds respectively.įor the case of the particle in the D Ring, the same approach used above can be followedĬircular velocity for the particle in D Ring: Where T is orbital period and r is the radius of the trajectory.Ĭase for the particle in the Encke Division: That can be done by means of the Kepler’s third law: To determine the circular velocity of both particles it is necessary to know the orbital period of each one. Where r is the radius of the trajectory and T is the orbital period The particle in the D ring is 1399 times faster than the particle in the Encke Division. Since the time that the ball reaches maximum point is almost t=2s that answer can not be possible, so the answer is t=3.333s for the ball to go up and down, passing the top of the building So, the ball pass the top of the building on its way down at 160 ft Now, we must know how much distance does it take to reach maximum point Since time can not be negative the answer is t=6.378sī) To find the time that takes the ball to pass the top of the building on its way down, we must find how much does it move tooįirst of all, we need to find the maximum hight and how much time does it take to reach it: If we want to know the time that takes the ball to hit the ground, we need to calculate it by doing s=0 that is the final hight.Ī) Solving for t, we are going to have two answers The equation that describes the motion of the ball is: The distance of each piece of mass dm from the axis is given by the variable x, as shown in the figure.The knowable variables are the initial hight and initial velocity We can therefore write dm = \(\lambda\)(dx), giving us an integration variable that we know how to deal with. Note that a piece of the rod dl lies completely along the x-axis and has a length dx in fact, dl = dx in this situation. We chose to orient the rod along the x-axis for convenience-this is where that choice becomes very helpful. If we take the differential of each side of this equation, we find ![]()
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