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Question1 ptsA 60 kg dancer applies a horizontal force of 800 N on the dance floor. The dancer'$ acceleration will be:0 A -7.5 m/s/sB. 7.5 m/s/sC.-13.33 m/s/sD...

Question

Question1 ptsA 60 kg dancer applies a horizontal force of 800 N on the dance floor. The dancer'$ acceleration will be:0 A -7.5 m/s/sB. 7.5 m/s/sC.-13.33 m/s/sD. 13.33 m/s/sQuestion 10Ipts

Question 1 pts A 60 kg dancer applies a horizontal force of 800 N on the dance floor. The dancer'$ acceleration will be: 0 A -7.5 m/s/s B. 7.5 m/s/s C.-13.33 m/s/s D. 13.33 m/s/s Question 10 Ipts



Answers

A 60.0 -kg male dancer leaps 0.32 m high. a. With what momentum does he reach the ground? b. What impulse is needed to stop the dancer? c. As the dancer lands, his knees bend, lengthening the stopping time to 0.050 s. Find the average force exerted on the dancer's body. d. Compare the stopping force with his weight.

So for this question we have a block of mass M equals to 10 kg. Which is kept in which is kept on the rough horizontal floor in the direction parallel to the floor at equals to eight seconds the body start moving. So it means before the second U. S. Equals to zero. So we have to determine at velocity at equals to 10 seconds. Okay, feet in meta per second. So we can calculate the impulse of the uh impulse from the draft. So f mercury delta T from eight seconds 10 seconds. So this will be equal to one by two, particular by one by 200 minus 80 and Marie claire by 10 minus eight. Okay, this 100 minus 80 value can be obtained because the slope is constant. So at 100 at 10 seconds the value for the forces 100. So at eight seconds this value will be 80. Okay, so from here, after solving this impulse evidente T is equal to 20 newton second and from the impulse momentum Ferrum impulse momentum theorem or relation. Okay, so we can write that evidente thi this is equal to mass manipulative, a change in velocity that is we minus you. So F identity. It is equal to 20 and mass is equal to this 10 kg and we mine issues equals to zero. So from here we get still be that is equal to two m per second. Okay, so from the given options option D. It is the correct answer for this problem. Okay, thank you.

Mhm Hi student. Welcome to the solution of the problem. Number 370. No suppose an object of Moss, five kg is sitting at the origin. A force of eight. Newton acts in the positive X direction. An air force Of six. Newton acts in the positive wild direction. We are asked to find the magnitude of acceleration of the object. Now the relation between folds and acceleration is F is equal to mass multiplied by acceleration. Now splitting this into equations of X. Axis and Y. Axis effects. Or the ex commander falls is equal to mass multiplied by X. In other words, mass multiplied by the X. Component of acceleration. Also F. Y is equal to mass multiplied by the why component of acceleration. Now here eight newton this effects and six newton S. F. Y. So and mars has already given to be five kilogram so we can reduce the situations or interpret these two equations to be eight is equal to five multiplied by X. So X will have a value of eight x 5 metaphor. Second Squire. Similarly, A.Y. have a value of six x 5 me double second Squire. So acceleration vector, can we return to be one x 5, multiplied by eight. I plus six J. In that case the magnitude of acceleration will be wrote off eight square by five square bless six Squire by five square will be one by five root off eight Squire plus six square which is 10. The wrote Gives the value of 10 so it will be 10 x five. So the answer will be two m 2nd race to -2 out of the given options. Option B corresponds to this answer. I hope the solution is clear to its. You acquired a mathematical calculation. Just always need to, as I always remind, you need to understand this equation of the false. It is a very fundamental equation that comes to solve many questions, and we also need to understand the vector analysis. I hope the solution is clear to you, and I hope the solution will help you to solve future problems. Also happy learning.

Solving party of this problem. So calculating F. One that is forced on two kg ball which is equal to M. And G, which is equal to two kg multiplication and 9.8 m per second squared which is equal to 19.6 Newton. Similarly, I can calculate the value of have to which is equal to M two G, which is equal to six multiplication, 9.8 which is equal to 58.8 newton as the photos on the second ball, which had must six kg. Now I am simplifying part B. So in part B. During free fall case every object fall with an acceleration of 9.8 m per second squared and it is independent of it is Indy pendant of the mass of the object mask of the object. That age. In freefall case every object falls with the actual reason 9.8 m per second squared, which is independent of the massive object. So both object board object have same acceleration, same acceleration, which is equal to 9.8 m per second squared

This question covers the concept of the northern 2nd law of motion and to solve our problems, fascinated to find the acceleration along X and Y direction. So from the kinetic motion, acceleration on the extradition is covering the changing velocity along the X direction upon the time. And similarly the acceleration along the wind direction is the change in velocity along the y direction upon the tank. So the acceleration along the X direction equals the changing speed along the X direction. And that is eight minus nine m per second upon the time. And that is eight seconds are the acceleration along the X direction is -0.12, 5 m or Second Square. And similarly the acceleration along the right direction is the final speed is 10 m per second, manage the initial speed along the wires, zero m per second Upon the time. eight seconds are the acceleration along the Y direction is uh 1.25 m for a second square. Now the component of the force, for part a the component of the force along the X direction equals the mass into the acceleration And the masses four Kg Into the acceleration. And that is -0.125 m for second step. All the component of the force along the extra action is Uh minus 0.5 newton. And similarly the component of the force along the right direction equals four kg Into 1.25 m 4 2nd sq uh F Y equals five. No. So we can write the net forces yeah -0.5 I plus five G. No. To and the magnitude of the net force equals square root off 0.5 square plus five square Norton. Are the magnitude of the Fools. Yeah. Uh for the second part mhm equals 5.03. No, buts right.


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