1) A 41-kg girl is bouncing on a trampoline.During a certain interval after leav
ID: 1681056 • Letter: 1
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1) A 41-kg girl is bouncing on a trampoline.During a certain interval after leaving the surface of thetrampoline, her kinetic energy decreases to 150 J from 490 J. Howhigh does she rise during this interval? Neglect airresistance. 2) A person pulls a toboggan for a distance of 43.6 m alongthe snow with a rope directed 33.8 ° above the snow. Thetension in the rope is 89.6 N. (a) How much workis done on the toboggan by the tension force? (b)How much work is done if the same tension is directed parallel tothe snow? 3) It takes 169 kJ of work to accelerate a car from 20.2 m/sto 27.8 m/s. What is the car's mass? Please i need the asnwers 1) A 41-kg girl is bouncing on a trampoline.During a certain interval after leaving the surface of thetrampoline, her kinetic energy decreases to 150 J from 490 J. Howhigh does she rise during this interval? Neglect airresistance. 2) A person pulls a toboggan for a distance of 43.6 m alongthe snow with a rope directed 33.8 ° above the snow. Thetension in the rope is 89.6 N. (a) How much workis done on the toboggan by the tension force? (b)How much work is done if the same tension is directed parallel tothe snow? 3) It takes 169 kJ of work to accelerate a car from 20.2 m/sto 27.8 m/s. What is the car's mass? Please i need the asnwersExplanation / Answer
490 J - 150 J = 340 J = potentialenergy increase = mgh,where m is mass in kg, g is 9.8 m/s2 and h is theincrease in height
solving for h, gives....h = 340 J/(41kg x 9.8m/s2) = 0.8462 meters Pulled distance S = 43.6 m Angle = 33.80 o The tension in the rope T = 89.60 N (a). Tension in snow direction T ' = T cos = 74.46 N work is done on thetoboggan by the tension force = T ' * S =3246.29 J (b). Tension in the same direction as snow then workdone W = T S = 3906.56 J
Pulled distance S = 43.6 m Angle = 33.80 o The tension in the rope T = 89.60 N (a). Tension in snow direction T ' = T cos = 74.46 N work is done on thetoboggan by the tension force = T ' * S =3246.29 J (b). Tension in the same direction as snow then workdone W = T S = 3906.56 J
The tension in the rope T = 89.60 N (a). Tension in snow direction T ' = T cos = 74.46 N work is done on thetoboggan by the tension force = T ' * S =3246.29 J (b). Tension in the same direction as snow then workdone W = T S = 3906.56 J
given energy = 169kJ intial velocity = vi = 20.2m/s final velocity = vf = 27.8m/s from work energy theorem W = change in kineticenergy 169*103J = 1/2 * m vf2-vi2 2 * 169*103/(27.8)2 -(20.2)2 = m solve we get m = ----------kg given energy = 169kJ intial velocity = vi = 20.2m/s final velocity = vf = 27.8m/s from work energy theorem W = change in kineticenergy 169*103J = 1/2 * m vf2-vi2 2 * 169*103/(27.8)2 -(20.2)2 = m solve we get m = ----------kg 2 * 169*103/(27.8)2 -(20.2)2 = m solve we get m = ----------kg
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