Assume m₁ = 1 kg. m₂ = 3 kg, and x₁ = 1 m, x2 = 5 m, where is the center of mass of these two objects?
Q: A 1.76 kg particle has the xy coordinates (-1.41 m, 0.619 m), and a 5.63 kg particle has the xy…
A: m1 = 1.76 kg r1 = (-1.41 , 0.619) m m2 = 5.63 kg r2 = (0.206 , -0.607) m m3 =3 kgrc.m= (-0.848 ,…
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A: m1 = 1.5 kg x1 = -2 m m2 = 2 kg x2 = 1.5m m3 = 3 kg x3 = 0m
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A: The formula for x-coordinate of the center of the mass is:
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A: Given that m1 = 4.73kgAnd its coordinates are (x1,y1)=(-1.50m, 0.794m)m2 = 5.13kgAnd its coordinate…
Q: A 1.34 kg particle has the xy coordinates (-1.54 m, 0.660 m), and a 2.30 kg particle has the xy…
A: m1 = 1.34 kg r1 = (-1.54 , 0.660) m m2 = 2.30 kg r2 = (0.896 , -0.181) m m3 =3.87 kgrc.m= (-0.571 ,…
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A: FINAL ANSWER: X-component of the centre of mass is =0m Y-component of the centre of mass is…
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A: position of center of mass rcm = (m1 r1 +m2 r2 )/(m1 + m2 )
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Q: A 45.0-kg skater is traveling due east at a speed of 2.50 m/s. A 70.0-kg skater is moving due south…
A: Page 1
Q: A 12.5 cm tall soft drink can has a mass of 17.7 g and contains 364 g of soda when full. Assuming…
A: Given :- m1 = 17.7 g x 1 = 6.25 m 2 = 364 g /2 = 182 (for half full) X 2 = 6.26 /2 = 3.125 (for…
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- Three masses are placed on the x-axis : 300 g at origin, 500 g at x = 40 cm and 400 g at x = 70 cm. Thedistance of the centre of mass from the origin is -(1) 45 cm (2) 50 cm (3) 30 cm (4) 40 cmTwo balls each have a mass of 4.50-kg. They are positioned on the y axis at 1.00 and 9.00 m, respectively; a third ball with a mass 2.30 kg is located at 6.00 m. Calculate the location of the center of mass of this system?Problem 3: A particle with mass ma = 3.00 kg is located at ra = (2.50 i + 3.50 j) m, and a second particle of mass m2B = 5.00 kg is located at rB = (1.50 i - 3.00 j) m. Find the location of the center of mass of the system relative to the point (1,1).
- Question is on the pictureAn 61-kg jogger is heading due east at a speed of 3.4 m/s. A 83-kg jogger is heading 52 ° north of east at a speed of 3.3 m/s. Find (a) the magnitude and (b) the direction of the sum of the momenta of the two joggers. Describe the direction as an angle with respect to due east.Find the center of mass (x̄, ȳ, z̄) of the block, given: L1 = 5 in, L2 = 4 in, L3 = 6 in, L4 = 5 in, L5 = 3 in, L6 = 8 in
- In the figure below, m1 3kg, m2= 4kg, and m3 = 8kg, and the scales on the axes are set by xs = 2m and ys = 2m. What are the x and y coordinates of the center of mass of the system?An 83-kg jogger is heading due east at a speed of 2.2 m/s. A 76-kg jogger is heading 18 ° north of east at a speed of 3.3 m/s. Find (a) the magnitude and (b) the direction of the sum of the momenta of the two joggers. Describe the direction as an angle with respect to due east.The mass of a particular skateboard plus its rider is m, = 76.6 kg, and the skateboarder is holding a book with mass mo = 3.27 kg. Initially atrest, the skateboarder tosses the textbook with a velocity of vb = 4.44 m/s at an angle 0 = 23.3° above the horizontal, and the book is caught by a friend.
- 11.The vector position of a 3.45 g particle moving in the xy plane varies in time according to i, - (3i + 35)t + 2jr? where t is in seconds and is in centimeters. At the same time, the vector position of a 5.20 g particle varies as i,- 31 - zir? - 6jt. (a) Determine the vector position (in cm) of the center of mass of the system at t = 2.90 s. x cm cm (b) Determine the linear momentum (in g cm/s) of the system at t = 2.90 s. 9. cm/s (e) Determine the velocity (in cm/s) of the center of mass at t = 2.90 s. cm/s (4) Determine the acceleration (in cm/s) of the center of mass at t = 2.90 s. cm/s? (e) Determine the net force (in µN) exerted on the two-particle system at t = 2.90 s. UN net