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physic of kot

# physic of kot - 200 km College of Science 300 km 30 deg 300...

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So (-2.75,-4.76)m So the distance between these points is 8.60m So the polar coordinates of its is ( 20 , -63.43 ° ) So the polar coordinates of its is ( 18 , -45 ° ) So r = 433 and y = 233 So (2.50m, 30 ° ) have Cartesian coordinates ( 53 , 4 54 )m So (3.80m, 120 ° ) have Cartesian coordinates ( - . , . 1 9 3 29 ) So the distance between this point is 4.54m 200 km 300 km 30 deg 300 sin30 o 300 cos30 o C B A θ R So the total distance is 486km and the angle north of west is 18deg So this vector has magnitude 6.12 units and angle of this vector is -112.5 ° with the positive x-axis So this vector has magnitude 1.78 units and angle of this vector is 22.5 ° with the x-axis So this vector has the magnitude 9.53 units and = . ° α 57 02 with the positive x-axis So this the magnitude of this resultant vector is R=7.92m Thus the + A B have component + = A B 2.59 i + 4.5 j and β = 60.0771 ° with x - axis Thus the + A B have component + = . - . A B 2 59i 1 5j and β1 = 329.9227º Thus the - B A have component - = B A -2.59 i + 1.5 j and β2 = 149.9227 So - - = . + - . A 2Bhave component A 2B 2 59i 4 5j and the angle isβ3=299.9229º So the magnitude of the displacement is R = 452.01ft So the distance from the end point to the start point is 2.337km Thus she must walk due to north 2.8km and dure east 1.31km to arrive at the same location So the displacement from the cave entrance is R= 358.468m Thus the magnitude of the B is 7.81 units and = . °, αx 59 19 = . °, αy 39 80 = . ° αz 67 41 So the magnitude of this vector is 40.31units and =- . ° θ 82 87 with the positive x - axis So the magnitude of the displacement is 788.12 mi Thus this vector have x = -25m and y = 43.30m and the vector in unit is C = - + . 25i 43 30j Thus C = + 5i 4j and the polar coordinate is C(8.01, 38.65º) Thus D = - + 1i 8j and the polar coordinate is D (6.47, 97.1250º) So the magnitude of the resultant vector is R = 7.21m and the direction of R is 56.30 ° from the axis Thus C has component C = 7.3 + . i 7 2j Thus the magnitude of the R is 6.21 blocks and direction of this vector is -69.56 ° from x-axis So the angle between A and B is 106.2602 So the magnitude of this vector is 5.91m and it have component (5,-1, -3) So the magnitude of this vector is 19.02 and it have component (4,-11, 15) Thus the angle between them is ( , ) A B = . ° 1 14 Thus the angle between them is ( , ) A B = arcos ( - + n2 1 n2 1 ) Thus the magnitude vector sum is R= 7.88N and the angle of R is α =97.8731 with the x-axis. the magnitude of this vector is R = 239.99 And angle is α = 57.23º with the negative x- axis or 237.23º with the positive x - axis So the total displacement is x = 179.583km So the average velocity is 69.524km/h Thus the average velocity of this time interval is 50m/s Thus the average velocity of this time interval is 41m/s Thus the velocity of this particle is 1.6m/s Thus the acceleration of this particle is a = -4m/s 2 Thus the position of the particle is x = 2m Thus the velocity at t = 3s is v = -3m/s Thus the acceleration of the particle is a = -2m/s 2 Thus the average speed is vx = 7m/s Thus the instantaneous speed is v 1 =10m/s and v 2 = 16m/s Thus the average acceleration is ax = 2.5m/s 2 Thus the average acceleration is a = 6m/s 2 Thus the acceleration of the car is a = 5.25m/s 2 Thus the distance in the first 8s is x = 504m Thus the speed of the car is = . / vf 94 5m s Thus the original speed is v = 4.70m/s Thus the acceleration is a = - . / 1 9 m s2 Thus the position at the instant it changes direction is x = 2.56m

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