If the forces that two interacting objects exert on each other are always exactly equal in magnitude and opposite in direction as stated by Newton’s third law, how is it possible for an object to undergo an acceleration?

D Question 1 Which of the following is true about the law of conservation of mass for a closed system? O Min mout O M₁-M₂ = Min-Mout Om₁=m₂ O min-m₁-mout O min=0 but mout>0

Estimate the magnitude of the force, in lbf, exerted on a 12-lb goose in a collision of duration 2 x 10-3s with an airplane taking off at 150 miles/h. Assume the bird's velocity is zero before the collision. Estimate the average acceleration magnitude, in ft/s², of the goose in the collision. |a| = ft/s2 Estimate the magnitude of the force, in lbf, exerted on the goose in the collision. |F| = lbf h

(a) A car with a mass of 850 kg is moving to the right with a constant speed of 1.44 m/s. What is the total force on the car? (b) What is the total force on the car if it is moving to the left? *

A 50 lb object is subjected to an applied upward force of 10 lbf. The only other force acting on the object is the force of gravity. The acceleration of gravity is 32.2 ft/s². * Your answer is incorrect. Determine the net acceleration of the object, in ft/s², with positive values being upward. 32.2 ft/s² a =

A simple pendulum consists of a 0.8- kg bob connected to a massless inextensible cord with a length L = 1.9 m. The bob is set into motion and its angular displacement is given by e(t) = 0.11cos(wt), where e is in radians and t is in seconds. Take g = 9.8 m/s^2, determine the mechanical energy of this pendulum.

The sea level value for the acceleration of gravity at 45° latitude in SI units is: 9.81 m/s². 32.2 m/s². O 32.2 ft/s². 9.81 ft/s².

Consider the case of a rotating wheel at rest and starting a clockwise rotation, meaning the negative direction of the angular velocity, and increasing (negatively) its value up to -12 rad/sec for 2 seconds. It then maintains a constant velocity for 2 seconds, and then uniformly reduces the magnitude of the velocity for 2 seconds until the wheel is momentarily stopped and restarts its rotation counter-clockwise with positive angular velocity, accelerating up to 20 rad/sec in 2 seconds and remaining at a constant rotation for 2 more seconds. Finally, the wheel stops gradually in 2 seconds. Next, you can see the graph of angular velocity versus time of this rotation: Apply the angular position equation.   with θo=0, wo=0, substituting the value of the angular acceleration in the range from 0 to 2 seconds obtained in question 2, perform the tabulation of values to fill the following table; describe the type of parabola and draw the graph: Equation:θ=f(t)   Concavity type:…

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A mass m = 1.0 kg moves under the action of the force given by F = 2i + 6tj, where t is in seconds. Yes, at t = 0 the mass starts from rest from the origin, determine (a) its acceleration, velocity and position as functions of t, (b) its linear momentum p, angular momentum L, as functions of t, (c) evaluate the previous sections for t = 2 s.