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klemol [59]
3 years ago
11

For an object in uniform circular motion, what can you say about the directions of the velocity, acceleration, and net force vec

tors? For an object in uniform circular motion, what can you say about the directions of the velocity, acceleration, and net force vectors? a) The velocity vector is perpendicular to the acceleration vector; the acceleration vector is perpendicular to the net force vector. b) The velocity vector is perpendicular to the acceleration vector; the acceleration vector is parallel to the net force vector. c) The velocity vector is parallel to the acceleration vector; the acceleration vector is perpendicular to the net force vector. d) The velocity vector is parallel to the acceleration vector; the acceleration vector is parallel to the net force vector.
Physics
1 answer:
zavuch27 [327]3 years ago
3 0

Answer: b) The velocity vector is perpendicular to the acceleration vector; the acceleration vector is parallel to the net force vector.

Explanation: A change in velocity creates an acceleration. As the object rotates through the circular path it is constantly changing direction, and hence accelerating, which causes a constant force to act upon the object. This Force acts towards the center of curvature, directly toward the axis of rotation in a direction parallel to the acceleration of the body along the path. Because the object is moving perpendicular to the force, the path followed by the object is a circular one. Hence the velocity of the object is perpendicular to the acceleration.

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Misha Larkins [42]

Hi there!

The maximum deformation of the bumper will occur when the car is temporarily at rest after the collision. We can use the work-energy theorem to solve.

Initially, we only have kinetic energy:

KE = \frac{1}{2}mv^2

KE = Kinetic Energy (J)
m = mass (1060 kg)
v = velocity (14.6 m/s)

Once the car is at rest and the bumper is deformed to the maximum, we only have spring-potential energy:

U_s = \frac{1}{2}kx^2

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Since energy is conserved:

E_I = E_f\\\\KE = U_s\\\\\frac{1}{2}mv^2 = \frac{1}{2}kx^2

We can simplify and solve for 'x'.

mv^2 = kx^2\\\\x = \sqrt{\frac{mv^2}{k}}

Plug in the givens and solve.

x = \sqrt{\frac{(1060)(14.6^2)}{(1.14*10^7)}} = \boxed{0.0198 m}

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