- Can You Please Help In Part F To Solve F Check The Previous Parts Solution It Related 1 (65.43 KiB) Viewed 34 times
can you please help in part f? To solve f check the previous parts solution it related.
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can you please help in part f? To solve f check the previous parts solution it related.
can you please help in part f? To solve f check the previousparts solution it related.
A 0.00600 kg bullet traveling horizontally with speed 1.00 x 10³ m/s strikes a 16.1 kg door, embedding itself 10.6 cm from the side opposite the hinges as shown in the figure below. The 1.00 m wide door is free to swing on its frictionless hinges. Hinge (a) Before it hits the door, does the bullet have angular momentum relative the door's axis of rotation? Ⓒ Yes O No (b) If so, evaluate this angular momentum (in kg - m²/s). (If not, enter zero.) 5.37 ✓kg.m²/s If not, explain why there is no angular momentum. Hence the angular momentum of the bullet relative to the door's axis is 0.636kg.m^2/s Score: 0.12 out of 0.12 Comment: (c) Is mechanical energy of the bullet-door system constant in this collision? Answer without doing a calculation. O Yes Ⓒ No (d) At what angular speed (in rad/s) does the door swing open immediately after the collision? 0.98 rad/s (e) Calculate the total energy of the bullet-door system and determine whether it is less than or equal to the kinetic energy of the bullet before the collision. (Enter your answers in J.) KE - 2.58 KE, = 3000 (f) What If? Imagine now that the door is hanging vertically downward, hinged at the top, so that the figure is a side view of the door and bullet during the collision. What is the maximum height (in cm) that the bottom of the door will reach after the collision? x cm
A 0.00600 kg bullet traveling horizontally with speed 1.00 x 10³ m/s strikes a 16.1 kg door, embedding itself 10.6 cm from the side opposite the hinges as shown in the figure below. The 1.00 m wide door is free to swing on its frictionless hinges. Hinge (a) Before it hits the door, does the bullet have angular momentum relative the door's axis of rotation? Ⓒ Yes O No (b) If so, evaluate this angular momentum (in kg - m²/s). (If not, enter zero.) 5.37 ✓kg.m²/s If not, explain why there is no angular momentum. Hence the angular momentum of the bullet relative to the door's axis is 0.636kg.m^2/s Score: 0.12 out of 0.12 Comment: (c) Is mechanical energy of the bullet-door system constant in this collision? Answer without doing a calculation. O Yes Ⓒ No (d) At what angular speed (in rad/s) does the door swing open immediately after the collision? 0.98 rad/s (e) Calculate the total energy of the bullet-door system and determine whether it is less than or equal to the kinetic energy of the bullet before the collision. (Enter your answers in J.) KE - 2.58 KE, = 3000 (f) What If? Imagine now that the door is hanging vertically downward, hinged at the top, so that the figure is a side view of the door and bullet during the collision. What is the maximum height (in cm) that the bottom of the door will reach after the collision? x cm