You turn off your electric drill and find that the time interval for the rotating bit to come to rest due to frictional torque in the drill is Δ t . You replace the bit with a larger one that results in a doubling of the moment of inertia of the drill’s entire rotating mechanism. When this larger bit is rotated at the same angular speed as the first and the drill is turned off, the frictional torque remains the same as that for the previous situation. What is the time interval for this second bit to come to rest? (a) 4Δ t (b) 2Δ t (c) Δ t (d) 0.5Δ t (e) 0.25Δ t (f) impossible to determine
You turn off your electric drill and find that the time interval for the rotating bit to come to rest due to frictional torque in the drill is Δ t . You replace the bit with a larger one that results in a doubling of the moment of inertia of the drill’s entire rotating mechanism. When this larger bit is rotated at the same angular speed as the first and the drill is turned off, the frictional torque remains the same as that for the previous situation. What is the time interval for this second bit to come to rest? (a) 4Δ t (b) 2Δ t (c) Δ t (d) 0.5Δ t (e) 0.25Δ t (f) impossible to determine
Solution Summary: The author explains that frictional torque opposes the rotation of the drill and causes to stop it.
You turn off your electric drill and find that the time interval for the rotating bit to come to rest due to frictional torque in the drill is Δt. You replace the bit with a larger one that results in a doubling of the moment of inertia of the drill’s entire rotating mechanism. When this larger bit is rotated at the same angular speed as the first and the drill is turned off, the frictional torque remains the same as that for the previous situation. What is the time interval for this second bit to come to rest? (a) 4Δt(b) 2Δt(c) Δt(d) 0.5Δt(e) 0.25Δt(f) impossible to determine
A 0.500 kg sphere moving with a velocity given by (2.00î – 2.60ĵ + 1.00k) m/s strikes another sphere of mass 1.50 kg moving with an initial velocity of (−1.00î + 2.00ĵ – 3.20k) m/s.
(a) The velocity of the 0.500 kg sphere after the collision is (-0.90î + 3.00ĵ − 8.00k) m/s. Find the final velocity of the 1.50 kg sphere.
R =
m/s
Identify the kind of collision (elastic, inelastic, or perfectly inelastic).
○ elastic
O inelastic
O perfectly inelastic
(b) Now assume the velocity of the 0.500 kg sphere after the collision is (-0.250 + 0.850ĵ - 2.15k) m/s. Find the final velocity of the 1.50 kg sphere.
✓ =
m/s
Identify the kind of collision.
O elastic
O inelastic
O perfectly inelastic
(c) Take the velocity of the 0.500 kg sphere after the collision as (−1.00ỉ + 3.40] + ak) m/s. Find the value of a and the velocity of the 1.50 kg sphere after an elastic collision. (Two values of a are possible, a positive value and a negative value. Report each with their
corresponding final velocities.)
a…
A cannon is rigidly attached to a carriage, which can move along horizontal rails, but is connected to a post by a large spring, initially unstretched and with force constant k = 1.31 x 104 N/m, as in the figure below. The cannon fires a 200-kg projectile at a velocity of 136 m/s directed 45.0°
above the horizontal.
45.0°
(a) If the mass of the cannon and its carriage is 5000 kg, find the recoil speed of the cannon.
m/s
(b) Determine the maximum extension of the spring.
m
(c) Find the maximum force the spring exerts on the carriage. (Enter the magnitude of the force.)
N
launch angle.
Passage Problems
Alice (A), Bob (B), and Carrie (C) all start from their dorm and head
for the library for an evening study session. Alice takes a straight path,
Chapter 10 Solutions
Physics for Scientists and Engineers, Technology Update, Hybrid Edition (with Enhanced WebAssign Multi-Term LOE Printed Access Card for Physics)
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