Q 1 If two particles have equal kinetic energies, are their momentanecessarily e
ID: 1724287 • Letter: Q
Question
Q 1
If two particles have equal kinetic energies, are their momentanecessarily equal?Explain. Marks 2
Q 2
Where is the center of mass of Earth's atmosphere? Marks2
Q 3
An airplane propeller is rotating at 1900 rev/min.Compute thepropeller's angular velocity inrad/s. Marks 3
Q 4
Two ice dancers are at rest on the ice, facing each other withtheir hands together. They push off on each other in order to seteach other in motion. The subsequent momentum change (magnitudeonly) of the two skaters will be ____. Also write it’s (theanswer you choose) detail reason. Marks5
A. greatest for the skater who is pushed upon with the greatestforce
B. greatest for the skater who pushes with the greatestforce
C. the same for each skater
D. greatest for the skater with the most mass
E. greatest for the skater with the leastmass
Q 5
How much work is done by an applied force to lift a 20-Newtonblock 2.0 meters vertically at a constantspeed? Marks 3
Q 6
A student with a mass of 50.0 kg runs up three flights of stairsin 10.0 sec. The student has gone a vertical distance of 8.0 m.Determine the amount of work done by the student to elevate hisbody to this height. Assume that her speed is constant. Marks5
Explanation / Answer
Question 4:
Answer: C In thissituation, the force on the first ice dancer is the same as theforce on the second ice dancer (Newton's third law of motion). Andthese forces act for the same amount of time to cause equalimpulses on each skater. Since impulse is equal to momentum change,both skaters must also have equal momentum changes. The mass of theindividual skaters will only effect the subsequent velocitychange.
Question 5:
To lift a 20-Newton block at constant speed, 20-Nof force must be applied to it (Newton's laws). Thus, W = (20 N) *(2 m) * cos (0 degrees) = 40Joules
Question 6:
The student weighs 490 N (Fgrav= 50 kg * 9.8m/s/s).
To lift a 490-Newton person at constant speed, 490 N offorce must be applied to it (Newton's laws). The force is up, thedisplacement is up, and so the angle theta in the work equation is0 degrees. Thus,
W = (490 N) * (8 m) * cos (0 degrees)= 3920 Joules
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