Portage Phys 165 Module 5 Exam
A 1475-kg car is traveling at 17.Zero m/s. What is the magnitude and direction of the internet
pressure appearing on the car even as it comes to a stop in 5.32 s? - ANS-Use
FdeltaT=mvf-mvi
Answer in Newtons
path is opposite the preliminary speed
A 192-kg red bumper vehicle is transferring to the East and towards a a hundred forty five-kg
inexperienced bumper car. The crimson bumper automobile is moving at 4.Seventy eight m/s
and the inexperienced one is at rest. If they collide head-on and the crimson bumper vehicle
keeps to move forward at 1.19 m/s, how rapid does the inexperienced automobile move after
the collision, and in what path does the inexperienced car flow? -
ANS-mv1initial+mv2initial=mv1final+mv2final
Solve for v2f answer in m/s heading East
A forty five,000-kg empty train vehicle is shifting at four.00 m/s and it collides with another
educate vehicle with a mass of 65,000 kg that is first of all at rest.
A. How rapid do they pass after they collide if they become coupled?
B. How much kinetic strength is lost because of the collision? - ANS-a)
mvcar1inital+mvcar2initial =mvcar1final+mvcar2final
remedy for vf=1.6m/s
b)
KEf=1/2mv^2 answer in MJ
KEinitial=1/2mv^2 solution in MJ
deltaKE=KEf=KEi answer in MJ
A bowling ball's rotational inertia can be found the usage of the subsequent equation: . Consider
a 4.Eighty five-kg bowling ball with a radius of 0.One zero one m spinning at a hundred and
fifteen rad/s. Calculate the following values:
a. Calculate the rotational inertia of the bowling ball.
B. Calculate the angular momentum of the bowling ball.
C. Calculate the rotational kinetic power of the bowling ball. - ANS-a) I=2/5mR^2 solution in
kgm^2
b) angular momentum=Iw=0.0198kgm^2*115rad/s=2.28kgm^2/s
c)KEr=1/2Iw^2=131J
During a 4.50-min journey, a carousel goes around at a fee of five.25 rpm. Answer the
subsequent questions on this trip, assuming its angular pace is constant.
A. What is the angular speed in rad/s?B. Through what angular displacement will the riders go
through for the duration of one 4.50-min ride?C. How many rotations will this be? - ANS-a)
w=5.25*2pirad/1rev*1 min/60s = 0.550 rad/s
b)theta=wt=zero.550rad/s*270s = 149 rad
A 1475-kg car is traveling at 17.Zero m/s. What is the magnitude and direction of the internet
pressure appearing on the car even as it comes to a stop in 5.32 s? - ANS-Use
FdeltaT=mvf-mvi
Answer in Newtons
path is opposite the preliminary speed
A 192-kg red bumper vehicle is transferring to the East and towards a a hundred forty five-kg
inexperienced bumper car. The crimson bumper automobile is moving at 4.Seventy eight m/s
and the inexperienced one is at rest. If they collide head-on and the crimson bumper vehicle
keeps to move forward at 1.19 m/s, how rapid does the inexperienced automobile move after
the collision, and in what path does the inexperienced car flow? -
ANS-mv1initial+mv2initial=mv1final+mv2final
Solve for v2f answer in m/s heading East
A forty five,000-kg empty train vehicle is shifting at four.00 m/s and it collides with another
educate vehicle with a mass of 65,000 kg that is first of all at rest.
A. How rapid do they pass after they collide if they become coupled?
B. How much kinetic strength is lost because of the collision? - ANS-a)
mvcar1inital+mvcar2initial =mvcar1final+mvcar2final
remedy for vf=1.6m/s
b)
KEf=1/2mv^2 answer in MJ
KEinitial=1/2mv^2 solution in MJ
deltaKE=KEf=KEi answer in MJ
A bowling ball's rotational inertia can be found the usage of the subsequent equation: . Consider
a 4.Eighty five-kg bowling ball with a radius of 0.One zero one m spinning at a hundred and
fifteen rad/s. Calculate the following values:
a. Calculate the rotational inertia of the bowling ball.
B. Calculate the angular momentum of the bowling ball.
C. Calculate the rotational kinetic power of the bowling ball. - ANS-a) I=2/5mR^2 solution in
kgm^2
b) angular momentum=Iw=0.0198kgm^2*115rad/s=2.28kgm^2/s
c)KEr=1/2Iw^2=131J
During a 4.50-min journey, a carousel goes around at a fee of five.25 rpm. Answer the
subsequent questions on this trip, assuming its angular pace is constant.
A. What is the angular speed in rad/s?B. Through what angular displacement will the riders go
through for the duration of one 4.50-min ride?C. How many rotations will this be? - ANS-a)
w=5.25*2pirad/1rev*1 min/60s = 0.550 rad/s
b)theta=wt=zero.550rad/s*270s = 149 rad
