Phys 165 Module 5 Exam
A 1475-kg automobile is travelling at 17.0 m/s. What is the magnitude and course of the internet
pressure acting on the auto whilst it involves a prevent in five.32 s? - ANS-Use FΔT=mvf-mvi
FdΔt= (1475) (0) - (1475) *(17)
F 5.32= zero-25075
F= -25075 /5
F= -4713.34
*F= -4710 N*
path is opposite the preliminary velocity
A 192-kg red bumper automobile is shifting to the East and toward a one hundred forty five-kg
green bumper automobile. The pink bumper vehicle is transferring at four.78 m/s and the green
one is at relaxation. If they collide head-on and the pink bumper automobile continues to move
forward at 1.19 m/s, how speedy does the inexperienced vehicle flow after the collision, and in
what route does the green vehicle circulate? - ANS-mv1initial+mv2initial=mv1final+mv2final
(192)x (4.78)+(one hundred forty five) x (0)= (192) x (1.19)+ (a hundred forty five) (V2f)
917.Seventy six= 228.48+ 145V2f
689.28 =145 V2f
*V2f= four.75 m/s East*
A forty five,000-kg empty teach automobile is shifting at four.00 m/s and it collides with another
train automobile with a mass of 65,000 kg that is first of all at rest.
A. How fast do they move once they collide in the event that they become coupled?
B. How tons kinetic strength is lost because of the collision? - ANS-a)
mvcar1inital+mvcar2initial =mvcar1final+mvcar2final
(forty five,000) x (four) + (sixty five,000) x (0) = (forty five,000 + sixty six,000) * Vf
180,00= one hundred ten,000V
1.64 m/s
*Vf=1.6m/s*
b)
KElost = Ki-Kf
(half m1v1^2 + 1/2m2v2^2)- (1/2 (m1+m2)v^2)
(half x 45000 x (four^2) +(half x 65000 x (0^2) - (1/2 x 11000) (0.611^2)
KElost= 360,000- 2053.2055
KE lost= 357946.7345
*KE lost = 360000J ------ (0.36 MJ)*
A bowling ball's rotational inertia can be discovered the use of the subsequent equation: .
Consider a 4.Eighty five-kg bowling ball with a radius of zero.101 m spinning at a hundred and
fifteen rad/s. Calculate the subsequent values:
a. Calculate the rotational inertia of the bowling ball.
A 1475-kg automobile is travelling at 17.0 m/s. What is the magnitude and course of the internet
pressure acting on the auto whilst it involves a prevent in five.32 s? - ANS-Use FΔT=mvf-mvi
FdΔt= (1475) (0) - (1475) *(17)
F 5.32= zero-25075
F= -25075 /5
F= -4713.34
*F= -4710 N*
path is opposite the preliminary velocity
A 192-kg red bumper automobile is shifting to the East and toward a one hundred forty five-kg
green bumper automobile. The pink bumper vehicle is transferring at four.78 m/s and the green
one is at relaxation. If they collide head-on and the pink bumper automobile continues to move
forward at 1.19 m/s, how speedy does the inexperienced vehicle flow after the collision, and in
what route does the green vehicle circulate? - ANS-mv1initial+mv2initial=mv1final+mv2final
(192)x (4.78)+(one hundred forty five) x (0)= (192) x (1.19)+ (a hundred forty five) (V2f)
917.Seventy six= 228.48+ 145V2f
689.28 =145 V2f
*V2f= four.75 m/s East*
A forty five,000-kg empty teach automobile is shifting at four.00 m/s and it collides with another
train automobile with a mass of 65,000 kg that is first of all at rest.
A. How fast do they move once they collide in the event that they become coupled?
B. How tons kinetic strength is lost because of the collision? - ANS-a)
mvcar1inital+mvcar2initial =mvcar1final+mvcar2final
(forty five,000) x (four) + (sixty five,000) x (0) = (forty five,000 + sixty six,000) * Vf
180,00= one hundred ten,000V
1.64 m/s
*Vf=1.6m/s*
b)
KElost = Ki-Kf
(half m1v1^2 + 1/2m2v2^2)- (1/2 (m1+m2)v^2)
(half x 45000 x (four^2) +(half x 65000 x (0^2) - (1/2 x 11000) (0.611^2)
KElost= 360,000- 2053.2055
KE lost= 357946.7345
*KE lost = 360000J ------ (0.36 MJ)*
A bowling ball's rotational inertia can be discovered the use of the subsequent equation: .
Consider a 4.Eighty five-kg bowling ball with a radius of zero.101 m spinning at a hundred and
fifteen rad/s. Calculate the subsequent values:
a. Calculate the rotational inertia of the bowling ball.
