Answer:
16250 kgm/s due south
Explanation:
Applying,
M = mv................. Equation 1
Where M = momentum, m = mass, v = velocity.
From the car,
Given: m = 1000 kg, v = 6.5 m/s
Substitute these values into equation 1
M = 1000(6.5)
M = 6500 kgm/s
For the truck,
Given: m = 3500 kg, v = 6.5 m/s
Substitute these values into equation 1
M' = 3500(6.5)
M' = 22750 kgm/s.
Assuming South to be negative direction,
From the question,
Total momentum of the two vehicles = (6500-22750)
Total momentum of the two vehicles = -16250 kgm/s
Hence the total momentum of the two vehicles is 16250 kgm/s due south
Answer:
Change in momentum will be -4.4 kgm/sec
So option (A) is correct option
Explanation:
Mass of the ball is given m = 0.10 kg
Initial velocity of ball 
And velocity after rebound 
We have to find the change in momentum
So change in momentum is equal to 
( here negative sign shows only direction )
So option (A) will be correct answer
Answer:
a. 21.68 rad/s b. 30.78 m/s c. 897 rev/min² d. 1085 revolutions
Explanation:
a. Its angular speed in radians per second ω = angular speed in rev/min × 2π/60 = 207 rev/min × 2π/60 = 21.68 rad/s
b. The linear speed of a point on the flywheel is gotten from v = rω where r = radius of flywheel = 1.42 m
So, v = rω = 1.42 m × 21.68 rad/s = 30.78 m/s
c. Using α = (ω₁ - ω)/t where α = angular acceleration of flywheel, ω = initial angular speed of wheel in rev/min = 21.68 rad/s = 207 rev/min, ω₁ = final angular speed of wheel in rev/min = 1410 rev/min = 147.65 rad/s, t = time in minutes = 80.5/60 min = 1.342 min
α = (ω₁ - ω)/t
= (1410 - 207)/(80.5/60)
= 60(1410 - 207)/80.5
= 60(1203)80.5
= 896.65 rev/min² ≅ 897 rev/min²
d. Using θ = ωt + 1/2αt²
where θ = number of revolutions of flywheel. Substituting the values of the variables from above, ω = 207 rev/min, α = 896.65 rev/min² and t = 80.5/60 min = 1.342 min
θ = ωt + 1/2αt²
= 207 × 1.342 + 1/2 × 896.65 × 1.342²
= 277.725 + 807.417
= 1085.14 revolutions ≅ 1085 revolutions
