3.5 seconds is 3500 milliseconds.
Let height of twin falls = x
height of seven falls = 2.5x
x + 2.5x = 420
3.5x = 420
x = 420/3.5 = 120
so twin falls = x = 120 ft
seven falls = 2.5x = 300 ft
Answer:
6858.5712 m/s
Explanation:
Given that:
Radius, r
R = 3.20 * 10^3.
Normal force = 0.5 * normal weight
Normal force = Fn ; Normal weight = Fg
Fn = 0.5Fg
Recall:
mv² / R = Fn + Fg
Fn = 0.5Fg
mv² / R = 0.5Fg + Fg
mv² /R = 1.5Fg
mv² = 1.5Fg * R
F = mg
mv² = 1.5* mg * R
v² = 1.5gR
v = sqrt(1.5gR)
V = sqrt(1.5 * 9.8 * 3.2 * 10^3)
V = sqrt(47.04^3)
V = 6858.5712 m/s
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
