Tony is driving a truck of mass 2.00 × 103 kilograms to the west with a velocity of 14.0 meters/second. he collides with a car of mass 1.20 × 103 kilograms moving to the south with velocity 21.0 meters/second. after the collision, they move together. find the velocity with which they move together.

11.7 m/s

Given:

M1 = 2,000kg, V1 = 14.0m/s

M2 = 1200kg, V2 = 21.0m/s

V3 = Velocity of M1, and M2 after colliding.

Momentum before = Momentum after.

M1×V1 + M2×V2 = M1*V3 + M2*V3.

2000×14  + 1200×21 = 2000V3 + 1200V3,

28,000+ 25,200= 3200V3,

Divide both sides by 3200:

V3 = 8.75 + 7.875 = -8.75 - 7.875i

  = 11.77 m/s

The common velocity will be  11.77m/s, in the south west direction.

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft

the answer is c. the lower legs are levers, and the knees are fulcrums. the ankles hold the loads.