How much momentum will a dumbbell of mass 10kg transfer to the floor if it falls from a height of 80cm (assume that the dumbbell stops after hitting the ground) ? Take g=10m/s^2 (Point stealers will be reported so better dont do it)

momentum = 40kgm/s

Explanation:

Given that the

Mass M = 10kg

Height H = 80 cm = 0.8m

g = 10m/s^2

The velocity of the dumbbell as it fall can be calculated by using the formula;

V^2 = U^2 + 2gH

Since the dumbbell fell from rest, initial velocity U = 0

Substitute g and H to get final velocity V

V^2 = 0 + 2 × 10 × 0.8

V^2 = 16

V = sqrt(16)

V = 4 m/s

Momentum is the product of mass and velocity. That is,

Momentum = MV

Substitute mass M and V into the formula

Momentum = 10 × 4 = 40kgm/s

Therefore, 40 kgm/s momentum will be transferred as the dumbbell fall to the floor.

the answer is a. the yacht traveled from the dock north to the 200-meter marker in the bay in less than 5 minutes, giving the passengers several more hours to fish.

im doing peak rn so sry if its wrong

pythagoras' theorem

square root (-5^2+9^2)

square root (25+81)

square root (106)

angle is tan^-1(9/5)

neg x, pos y