Standing at the edge of a cliff 32.5m high, you do a ball. Later, you throw a second ball downward with an initial speed of 11m/s. Calculate the velocity if each of the balls when it hits the ground

The displacement of the ball

ℎ =

2 −

2

2

So, final speed of the first ball

1 = √2ℎ = √2 × 9.8 × 32.5 = 25.24

m

s

The change of it speed

∆1 = 25.24 − 0 = 25.34

m

s

The final speed of the second ball

2 = √

2 + 2ℎ = 27.53

m

s

The change of the second ball speed

∆2 = 27.53 − 11 = 16.53

m

s

Thus

∆1 > ∆2

∆1 > ∆2

Explanation:

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards

305.7

explanation:

density =m/v

density =917/3.00

density = 305.

approximately = 305.7

in four significant figures