An Olympic diver is on a diving platform 5.70 m above the water. To start her dive, she runs off of the platform with a speed of 1.27 m/s in the horizontal direction. What is the diver's speed, in m/s, just before she enters the water?

10.65 m/s

Explanation:

Given that an  Olympic diver is on a diving platform 5.70 m above the water. To start her dive, she runs off of the platform with a speed of 1.27 m/s in the horizontal direction.

Her initial velocity U = 1.27 m/s

The height H = 5.7 m

The final velocity V will be calculated by using the formula;

V^2 = U^2 + 2gH

Substitute U, g and H into the formula

V^2 = 1.27^2 + 2 × 9.8 × 5.7

V^2 = 1.6129 + 111.72

V^2 = 113.33

V = sqrt ( 113.33)

V = 10.65 m/s

Therefore, the diver's speed, in m/s, just before she enters the water is 10.65

electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j