n Olympic diver is on a diving platform 3.60 m above the water. To start her dive, she runs off of the platform with a speed of 1.3 m/s in the horizontal direction. What is the diver's speed, in m/s, just before she enters the water

8.5m/s

Explanation:

Using the equation of motion

v² = u² + 2gH

v is the final speed of the diver

u  is the initial speed of the diver

g is the acceleration due to gravity

H is the height of the object

Given the following

V = 1.3m/s

H = 3.60m

g = 9.8m/s²

Required

Initial speed of the diver u

Substitute the given values into the formula:

v² = u² -2gH

1.3² = u²  - 2(9.8)(3.60)

1.69 = u²-70.56

u² = 1.69+70.56

u² = 72.25

u = √72.25

u = 8.5m/s

Hence the diver's speed, in m/s, just before she enters the water is 8.5m/s

absorption lines are where light has been absorbed by the atom thus you see a dip in the spectrum whereas.

emission spectra have spikes in the spectra due to atoms releasing photons at those wavelengths.

explanation:

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west