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v = 4.02 m/s angle = 54.9° above the horizontal (which is 35.1° from the vertical) That is .552 m above the board, or 4.552 m above the water v = 9.72 m/s angle = 76.6° below the horizontal (which is 13.4° from the vertical)

Explanation:

write and understand the given problem

a)

Solve the x component

d = vt

3 = v(1.3)

v = 2.308 m/s in the x direction

Then for the y...

d = vot + 0.5 at^2

-4 = v (1.3) + (0.5) (-9.8) (1.3) ^2

v = 3.29 m/s

Now we can solve a)

The total initial velocity is from the Pythagorean theorem

v2 = (3.29) ^2 + (2.308) ^2

v = 4.02 m/s

b)

The direction from the tangent function

tan (angle) = 3.29 / 2.308

angle = 54.9° above the horizontal (which is 35.1° from the vertical)

c)

Apply vf  ^2 = vo ^2 + 2ad in the y direction

0 = 3.292 + 2(9.8) (d)

d = 0.552 m

That is .552 m above the board, or 4.552 m above the water

d)

to find the y direction

vf^ 2 = vo^ 2 + 2ad  

vf^ 2 = (3.29) ^2 + 2 (9.8) (4)

v = 9.45 m/s

Find the net by the Pythagorean Theorem

v^ 2 = (9.45) ^2 + (2.308) ^2

v = 9.72 m/s

e)

Direction by the tangent function

tan (angle) = 9.72 / 2.308

angle = 76.6° below the horizontal (which is 13.4° from the vertical)

1.11×10⁸  Electrons.

Explanation:

Applying,

Q = CV Equation 1

Where Q = Total charge on the capacitor, C = Capacitance of the capacitor, V = Voltage of the battery.

Given: C = 1.97 pF = 1.97×10⁻¹² F, V = 9.0 V

Substitute this value into equation 1

Q = 1.97×9×10⁻¹²

Q = 1.773×10⁻¹¹ C.

If the charge on one electron = 1.602×10⁻¹⁹ C.

Therefore number of electron = 1.773×10⁻¹¹ /1.602×10⁻¹⁹

Number of electron = 1.11×10⁸