For this problem, we use the Coulomb's law written in equation as:
F = kQ₁Q₂/d²
where
F is the electrical force
k is a constant equal to 9×10⁹
Q₁ and Q₂ are the charge of the two objects
d is the distance between the two objects
Substituting the values:
F = (9×10⁹)(-22×10⁻⁹ C)(-22×10⁻⁹ C)/(0.10 m)²
F = 0.0004356 N
Complete Question
A Ferris wheel on a California pier is 27 m high and rotates once every 32 seconds in the counterclockwise direction. When the wheel starts turning, you are at the very top.
What is your angular position 75 seconds after the wheel starts turning, measured counterclockwise from the top? Express your answer as an angle between 0∘ and 360∘. Express your answer in degrees.
Answer:

Explanation:
From the question we are told that:
Height 
Period 
Time 
Generally the equation for angular velocity is mathematically given by



Therefore



Therefore


Answer:
The tank is losing

Explanation:
According to the Bernoulli’s equation:
We are being informed that both the tank and the hole is being exposed to air :
∴ P₁ = P₂
Also as the tank is voluminous ; we take the initial volume
≅ 0 ;
then
can be determined as:![\sqrt{[2g (h_1- h_2)]](https://tex.z-dn.net/?f=%5Csqrt%7B%5B2g%20%28h_1-%20h_2%29%5D)
h₁ = 5 + 15 = 20 m;
h₂ = 15 m
![v_2 = \sqrt{[2*9.81*(20 - 15)]](https://tex.z-dn.net/?f=v_2%20%3D%20%5Csqrt%7B%5B2%2A9.81%2A%2820%20-%2015%29%5D)
![v_2 = \sqrt{[2*9.81*(5)]](https://tex.z-dn.net/?f=v_2%20%3D%20%5Csqrt%7B%5B2%2A9.81%2A%285%29%5D)
as it leaves the hole at the base.
radius r = d/2 = 4/2 = 2.0 mm
(a) From the law of continuity; its equation can be expressed as:
J = 
J = πr²
J =
J =
b)
How fast is the water from the hole moving just as it reaches the ground?
In order to determine that; we use the relation of the velocity from the equation of motion which says:
v² = u² + 2gh
₂
v² = 9.9² + 2×9.81×15
v² = 392.31
The velocity of how fast the water from the hole is moving just as it reaches the ground is : 

Answer:
The distance between the camera and the rock is 836.6 cm
Explanation:
A right triangle is formed where the hypotenuse (h) is the distance between the rock and the camera. One of the leg (l) is the distance between the camera and the surface. The angle between the hypotenuse and this leg is α = 90° - 13.69° = 76.31°. By definition:
cos α = adjacent/hypotenuse
cos(76.31) = 198.0/h
h = 198.0/cos(76.31)
h = 836.6 cm