Answer:
Evaporation
Explanation:
Evaporation is a form of mass tranfer phenomena where by water are moved from the earth surface into the atmosphere as vapours,it is path of the water cycle a decription of the path moved by land water until it turns into rain, humidity,air and temperature are factors that influence evaporation though evaporation can happen at all temperature
The color white is what you'd see when every color of light is combined.
Answer:
14m/s
Explanation:
Given parameters:
Radius of the curve = 50m
Centripetal acceleration = 3.92m/s²
Unknown:
Speed needed to keep the car on the curve = ?
Solution:
The centripetal acceleration is the inwardly directly acceleration needed to keep a body along a curved path.
It is given as;
a =
a is the centripetal acceleration
v is the speed
r is the radius
Now insert the parameters and find v;
v² = ar
v² = 3.92 x 50 = 196
v = √196 = 14m/s
it is just a matter of integration and using initial conditions since in general dv/dt = a it implies v = integral a dt
v(t)_x = integral a_{x}(t) dt = alpha t^3/3 + c the integration constant c can be found out since we know v(t)_x at t =0 is v_{0x} so substitute this in the equation to get v(t)_x = alpha t^3 / 3 + v_{0x}
similarly v(t)_y = integral a_{y}(t) dt = integral beta - gamma t dt = beta t - gamma t^2 / 2 + c this constant c use at t = 0 v(t)_y = v_{0y} v(t)_y = beta t - gamma t^2 / 2 + v_{0y}
so the velocity vector as a function of time vec{v}(t) in terms of components as[ alpha t^3 / 3 + v_{0x} , beta t - gamma t^2 / 2 + v_{0y} ]
similarly you should integrate to find position vector since dr/dt = v r = integral of v dt
r(t)_x = alpha t^4 / 12 + + v_{0x}t + c let us assume the initial position vector is at origin so x and y initial position vector is zero and hence c = 0 in both cases
r(t)_y = beta t^2/2 - gamma t^3/6 + v_{0y} t + c here c = 0 since it is at 0 when t = 0 we assume
r(t)_vec = [ r(t)_x , r(t)_y ] = [ alpha t^4 / 12 + + v_{0x}t , beta t^2/2 - gamma t^3/6 + v_{0y} t ]
To explain, I will use the equations for kinetic and potential energy:

<h3>Potential energy </h3>
Potential energy is the potential an object has to move due to gravity. An object can only have potential energy if 1) <u>gravity is present</u> and 2) <u>it is above the ground at height h</u>. If gravity = 0 or height = 0, there is no potential energy. Example:
An object of 5 kg is sitting on a table 5 meters above the ground on earth (g = 9.8 m/s^2). What is the object's gravitational potential energy? <u>(answer: 5*5*9.8 = 245 J</u>)
(gravitational potential energy is potential energy)
<h3>Kinetic energy</h3>
Kinetic energy is the energy of an object has while in motion. An object can only have kinetic energy if the object has a non-zero velocity (it is moving and not stationary). An example:
An object of 5 kg is moving at 5 m/s. What is the object's kinetic energy? (<u>answer: 5*5 = 25 J</u>)
<h3>Kinetic and Potential Energy</h3>
Sometimes, an object can have both kinetic and potential energy. If an object is moving (kinetic energy) and is above the ground (potential), it will have both. To find the total (mechanical) energy, you can add the kinetic and potential energies together. An example:
An object of 5 kg is moving on a 5 meter table at 10 m/s. What is the objects mechanical (total) energy? (<u>answer: KE = .5(5)(10^2) = 250 J; PE = (5)(9.8)(5) = 245 J; total: 245 + 250 = 495 J</u>)