Current = charge per second
2 Coulombs per second = 2 Amperes
Potential difference = (current)x(resistance) in volts.
That's (2 Amperes) x (2 ohms).
That's how to do it.
I think you can find the answer now.
Answer:
Implicit memory
Explanation:
Implicit memory stores skill-related data by repeating an activity that always follows the same pattern. It includes all motor, sensory, and intellectual skills, as well as every form of conditioning. The capacity thus acquired does not depend on consciousness. We are able to perform sometimes complex tasks with our thinking turned to something completely different. This explains why Adam types so fast without even remembering the exact location of the letters on the keyboard.
Answer:
a = 2 [m/s^2]
Explanation:
To solve this problem we must use the expressions of kinematics, we must bear in mind that when a body is at rest its velocity is zero.

where:
Vf = final velocity = 0
Vi = initial velocity = 60 [m/s]
a = desacceleration [m/s^2]
t = time = 30 [s]
Note: the negative sign of the above equation means that the car is slowing down, i.e. its speed decreases.
0 = 60 - (a*30)
a = 2 [m/s^2]
Answer:
The high of the ramp is 2.81[m]
Explanation:
This is a problem where it applies energy conservation, that is part of the potential energy as it descends the block is transformed into kinetic energy.
If the bottom of the ramp is taken as a potential energy reference point, this point will have a potential energy value equal to zero.
We can find the mass of the box using the kinetic energy and the speed of the box at the bottom of the ramp.
![E_{k}=0.5*m*v^{2}\\\\where:\\E_{k}=3.8[J]\\v = 2.8[m/s]\\m=\frac{E_{k}}{0.5*v^{2} } \\m=\frac{3.8}{0.5*2.8^{2} } \\m=0.969[kg]](https://tex.z-dn.net/?f=E_%7Bk%7D%3D0.5%2Am%2Av%5E%7B2%7D%5C%5C%5C%5Cwhere%3A%5C%5CE_%7Bk%7D%3D3.8%5BJ%5D%5C%5Cv%20%3D%202.8%5Bm%2Fs%5D%5C%5Cm%3D%5Cfrac%7BE_%7Bk%7D%7D%7B0.5%2Av%5E%7B2%7D%20%7D%20%5C%5Cm%3D%5Cfrac%7B3.8%7D%7B0.5%2A2.8%5E%7B2%7D%20%7D%20%5C%5Cm%3D0.969%5Bkg%5D)
Now applying the energy conservation theorem which tells us that the initial kinetic energy plus the work done and the potential energy is equal to the final kinetic energy of the body, we propose the following equation.
![E_{p}+W_{f}=E_{k}\\where:\\E_{p}= potential energy [J]\\W_{f}=23[J]\\E_{k}=3.8[J]\\](https://tex.z-dn.net/?f=E_%7Bp%7D%2BW_%7Bf%7D%3DE_%7Bk%7D%5C%5Cwhere%3A%5C%5CE_%7Bp%7D%3D%20potential%20energy%20%5BJ%5D%5C%5CW_%7Bf%7D%3D23%5BJ%5D%5C%5CE_%7Bk%7D%3D3.8%5BJ%5D%5C%5C)
And therefore
![m*g*h + W_{f}=3.8\\ 0.969*9.81*h - 23= 3.8\\h = \frac{23+3.8}{0.969*9.81}\\ h = 2.81[m]](https://tex.z-dn.net/?f=m%2Ag%2Ah%20%2B%20W_%7Bf%7D%3D3.8%5C%5C%200.969%2A9.81%2Ah%20-%2023%3D%203.8%5C%5Ch%20%3D%20%5Cfrac%7B23%2B3.8%7D%7B0.969%2A9.81%7D%5C%5C%20h%20%3D%202.81%5Bm%5D)