Speed can be found through the application of concepts related to potential energy and kinetic energy, for which you have
Where,
m = mass
v = Velocity
g = Gravitational acceleration
h = Height
Re-arrange to find the velocity we have,
Therefore the speed at which water squirts out of the hole is .3253m/s
Answer:
89.45 v/v
Explanation:
Let's take the data:
First draw the amplifier circuit.
After the circuit, the voltage division rule can be used to compute the parameters:
The input section is computed like this:
The output section is computed like this
The product A gives
A = A××
Computing gives output voltage = 89.45 v/v
Answer:
This can be translated to:
"find the electrical charge of a body that has 1 million of particles".
First, it will depend on the charge of the particles.
If all the particles have 1 electron more than protons, we will have that the charge of each particle is q = -e = -1.6*10^-19 C
Then the total charge of the body will be:
Q = 1,000,000*-1.6*10^-19 C = -1.6*10^-13 C
If we have the inverse case, where we in each particle we have one more proton than the number of electrons, the total charge will be the opposite of the one of before (because the charge of a proton is equal in magnitude but different in sign than the charge of an electron)
Q = 1.6*10^-13 C
But commonly, we will have a spectrum with the particles, where some of them have a positive charge and some of them will have a negative charge, so we will have a probability of charge that is peaked at Q = 0, this means that, in average, the charge of the particles is canceled by the interaction between them.
Answer:
vB = 15.4 m/s
Explanation:
Principle of conservation of energy:
Because there is no friction the mechanical energy is conserve
ΔE = 0
ΔE : mechanical energy change (J)
K : Kinetic energy (J)
U: Potential energy (J)
K = (1/2)mv²
U = m*g*h
Where :
m: mass (kg)
v : speed (m/s)
h : hight (m)
Ef - Ei = 0
(K+U)final - (K+U)initial =0
(K+U)final = (K+U)initial
((1/2)mv²+m*g*h)final = ((1/2)mv²+m*g*h)initial , We divided by m both sides of the equation:
((1/2)vB² + g*hB = (1/2 )vA²+ g*hA
(1/2) (vB)² + (9.8)*(14.7) = 0 + (9.8)(26.8 )
(1/2) (vB)² = (9.8)(26.8 ) - (9.8)*(14.7)
(vB)² = (2)(9.8)(26.8 - 14.7)
(vB)² = 237.16
vB = 15.4 m/s : speed of the cart at B