The current is defined as the ratio between the charge Q flowing through a certain point of a wire and the time interval,
:
First we need to find the net charge flowing at a certain point of the wire in one second,
. Using I=0.92 A and re-arranging the previous equation, we find
Now we know that each electron carries a charge of
, so if we divide the charge Q flowing in the wire by the charge of one electron, we find the number of electron flowing in one second:
Explanation:
20 joule is your answer
Answer:
here
mass m =100kg
distance d=50m
acceleration due to gravity a =10m/s²
work =force×displacement
= ma/d=100×10/50=20joule
Answer:
v = 5.34[m/s]
Explanation:
In order to solve this problem, we must use the theorem of work and energy conservation. This theorem tells us that the sum of the mechanical energy in the initial state plus the work on or performed by a body must be equal to the mechanical energy in the final state.
Mechanical energy is defined as the sum of energies, kinetic, potential, and elastic.
E₁ = mechanical energy at initial state [J]
In the initial state, we only have kinetic energy, potential energy is not had since the reference point is taken below 1.5[m], and the reference point is taken as potential energy equal to zero.
In the final state, you have kinetic energy and potential since the car has climbed 1.5[m] of the hill. Elastic energy is not available since there are no springs.
E₂ = mechanical energy at final state [J]
Now we can use the first statement to get the first equation:
where:
W₁₋₂ = work from the state 1 to 2.
where:
h = elevation = 1.5 [m]
g = gravity acceleration = 9.81 [m/s²]
![58 = v^{2} +29.43\\v^{2} =28.57\\v=\sqrt{28.57}\\v=5.34[m/s]](https://tex.z-dn.net/?f=58%20%3D%20v%5E%7B2%7D%20%2B29.43%5C%5Cv%5E%7B2%7D%20%3D28.57%5C%5Cv%3D%5Csqrt%7B28.57%7D%5C%5Cv%3D5.34%5Bm%2Fs%5D)
The crate’s acceleration is gravity= 9.81m/s^2
The ball’s acceleration is also gravity=9.81m/s^2
There’s not enough info here i’m sorry
