Hey ／人 ◕ ‿‿ ◕ 人＼
The answer is transmission
uses a series of gears to transmit power to facilitate changes in speed .
GLAD TO HELP
~~~ ╔͎═͓═͙╗
~~~ ╚̨̈́═̈́﴾ ̥̂˖̫˖̥ ̂ )
Answer:
F = 768 N
Explanation:
It is given that,
Speed of the elevator, v = 3.2 m/s
Grain drops into the car at the rate of 240 kg/min, 
We need to find the magnitude of force needed to keep the car moving constant speed. The relation between the momentum and the force is given by :
Since, the speed is constant,
F = 768 N
So, the magnitude of force need to keep the car is 768 N. Hence, this is the required solution.
It is transferred by direct contact, say you touched it, you would feel the heat
I don't completely understand your drawing, although I can see that you certainly
did put a lot of effort into making it. But calculating the moment is easy, and we
can get along without the drawing.
Each separate weight has a 'moment'.
The moment of each weight is:
(the weight of it) x (its distance from the pivot/fulcrum) .
That's all there is to a 'moment'.
The lever (or the see-saw) is balanced when (the sum of all the moments
on one side) is equal to (the sum of the moments on the other side).
That's why when you're on the see-saw with a little kid, the little kid has to sit
farther away from the pivot than you do. The kid has less weight than you do,
so he needs more distance in order for his moment to be equal to yours.
Given Information:
Current = I = 20 A
Diameter = d = 0.205 cm = 0.00205 m
Length of wire = L = 1 m
Required Information:
Energy produced = P = ?
Answer:
P = 2.03 J/s
Explanation:
We know that power required in a wire is
P = I²R
and R = ρL/A
Where ρ is the resistivity of the copper wire 1.68x10⁻⁸ Ω.m
L is the length of the wire and A is the area of the cross-section and is given by
A = πr²
A = π(d/2)²
A = π(0.00205/2)²
A = 3.3x10⁻⁶ m²
R = ρL/A
R = 1.68x10⁻⁸*(1)/3.3x10⁻⁶
R = 5.09x10⁻³ Ω
P = I²R
P = (20)²*5.09x10⁻³
P = 2.03 Watts or P = 2.03 J/s
Therefore, 2.03 J/s of energy is produced in 1.00 m of 12-gauge copper wire carrying a current of 20 A
