Place the next vector with its tail at the previous vector's head. ... To subtract vectors, proceed as if adding the two vectors, but flip the vector to be subtracted across the axes and then join it tail to head as if adding. Adding or subtracting any number of vectors yields a resultant vector.
Explanation:
Data is inappropriate
here, we need gauge of the wire i.e., diameter of the wire, so that we calculate the resistance by using the formula
R = ρl/A
where R= resistance ; Ω
l = length of wire ; m
A = area of wire ; m²
ρ = resistivity ; Ω-m
But in general ohms law is
V = I R
R = V/I ;
but here we also calculate "R" from length of wire in which the current is flowing.
I hope it is helpful to you.
Answer:
We cannot place three forces of 5g, 6g, and 12g in equilibrium.
Explanation:
Equilibrium means their sum must be zero.
Here the forces are 5g, 6g, and 12g.
For number of forces to be in equilibrium the magnitude of largest vector should be less than sum of the magnitude of other vectors.
Here
Magnitude of largest force = 12 g
Sum of magnitudes of other forces = 5g + 6g = 11g
Magnitude of largest force > Sum of magnitudes of other forces
So this forces cannot form equilibrium.
We cannot place three forces of 5g, 6g, and 12g in equilibrium.
The question doesn't give us enough information to answer.
The answer depends on the mass of the object, how long the force
acts on the object, the OTHER forces on the object, and whether the
object is free to move.
-- If you increase the force with which you push on a brick wall,
the amount of work done remains unchanged, namely Zero.
-- If you push on a pingpong ball with a force of 1 ounce for 1 second,
the ball accelerates substantially, it moves a substantial distance, and
so the work done is substantial.
-- But if you push on a battleship, even with a much bigger force ...
let's say 1 pound ... and keep pushing for a month ... the ship accelerates
microscopically, moves a microscopic distance, and the work done by
your force is microscopic.
Answer:
a) w = 4.24 rad / s
, b) α = 8.99 rad / s²
Explanation:
a) For this exercise we use the conservation of kinetic energy,
Initial. Vertical bar
Emo = U = m g h
Final. Just before touching the floor
Emf = K = ½ I w2
As there is no friction the mechanical energy is conserved
Emo = emf
mgh = ½ m w²
The moment of inertial of a point mass is
I = m L²
m g h = ½ (m L²) w²
w = √ 2gh / L²
The initial height h when the bar is vertical is equal to the length of the bar
h = L
w = √ 2g / L
Let's calculate
w = RA (2 9.8 / 1.09)
w = 4.24 rad / s
b) Let's use Newton's equation for rotational motion
τ = I α
F L = (m L²) α
The force applied is the weight of the object, which is at a distance L from the point of gro
mg L = m L² α
α = g / L
α = 9.8 / 1.09
α = 8.99 rad / s²
