At a construction site, the site manager notices that a crane takes 20 seconds to lift a 500kg steel beam up to a height of 15 m
1 answer:
You might be interested in
Answer:
Ф,
Ф
Explanation:
Now find the components NxNxN_x and NyNyN_y of N⃗ N→N_vec in the tilted coordinate system of Part B. Express your answer in terms of the length of the vector NNN and the angle θθtheta, with the components separated by a comma.
Vectors are quantities that have both magnitude and direction while scalar quantities have only magnitude but no direction.
This a vector quantity
from the diagram the horizontal component of the length of the vector will be
Ф
the vertical component will be
Ф
this is in the opposite direction because the x can be extrapolated to the negative axis
relation between potential difference and electric field is given as
so here we know that
d = 3 cm
So now when plates are separated to 4 cm distance carefully
the potential difference between them will change but the electric field between them will remain constant
So at distance of 4 cm also the electric field will be E = 1000 N/C
Answer:
L = μ₀ n r / 2I
Explanation:
This exercise we must relate several equations, let's start writing the voltage in a coil
= - L dI / dt
Let's use Faraday's law
E = - d Ф_B / dt
in the case of the coil this voltage is the same, so we can equal the two relationships
- d Ф_B / dt = - L dI / dt
The magnetic flux is the sum of the flux in each turn, if there are n turns in the coil
n d Ф_B = L dI
we can remove the differentials
n Ф_B = L I
magnetic flux is defined by
Ф_B = B . A
in this case the direction of the magnetic field is along the coil and the normal direction to the area as well, therefore the scalar product is reduced to the algebraic product
n B A = L I
the loop area is
A = π R²
we substitute
n B π R² = L I (1)
To find the magnetic field in the coil let's use Ampere's law
∫ B. ds = μ₀ I
where B is the magnetic field and s is the current circulation, in the coil the current circulates along the length of the coil
s = 2π R
we solve
B 2ππ R = μ₀ I
B = μ₀ I / 2πR
we substitute in
n ( μ₀ I / 2πR) π R² = L I
n μ₀ R / 2 = L I
L = μ₀ n r / 2I