In step 1, to increase the potential energy, the iron will move towards the electromagnet.
In step 2, to increase the potential energy, the iron will move towards the electromagnet.
<h3>Potential energy of a system of magnetic dipole</h3>
The potential energy of a system of dipole depends on the orientation of the dipole in the magnetic field.
where;
is the dipole moment- B is the magnetic field
Increase in the distance (r) reduces the potential energy. Thus, we can conclude the following;
- In step 1, to increase the potential energy, the iron will move towards the electromagnet.
- In step 2, when the iron is rotated 180, it will still maintain the original position, to increase the potential energy, the iron will move towards the electromagnet.
Learn more about potential energy in magnetic field here: brainly.com/question/14383738
<span>Newton's Third Law of Action-Reaction is that for each and every action that happens, there is an equal and opposite reaction to it. In the scenario of a roller coaster, this is when you push down on the seat of the roller coaster as it flies along and the seat pushes back against you.</span>
Answer: 1.176×10^-3 s
Explanation: The time constant formulae for an RC circuit is given below as
t =RC
Where t = time constant , R = magnitude of resistance = 21 ohms , C = capacitance of capacitor = 56 uf = 56×10^-6 F
t = 56×10^-6 × 21
t = 1176×10^-6
t = 1.176×10^-3 s
Answer:
the shooting angle ia 18.4º
Explanation:
For resolution of this exercise we use projectile launch expressions, let's see the scope
R = Vo² sin (2θ) / g
sin 2θ = g R / Vo²
sin 2θ = 9.8 75/35²
2θ = sin⁻¹ (0.6)
θ = 18.4º
To know how for the arrow the tree branch we calculate the height of the arrow at this point
X2 = 75/2 = 37.5 m
We calculate the time to reach this point since the speed is constant on the X axis
X = Vox t
t2 = X2 / Vox = X2 / (Vo cosθ)
t2 = 37.5 / (35 cos 18.4)
t2 = 1.13 s
With this time we calculate the height at this point
Y = Voy t - ½ g t²
Y = 35 sin 18.4 1.13 - ½ 9.8 1,13²
Y = 6.23 m
With the height of the branch is 3.5 m and the arrow passes to 6.23, it passes over the branch
Answer:
a) The current density ,J = 2.05×10^-5
b) The drift velocity Vd= 1.51×10^-15
Explanation:
The equation for the current density and drift velocity is given by:
J = i/A = (ne)×Vd
Where i= current
A = Are
Vd = drift velocity
e = charge ,q= 1.602 ×10^-19C
n = volume
Given: i = 5.8×10^-10A
Raduis,r = 3mm= 3.0×10^-3m
n = 8.49×10^28m^3
a) Current density, J =( 5.8×10^-10)/[3.142(3.0×10^-3)^2]
J = (5.8×10^-10) /(2.83×10^-5)
J = 2.05 ×10^-5
b) Drift velocity, Vd = J/ (ne)
Vd = (2.05×10^-5)/ (8.49×10^28)(1.602×10^-19)
Vd = (2.05×10^-5)/(1.36 ×10^10)
Vd = 1.51× 10^-5
