Answer:
Technician B
Explanation:
Both AWG and metric are sized by cross-sectional area.
Technician A is wrong: 12 gauge wire is larger diameter rated for 20 amps in free air. 18 awg is smaller diameter and typically used for speaker wiring, Class II or low voltage and sub-circuits within appliances.
Answer:
The correct answer will be "400.4 N". The further explanation is given below.
Explanation:
The given values are:
Mass of truck,
m = 600 kg
g = 9.8 m/s²
On equating torques at the point O,
⇒
So that,
On putting the values, we get
⇒ 
⇒ 
Answer:
a) V(t) = Ldi(t)/dt
b) If current is constant, V = 0
Explanation:
a) The voltage, V(t), across an inductor is proportional to the rate of change of the current flowing across it with time.
If V represents the Voltage across the inductor
and i(t) represents the current across the inductor in time, t.
V(t) ∝ di(t)/dt
Introducing a proportionality constant,L, which is the inductance of the inductor
The general equation describing the voltage across the inductor of inductance, L, as a function of time when a current flows through it is shown below.
V(t) = Ldi(t)/dt ..................................................(1)
b) If the current flowing through the inductor is constant i.e. does not vary with time
di(t)/dt = 0 and hence the general equation (1) above becomes
V(t) = 0
The thickness of aluminium needed to stop the beam electrons, protons and alpha particles at the given dfferent kinetic energies is 1.5 x 10⁻¹⁴ m.
<h3>
Thickness of the aluminum</h3>
The thickness of the aluminum can be determined using from distance of closest approach of the particle.

where;
- Z is the atomic number of aluminium = 13
- e is charge
- r is distance of closest approach = thickness of aluminium
- k is Coulomb's constant = 9 x 10⁹ Nm²/C²
<h3>For 2.5 MeV electrons</h3>

<h3>For 2.5 MeV protons</h3>
Since the magnitude of charge of electron and proton is the same, at equal kinetic energy, the thickness will be same. r = 1.5 x 10⁻¹⁴ m.
<h3>For 10 MeV alpha-particles</h3>
Charge of alpah particle = 2e

Thus, the thickness of aluminium needed to stop the beam electrons, protons and alpha particles at the given dfferent kinetic energies is 1.5 x 10⁻¹⁴ m.
Learn more about closest distance of approach here: brainly.com/question/6426420