Answer: the half-angle "alpha" of the Mach cone = 30⁰
Explanation:
To calculate the half-angle "alpha" of the Mach cone.
we say ;
Sin∝ = 1 / Ma
given that Ma = 2
now we substitute
Sin∝ = 1 / 2
Sin∝ = 0.5
∝ = Sin⁻¹ 0.5
∝ = 30⁰
Therefore, the half-angle "alpha" of the Mach cone is 30⁰
emf generated by the coil is 1.57 V
Explanation:
Given details-
Number of turns of wire- 1000 turns
The diameter of the wire coil- 1 cm
Magnetic field (Initial)= 0.10 T
Magnetic Field (Final)=0.30 T
Time=10 ms
The orientation of the axis of the coil= parallel to the field.
We know that EMF of the coil is mathematically represented as –
E=N(ΔФ/Δt)
Where E= emf generated
ΔФ= change inmagnetic flux
Δt= change in time
N= no of turns*area of the coil
Substituting the values of the above variables
=1000*3.14*0.5*10-4
=.0785
E=0.0785(.2/10*10-3)
=1.57 V
Thus, the emf generated is 1.57 V
Answer:
EH buddy use a sparkplug use a drill through a hose im from da bronx
Explanation:
