Answer:
Check the explanation
Explanation:
This is the step by step explanation to the above question:
![v_i = v [ f_L *(v - v_b) - f_s*(v + v_b)] / [f_L * (v - v_b) + f_s*(v +v_b)]](https://tex.z-dn.net/?f=v_i%20%3D%20v%20%5B%20f_L%20%2A%28v%20-%20v_b%29%20-%20f_s%2A%28v%20%2B%20v_b%29%5D%20%2F%20%5Bf_L%20%2A%20%28v%20-%20v_b%29%20%2B%20f_s%2A%28v%20%2Bv_b%29%5D)
= v * (83.1 * (v-4.3) - 80.7 ( v+4.3))/ [83.1 *(v - 4.3) + 80.7*(v + 4.3)]
v = 344 m/s
vi = 344 * ( 83.1* (344-4.3) - 80.7*(344+4.3) ) / (83.1 *(344 - 4.3) + 80.7*(344 + 4.3))
= 0.74 m/s
Answer: I = 111.69 pA
Explanation: The hall effect is all about the fact that when a semiconductor is placed perpendicularly to a magnetic field, a voltage is generated which could be measured at right angle to the current path. This voltage is known as the hall voltage.
The hall voltage of a semiconductor sensor is given below as
V = I×B/qnd
Where V = hall voltage = 1.5mV =1.5/1000=0.0015V
I = current =?,
n= concentration of charge (electron density) = 5.8×10^20cm^-3 = 5.8×10^20/(100)³ = 5.8×10^14 m^-3
q = magnitude of an electronic charge=1.609×10^-19c
B = strength of magnetic field = 5T
d = thickness of sensor = 0.8mm = 0.0008m
By slotting in the parameters, we have that
0.0015 = I × 5/5.8×10^14 × 1.609×10^-19×0.0008
0.0015 = I×5/7.446×10^-8
I = (0.0015 × 7.446×10^-8)/5
I = 111.69*10^(-12)
I = 111.69 pA
A and B are equivalent. That's one way instruments are often grouped. (the "sopranos", the "altos", the "bass")
C is another way instruments are often grouped; (the "woods", the "brass")
D is another way instruments are often grouped; (the "strings", the "percussions")
Answer:
E = 10⁵ J
Explanation:
given,
Power, P = 100 TW
= 100 x 10¹² W
time, t = 1 ns
= 1 x 10⁻⁹ s
The energy of a single pulse is:-
Energy = Power x time
E = P t
E = 100 x 10¹² x 1 x 10⁻⁹
E = 10⁵ J
The energy contained in a single pulse is equal to 10⁵ J
