Answer:
The combination of elements most likely to comprise the circuit are resistor, inductor and capacitor
Explanation:
The impedance of an LCR circuit shown as
Z = √R² + (X↓l - X↓c)²
Z = √R² + (2π∨L - 1/2π∨c)²
Variation of Z with respect to υ is shown in the figure.
As υ increases, Z decreases and so the current increases.
At υ = υ↓r
Z is minimum, current is maximum. Beyond
υ = υ↓r
Z increases and so current decreases.
so the combination of circuit elements that is most suitable to comprise
the circuit is R, L and C.
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Same temperature and difference in air pressure
Answer:
The answer is 4N or B
Explanation:
Just the equation W = F x D.
We have W = 8 J and D = 2 m using algebra ....
8J/2m = F ... F = 4.
Answer:
life (N) of the specimen is 117000 cycles
Explanation:
given data
ultimate strength Su = 120 kpsi
stress amplitude σa = 70 kpsi
solution
we first calculate the endurance limit of specimen Se i.e
Se = 0.5× Su .............1
Se = 0.5 × 120
Se = 60 kpsi
and we know strength of friction f = 0.82
and we take endurance limit Se is = 60 kpsi
so here coefficient value (a) will be
a = 
......................1
put here value and we get
a = 
a = 161.4 kpsi
so coefficient value (b) will be
b = 
b = 
b = −0.0716
so here number of cycle N will be
N = 
put here value and we get
N = 
N = 117000
so life (N) of the specimen is 117000 cycles
<span>Cobalt-60 is undergoing a radioactivity decay.
The formula of the decay is n=N(1/2)</span>∧(T/t).
<span>Where N </span>⇒ original mass of cobalt
<span> n </span>⇒ remaining mass of cobalt after 3 years
T ⇒ decaying period
t ⇒ half-life of cobalt.
So,
0.675 = 1 × 0.5∧(3/t)
log 0.675 = log 0.5∧(3/t)
3/t = log 0.675 ÷log 0.5
3/t= 0.567
t = 3÷0.567
= 5.290626524
the half-life of Cobalt-60 is 5.29 years.
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