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tatyana61 [14]
4 years ago
12

Identify each process labeled in the diagram

Physics
1 answer:
tatuchka [14]4 years ago
5 0
A evorpation b transportation c      accumlation                     d condesation   e percipation 
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At resonance, what is impedance of a series RLC circuit?
Elis [28]

Answer:

c) Equal to R

b) the width of the resonance

Explanation: In general the impedance Z of an electrc circuit is:

Z = R + jX

Now when the circuit is capacitive, the above mentioned relation become  Z = R + 1/jwc  where ( w = 2πf  )

And when the circuit is inductive Z becomes

Z  =  R + j wl

Resonance  condition implies that the reactance created by capacitors are equal at the inductances produced by inductors, in other words the circuit will behaves as it were resistive.

The impedance will be equal to R

The Q factor is[

Q = X/R  in which X is the module of either the capacitive or inductive reactance.

Q has an inverse relation with the band width.  

3 0
4 years ago
A cyclist is travelling at 4.0 m/s. She speeds up to 16 m/s in a time of 5.6 s. Calculate her acceleration.​
adell [148]

Answer:

Initial velocity, u=15m/s

Initial velocity, u=15m/sFinal velocity, v=0m/s

Initial velocity, u=15m/sFinal velocity, v=0m/sDistance, s=18m

Initial velocity, u=15m/sFinal velocity, v=0m/sDistance, s=18mAcceleration, a=?

Initial velocity, u=15m/sFinal velocity, v=0m/sDistance, s=18mAcceleration, a=?Using relation, v

Initial velocity, u=15m/sFinal velocity, v=0m/sDistance, s=18mAcceleration, a=?Using relation, v 2

Initial velocity, u=15m/sFinal velocity, v=0m/sDistance, s=18mAcceleration, a=?Using relation, v 2 −u

Initial velocity, u=15m/sFinal velocity, v=0m/sDistance, s=18mAcceleration, a=?Using relation, v 2 −u 2

Initial velocity, u=15m/sFinal velocity, v=0m/sDistance, s=18mAcceleration, a=?Using relation, v 2 −u 2 =2as

Initial velocity, u=15m/sFinal velocity, v=0m/sDistance, s=18mAcceleration, a=?Using relation, v 2 −u 2 =2as0

Initial velocity, u=15m/sFinal velocity, v=0m/sDistance, s=18mAcceleration, a=?Using relation, v 2 −u 2 =2as0 2

Initial velocity, u=15m/sFinal velocity, v=0m/sDistance, s=18mAcceleration, a=?Using relation, v 2 −u 2 =2as0 2 −(15)

Initial velocity, u=15m/sFinal velocity, v=0m/sDistance, s=18mAcceleration, a=?Using relation, v 2 −u 2 =2as0 2 −(15) 2

Initial velocity, u=15m/sFinal velocity, v=0m/sDistance, s=18mAcceleration, a=?Using relation, v 2 −u 2 =2as0 2 −(15) 2 =2a×18

Initial velocity, u=15m/sFinal velocity, v=0m/sDistance, s=18mAcceleration, a=?Using relation, v 2 −u 2 =2as0 2 −(15) 2 =2a×18−225=36a

Initial velocity, u=15m/sFinal velocity, v=0m/sDistance, s=18mAcceleration, a=?Using relation, v 2 −u 2 =2as0 2 −(15) 2 =2a×18−225=36aa=

Initial velocity, u=15m/sFinal velocity, v=0m/sDistance, s=18mAcceleration, a=?Using relation, v 2 −u 2 =2as0 2 −(15) 2 =2a×18−225=36aa= 36

Initial velocity, u=15m/sFinal velocity, v=0m/sDistance, s=18mAcceleration, a=?Using relation, v 2 −u 2 =2as0 2 −(15) 2 =2a×18−225=36aa= 36−225

Initial velocity, u=15m/sFinal velocity, v=0m/sDistance, s=18mAcceleration, a=?Using relation, v 2 −u 2 =2as0 2 −(15) 2 =2a×18−225=36aa= 36−225

Initial velocity, u=15m/sFinal velocity, v=0m/sDistance, s=18mAcceleration, a=?Using relation, v 2 −u 2 =2as0 2 −(15) 2 =2a×18−225=36aa= 36−225 =−6.25m/s

Initial velocity, u=15m/sFinal velocity, v=0m/sDistance, s=18mAcceleration, a=?Using relation, v 2 −u 2 =2as0 2 −(15) 2 =2a×18−225=36aa= 36−225 =−6.25m/s 2

Initial velocity, u=15m/sFinal velocity, v=0m/sDistance, s=18mAcceleration, a=?Using relation, v 2 −u 2 =2as0 2 −(15) 2 =2a×18−225=36aa= 36−225 =−6.25m/s 2

Initial velocity, u=15m/sFinal velocity, v=0m/sDistance, s=18mAcceleration, a=?Using relation, v 2 −u 2 =2as0 2 −(15) 2 =2a×18−225=36aa= 36−225 =−6.25m/s 2 So, deceleration is 6.25m/s

Initial velocity, u=15m/sFinal velocity, v=0m/sDistance, s=18mAcceleration, a=?Using relation, v 2 −u 2 =2as0 2 −(15) 2 =2a×18−225=36aa= 36−225 =−6.25m/s 2 So, deceleration is 6.25m/s 2

Initial velocity, u=15m/sFinal velocity, v=0m/sDistance, s=18mAcceleration, a=?Using relation, v 2 −u 2 =2as0 2 −(15) 2 =2a×18−225=36aa= 36−225 =−6.25m/s 2 So, deceleration is 6.25m/s 2 .

5 0
2 years ago
Read 2 more answers
What is shown by the pattern of iron filings around these two magnets?
nadya68 [22]

Answer:c

Explanation:

8 0
3 years ago
Read 2 more answers
IF U ANSWER I WILL PUT U AS BRAINIEST ANSWER AND GIVE U A THANK U
Bingel [31]
<span>
At the Earth's surface, warm air expands and rises, creating
what is known as an area of low pressure.

Cold air is dense and sinks to the surface to create what is
known as an area of high pressure.</span>


8 0
4 years ago
Read 2 more answers
Un automovil de 900 kg toma una curva de radio de 40 m con una rapidez constante de 50 km/h. Cual es la fuerza neta necesaria pa
Llana [10]

Answer:

Fc = 4340,93 Newton

Explanation:

Dados los siguientes datos;

Masa = 900 kg

Velocidad, V = 50 km/h a metros por segundo = (50 * 1000)/(60 * 60) = 50000/3600 = 13,89 m/s

Radio, r = 40 m

Para encontrar la fuerza centrípeta;

Fc = mv² / r

Fc = (900 * 13,89²)/40

Fc = (900 * 192,93)/40

Fc = 173637/40

Fc = 4340,93 Newton

4 0
3 years ago
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