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muminat
2 years ago
10

the inductive reactance of a 20.0-mh inductor at a certain frequency is 120 ω. what is the frequency in hz?

Physics
2 answers:
7nadin3 [17]2 years ago
7 0

Answer:

{ \rm{X _{L}} = 2\pi fL}

XL is the inductive reactance

f is frequency

L is the inductance

{ \rm{f =  \frac{{X _{L}}}{2\pi} }} \\  \\ { \rm{f =  \frac{120 }{2\pi} }} \\  \\{ \rm{frequency = 19.1 \: hz}}

QveST [7]2 years ago
6 0

The inductive reactance of a 20.0-mh inductor at a certain frequency is 120 ω. 955hz is the frequency in hz

In physics, frequency is the number of waves that pass a fixed point in a unit of time as well as the number of cycles or vibrations that a body in periodic motion experiences in a unit of time.

How frequently something occurs is what is meant by the word frequency. A person blinking their eyelids 47 times in a minute is an example of frequency. the quality or state of happening frequently.

Frequency is a numerical value that expresses how frequently a specific item appears in the supplied data set. Grouped frequency distribution and Ungrouped In physics, frequency is a numerical value that expresses how frequently a specific item appears in the supplied data set. Distribution are the two different types of frequency tables.

To learn more about frequency please visit brainly.com/question/5102661
#SPJ4

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A small object moves along the x-axis with acceleration ax(t) = −(0.0320m/s3)(15.0s−t). At t = 0 the object is at x = -14.0 m an
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Complete Question

A small object moves along the x-axis with acceleration ax(t) = −(0.0320m/s3)(15.0s−t)−(0.0320m/s3)(15.0s−t). At t = 0 the object is at x = -14.0 m and has velocity v0x = 7.10 m/s. What is the x-coordinate of the object when t = 10.0 s?

Answer:

The position of the object at t = 10s is  X  =  38.3 \  m

Explanation:

From the question we are told that

The acceleration along the x axis is  a_{x}t  =  -(0.0320\ m/s^3)(15.0 s- t)- (0.0320\ m/s^3)

  The position of the object at t = 0 is  x = -14.0 m

  The velocity at t = 0 s is  v_{0}x = 7.10 m/s

Generally from the equation for acceleration along x axis we have that

     a_x = \frac{dV_{x}}{dt}  = -0.032 (15- t)

=>   \int\limits  {dV_{x}} \, = \int\limits  {-0.032(15- t)} \, dt

=>   V_{x} = -0.032 [15t - \frac{t^2 }{2} ]+ K_1

At  t =0  s   and  v_{0}x = 7.10 m/s

=>   7.10  = -0.032 [15(0) - \frac{(0)^2 }{2} ]+ K_1

=>   K_1 = 7.10      

So  

      \frac{dX}{dt}  = -0.032 [15t - \frac{t^2 }{2} ]+ K_1

=>  \int\limits dX  = \int\limits [-0.032 [15t - \frac{t^2 }{2} ]+ K_1] }{dt}

=>  X  =  -0.032 [ 15\frac{t^2}{2}  - \frac{t^3 }{6} ]+ K_1t +K_2

At  t =0  s   and   x = -14.0 m

  -14  =  -0.032 [ 15\frac{0^2}{2}  - \frac{0^3 }{6} ]+ K_1(0) +K_2

=>   K_2 = -14

So

     X  =  -0.032 [ 15\frac{t^2}{2}  - \frac{t^3 }{6} ]+ 7.10 t -14

At  t = 10.0 s

      X  =  -0.032 [ 15\frac{10^2}{2}  - \frac{10^3 }{6} ]+ 7.10 (10) -14

=>   X  =  38.3 \  m

             

     

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3 years ago
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