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Nataly [62]
3 years ago
10

A cassette recorder uses a plug-in transformer to convert 120 V to 12.0 V, with a maximum current output of 200 mA. What is the

current input?
Physics
1 answer:
goldfiish [28.3K]3 years ago
4 0

Answer:

20 mA

Explanation:

We have given that the transformer convert 120 V to 12 V means it is a step down transformer

The turn ratio of the transformer is given as a=\frac{N_1}{N_2}=\frac{V_1}{V_2}=\frac{120}{12}=10 here V_! is primary voltage and V_2 is secondary voltage

As the voltage decreases from primary to secondary so current will increase from primary to secondary

We have given output current that is secondary current so primary that is input current will be less

So input current =\frac{200}{10}=20mA

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Hahahahha ok it’s B or C or it B
5 0
2 years ago
How long will it take to go 150000m traveling at 50km/hr
Tom [10]

Answer:

3000 hurs

Explanation:  just divide 150000 by 50 and get 3000

4 0
3 years ago
What is the strength of the electric field 0.1 mmmm below the center of the bottom surface of the plate
Aleksandr [31]

Complete Question

A thin, horizontal, 12-cm-diameter copper plate is charged to 4.4 nC . Assume that the electrons are uniformly distributed on the surface. What is the strength of the electric field 0.1 mm above the center of the top surface of the plate?

Answer:

The  values is  E =248.2 \  N/C

Explanation:

From the question we are told that

   The  diameter is  d =  12 \  cm  =  0.12 \ m

    The charge  is  Q =  4.4 nC  =  4.4 *10^{-9} \  C

    The  distance from the center  is  k =  0.1 \ mm   =  1*10^{-4} \  m

Generally the radius is mathematically represented as

        r =  \frac{d}{2}

=>     r =  \frac{0.12}{2}

=>       r =  0.06 \  m

Generally electric field is mathematically represented as  

       E =  \frac{Q}{ 2\epsilon_o } [1 - \frac{k}{\sqrt{r^2 +  k^2 } } ]

substituting values  

      E =  \frac{4.4 *10^{-9}}{ 2* (8.85*10^{-12}) } [1 - \frac{(1.00 *10^{-4})}{\sqrt{(0.06)^2 +  (1.0*10^{-4})^2 } } ]

     E =248.2 \  N/C

4 0
3 years ago
A 1.0 meter long rod is held horizontally in the east-west direction on a distant planet and is dropped from a height of 1.23 m.
enyata [817]

Answer:

The induced emf between two end is 34.02 \times 10^{-5} V

Explanation:

Given:

Length of rod l = 1 m

Height h = 1.23 m

Magnetic field B = 6.93 \times 10^{-5} T

For finding induced emf,

  \epsilon = Blv

Where v = velocity of rod,

For finding the velocity of rod.

From kinematics equation,

 v^{2} = v_{o} ^{2} + 2gh

Where v_{o} = initial velocity, g = 9.8 \frac{m}{s^{2} }

   v = \sqrt{2gh}

   v = \sqrt{2 \times 9.8 \times 1.23}

   v = 4.91 \frac{m}{s}

Put the velocity in above equation,

   \epsilon = 6.93 \times 10^{-5} \times 1 \times 4.91

   \epsilon = 34.02 \times 10^{-5} V

Therefore, the induced emf between two end is 34.02 \times 10^{-5} V

6 0
3 years ago
How can you increase the force required to move an object without changing the mass of the object?
Effectus [21]

To increase the force required to move an object without change mass is : To increase the acceleration of the object

<h3>Force acting on an object </h3>

Given that Force = Mass * acceleration

An increase in the acceleration of an object in motion will result in a proportional increase in the Force required to move the object becasue force of an object is directly proportional to the acceleration of an object.

Hence we can conclude that To increase the force required to move an object without change mass is To increase the acceleration of the object.

Learn more about Force : brainly.com/question/25239010

8 0
2 years ago
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