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OLEGan [10]
4 years ago
5

What is the voltage in a circuit that has a current of 10.0 amps and a resistance of 28.5 ohms?

Physics
2 answers:
patriot [66]4 years ago
5 0
<span>Ohm's law deals with the relation between voltage and current in an ideal conductor. It states that: Potential difference across a conductor is proportional to the current that pass through it. It is expressed as V=IR.

V = 10.0 A (28.5 ohms) = 285 V </span>
Stella [2.4K]4 years ago
3 0

V  =  I · R  =  (10 A) · (28.5 Ω)  =  285 volts .
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1.) This person
valina [46]

Answer: 0.78m/s^2

Explanation:

Force is the product of mass of an object and its acceleration.

i.e Force = Mass x Acceleration

In this case,

Force applied on cart = 35N

Mass of cart and contents = 45Kg

Acceleration = ?

Then, Force = Mass x Acceleration

35N = 45Kg x Acceleration

Acceleration = (35N/45kg)

Acceleration = 0.78 m/s^2 (The unit of acceleration is metre per second square)

Thus, the shopping cart moved with an accelerations of 0.78m/s^2

8 0
3 years ago
A frictionless pendulum of length of 3 m swings with an amplitude of 10o. At its maximum displacement, the potential energy of t
olchik [2.2K]
<h2>Option A is the correct answer.</h2>

Explanation:

For a simple pendulum we have at any position

             Total energy = Constant        

             Kinetic energy + Potential energy = Constant

At maximum displacement of pendulum, velocity is zero, hence kinetic energy is zero.

At maximum displacement the pendulum only have potential energy.

Given that maximum potential energy is 10 J.

 That is at any position                

                     Kinetic energy + Potential energy = 10 J

Now we need to find kinetic energy when potential energy is 5 J.

                     Kinetic energy + 5 J = 10 J

                     Kinetic energy = 5 J

Option A is the correct answer.

6 0
3 years ago
A beam of protons is directed in a straight line along the positive zz ‑direction through a region of space in which there are c
Zinaida [17]

Answer:

The magnitude is B = \frac{450}{7.9* 10^5}

The direction is  the positive x axis

Explanation:

From the question we are told that

   The  electric field is  E = 450 V/m in the negative y ‑direction

   The speed of the proton is  v= 7.9* 10^5\  m/s in the positive z direction

Generally the overall force acting on  the proton is mathematical represented as

              F_E =  q(\vec E + \vec v  * \vec B)

Now for the beam of protons to continue un-deflected along its straight-line trajectory then  F_E =0

So

          0  =  q( E (-y) + v(z)  * \vec B)

=>      E\^y = v \^ z  * \vec B

Generally from unit vector cross product vector multiplication

         \^ z  \ *  \  \^ x  =  \^  y

So the direction of  B (magnetic field must be in the positive x -axis )

So

       E\^y = v \^ z  *  B\^ x

=>     E\^y = vB ( \^ z  *  \^ x)

=>     E\^y = vB ( \^y)        

=>      E = vB  

=>      B = \frac{450}{7.9* 10^5}

=>      B =  0.0005696 \ T  

   

   

6 0
4 years ago
The armature windings of a dc motor have a resistance of 3.0 Ω. The motor is connected to a 120V line, and when the motor reach
svlad2 [7]

Answer:

0.6 A

Explanation:

As the motor gears towards full speed, the voltage of the circuit tends to become the difference that exists between the line voltage and the that of the back emf. Remembering Ohm's law, we then apply it to get the final, lower current that is based on the reduced voltage Ef. We use the provided resistance in the question, that is 3 ohms.

Ef = (120 V) - (117 V) = 3 V

If = Ef/R = (3 V) / (5.0 Ω) = 0.6 A

6 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
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