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Alex_Xolod [135]
2 years ago
15

Which property does not apply to a motor? A. It uses an electromagnet. B. The direction of current changes twice during the rota

tion of the coil. C. It converts electrical energy into mechanical energy. D. It converts mechanical energy into electrical energy. E. It uses an outside source of electrical c
Physics
2 answers:
vitfil [10]2 years ago
5 0

Answer:

D. It converts mechanical energy into electrical energy.

Explanation:

Ok lets disregard the whole "oh yay physics explainations" and look at this with just a little common sense and use some elimination and logic.

C and D are pretty similar so just look at those two.

C. It converts electrical energy into mechanical energy.

D. It converts mechanical energy into electrical energy.

Think about what a motor is in something like a fan. How is it powered? ELECTRICITY! So it must convert electricity to mechnical neergy so therefore D must be incorrect.

Rufina [12.5K]2 years ago
3 0

Answer:  

     

Explanation:

A motor converts electrical energy to mechanical energy. A current carrying coil is placed between electromagnets. A magnetic force is exerted on the coil which makes it rotate. The direction of current changes twice during the rotation of the coil. An outside source of electric current is used.

Thus, the wrong statement is D. It converts mechanical energy into electrical energy. It is the generator which converts mechanical energy into electrical energy.

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The train passes point A with a speed of 30 m/s and begins to decrease its speed at a constant rate of at = - 0.25 m/s^2. Determ
prisoha [69]

Explanation:

At point B, the velocity speed of the train is as follows.

          \nu^{2}_{B} = \nu^{2}_{A} + 2a_{t} (s_{B} - s_{A})

                           = (30)^{2} + 2(-0.25(412 - 0))

                           = 26.34 m/s

Now, we will calculate the first derivative of the equation of train.

          y = 200 e^{\frac{x}{1000}}

      \frac{dy}{dx} = 0.2 e^{\frac{x}{1000}}

Now, second derivative of the train is calculated as follows.

         \frac{dy}{dx} = 0.2 e^{\frac{x}{1000}}      

       \frac{d^{2}y}{dx^{2}} = 0.2 (10^{-3}) e^{\frac{x}{1000}}    

Radius of curvature of the train is as follows.  

   \rho = \frac{[1 + (\frac{dy}{dx})^{2}]^{\frac{3}{2}}}{\frac{d^{2}y}{dx^{2}}}

               = \frac{[1 + 0.2e^{\frac{400}{1000}}^{2}]^{\frac{3}{2}}}{0.2(10^{-3})e^{\frac{400}{1000}}}

              = 3808.96 m

Now, we will calculate the normal component of the train as follows.

            a_{n} = \frac{\nu^{2}_{B}}{\rho}

                        = \frac{(26.34)^{2}}{3808.96}

                        = 0.1822 m/s^{2}

The magnitude of acceleration of train is calculated as follows.

            a = \sqrt{(a_{t})^{2} + (a_{n})^{2}}

               = \sqrt{(-0.25)^{2} + (0.1822)^{2}}

              = 0.309 m/s^{2}

Thus, we can conclude that magnitude of the acceleration of the train when it reaches point B, where sAB = 412 m is 0.309 m/s^{2}.

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3 years ago
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