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Kruka [31]
2 years ago
10

What is the main difference between a generator and an electric motor?

Physics
2 answers:
Amanda [17]2 years ago
7 0
I got B
Hoped it helped
Novosadov [1.4K]2 years ago
4 0

Answer:

A

Explanation:

an electric motor is a machine that converts electrical energy to mechanical energy. a generator is a machine that converts mechanical energy to electrical energy

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Two horizontal metal plates, each 10.0 cm square, are aligned 1.00 cm apart with one above the other. They are given equal-magni
Ber [7]

Answer:

a) motion is PARABOLIC, b) positive particle is accelerated towards the negative plate,  c)  x = 6.19 10⁹ m

Explanation:

This exercise looks at the motion of a positively charged particle in an electric field.

a) Since the field is vertical the acceleration in this direction is

            F = m a

the electric force is

           F = q E

we substitute

          q E = m a

           a = qE / m

the mass of the particle is m = 2.00 10-16 kg

           a = 1.6 10⁻¹⁹ 2.02 10³ / 2.00 10⁻¹⁶ kg

           a = 1,616 m / s²

           

on the x-axis there are no relationships because there are no forces.

Since the particle has velocities in both axes, its motion is PARABOLIC,

b) the positive particle is accelerated towards the negative plate,

The field is descending, for which the event is down

c) where  hit the particle on the x-axis

they indicate that the particle leaves the center of the negative plate, for which we will fix our reference system at this point.

Let's find the components of the initial velocity.

           sin θ = v_{oy} / v

           cos θ = v₀ₓ / v

           v_{oy} = v₀ sin θ

           v₀ₓ = v₀ cos θ

           v_{oy) = 1.02 10⁵ sin 37 = 0.6139 10⁵ m / s

           v₀ₓ = 1.02 10⁵ cos 37 = 0.8146 10⁵ m / s

Let's find the time it takes to hit the negative plate

            y = y₀I + v_{oy} t + ½ a and t2

in this case the positions are y = y₀ = 0 and the accelerations

a = - 1,616m/s2,

we substitute

            0 = 0 + v_{oy} t - ½ a_y t²

            v_{oy}= ½ a_y t

            t = 2v_{oy} / a_y

let's calculate

           t = 2 0.6139 10⁵ / 1.616

           t = 7.597 10⁴s

in this time the particle travels a horizontal distance

           x = v₀ₓ t

           x = 0.8145 10⁵ 7.597 10⁴

           x = 6.19 10⁹ m

the particle falls off the plate

4 0
2 years ago
A child swings on a playground swing. How many times does the child swing through the swing's equilibrium position during the co
Karolina [17]
<h2>The child swing through the swing's equilibrium position 6 times during the course of 3 periods.</h2>

Explanation:

One period means time taken to complete one revolution.

In case of swings in one period time it travels the same position through two times.

Here we need to find how many times does the child swing through the swing's equilibrium position during the course of 3 period(s) of motion.

For 1 period = 2 times

For 3 periods = 3 x For 1 period

For 3 periods = 3 x 2 times

For 3 periods = 6 times

The child swing through the swing's equilibrium position 6 times during the course of 3 periods.

3 0
3 years ago
Si un ciclista se mueve a una velocidad de 5 m/s y acelera 1 m/s2, a los 10 segundos su velocidad será
MrRa [10]

Answer:

A los 10 segundos su velocidad será 15 \frac{m}{s}

Explanation:

La aceleración de un objeto es una magnitud que indica cómo cambia la velocidad del objeto en una unidad de tiempo.

En otras palabras, la aceleración relaciona los cambios de la velocidad con el tiempo en el que se producen, es decir que mide cómo de rápidos son los cambios de velocidad:

  • Una aceleración grande significa que la velocidad cambia rápidamente.
  • Una aceleración pequeña significa que la velocidad cambia lentamente.
  • Una aceleración cero significa que la velocidad no cambia.

La aceleración "a" puede ser calculada mediante la expresión:

a=\frac{vfinal - vinicial}{tiempo}

En este caso:

  • a= 1 \frac{m}{s^{2} }
  • vfinal= ?
  • vinicial= 5 \frac{m}{s}
  • tiempo= 10 s

Reemplazando:

1\frac{m}{s^{2} }=\frac{vfinal - 5\frac{m}{s} }{10 s}

Resolviendo se obtiene:

1 \frac{m}{s^{2} } *10 s= vfinal - 5 \frac{m}{s}

10 \frac{m}{s} = vfinal - 5 \frac{m}{s}

10 \frac{m}{s} + 5 \frac{m}{s} = vfinal

15 \frac{m}{s} = vfinal

<u><em>A los 10 segundos su velocidad será 15 </em></u>\frac{m}{s}<u><em></em></u>

5 0
2 years ago
On a highway curve with a radius of 46 meters, the maximum force of static friction that can act on a 1,200 kg car going around
Mekhanik [1.2K]

Answer:

v\approx 16.956\,\frac{m}{s}

Explanation:

The motion of the vehicule on a highway curve can be modelled by the following equation of equilibrium:

\Sigma F = f = m\cdot \frac{v^{2}}{R}

The maximum speed is:

v = \sqrt{\frac{f\cdot R}{m} }

v = \sqrt{\frac{(7500\,N)\cdot (46\,m)}{1200\,kg} }

v\approx 16.956\,\frac{m}{s}

7 0
3 years ago
For an object with a given mass on Earth, calculate the weight of the object with the mass equal in magnitude to the number repr
leonid [27]

<u>Answer:</u> The weight of the object is 29.4 N

<u>Explanation:</u>

To calculate the weight of the object, we use the equation:

W=m\times g

where,

m = mass of the object = 3 kg

g = acceleration due to gravity = 9.8m/s^2

Putting values in above equation, we get:

W=3kg\times 9.8m/s^2\\\\W=29.4N

Hence, the weight of the object is 29.4 N

6 0
3 years ago
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