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3 years ago

14

Which type of motor is shown below?

2 answers:

Viefleur [7K]3 years ago

5 0

Answer:

AC motor........A

Explanation:

Plato, an AC motor is much simpler than a DC motor. As alternating current passes through the coil, it becomes magnetized. Since the current is alternating, the north and south ends keep switching.

azamat3 years ago

3 0

Answer:

magnetic

Explanation:

PLATO

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A circuit contains two resistors in series. The voltage drop across the first is 10 V. The voltage drop across the second is als

iragen [17]

The voltage provided by the power supply = 20 V

<h3>Resistors in series </h3>

The Voltage across resistors in series in a circuit is equal to the sum of the voltage drops across each resistor.

Therefore for two resistors in series the total voltage provided by the power supply is equivalent to the summation of the voltage drops acroos each resistor ( i.e 10 V + 10 V = 20 V )

Hence we can conclude that the voltage provided by the power supply is 20 V

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2 years ago

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Ber [7]

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4 years ago

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Calculate the work done by a 47 n force pushing a 0.025 kg pencil 0.25 m against a force of 23.

inna [77]

Fnet=fa-fr
=47-23
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3 years ago

Scientists want to place a 3400.0 kg satellite in orbit around Mars. They plan to have the satellite orbit at a speed of 2480.0

BaLLatris [955]

Answer:

Part a)

$r = 6.96 \times 10^6 m$

Part b)

$F_g = 3.004 \times 10^3 N$

Part c)

$a = 0.88 m/s^2$

Part d)

$v = \frac{2480}{2} = 1240 m/s$

Explanation:

Part a)

As we know that the orbital speed of the satellite is given as

$v = 2480 m/s$

now we will have

$v = \sqrt{\frac{GM_{mars}}{r}}$

now we have

$M_{mars} = 6.4191 \times 10^{23} kg$

$R_{mars} = 3.397 \times 10^6 m$

now we have

$2480 = \sqrt{\frac{(6.67 \times 10^{-11})(6.4191 \times 10^{23})}{r}}$

$r = 6.96 \times 10^6 m$

Part b)

Here force between mars and satellite is the gravitational attraction force which is given as

$F_g = \frac{GM_{mars} m}{r^2}$

$F_g = \frac{(6.67\times 10^{-11})(6.4191 \times 10^{23})(3400)}{(6.96\times 10^6)^2}$

$F_g = 3.004 \times 10^3 N$

Part c)

Acceleration of satellite is the ratio of gravitational force and mass of the satellite

$a = \frac{F_g}{m}$

$a = \frac{3004}{3400}$

$a = 0.88 m/s^2$

Part d)

As we know by III law of kepler

$\frac{T_1^2}{T_2^2} = \frac{r_1^3}{r_2^3}$

here we know that T2 = 8 T1

$(\frac{1}{8})^2= \frac{r_1^3}{r_2^3}$

$\frac{r_1}{r_2} = (\frac{1}{2})^2$

so we have

$r_2 = 4r_1$

as we know that speed is given as

$v = \sqrt{\frac{GM}{r}}$

so we can say since radius is orbit becomes 4 times so the orbital speed must be half

$v = \frac{2480}{2} = 1240 m/s$

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spayn [35]

Answer:

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