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liubo4ka [24]
3 years ago
10

An inductor is connected across an ac source. suppose the frequency of the source is doubled. what happens to the inductive reac

tance of the inductor?
Physics
1 answer:
MaRussiya [10]3 years ago
8 0
<span>If an inductor is connected across an ac source and suppose the frequency of the source is doubled, then t</span>he inductive reactance of the inductor is also doubled. The inductive reactance (XL) is the t<span>he opposition to current flowing through a coil in an AC circuit, the </span>impedance measured in Ohms and can be calculated with the following formula:
XL=2*pi*f*L,
where f is the frequency. So, if the frequency is doubled than also the inductive reactance is doubled.
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Astronaut mark uri is space-traveling from planet x to planet y at a speed of relative to the planets, which are at rest relativ
lyudmila [28]
<span>From the point of view of the astronaut, he travels between planets with a speed of 0.6c. His distance between the planets is less than the other bodies around him and so by applying Lorentz factor, we have 2*</span>√1-0.6² = 1.6 light hours. On the other hand, from the point of view of the other bodies, time for them is slower. For the bodies, they have to wait for about 1/0.6 = 1.67 light hours while for him it is 1/(0.8) = 1.25 light hours. The remaining distance for the astronaut would be 1.67 - 1.25 = 0.42 light hours. And then, light travels in all frames and so the astronaut will see that the flash from the second planet after 0.42 light hours and from the 1.25 light hours is, 1.25 - 0.42 = 0.83 light hours or 49.8 minutes.
5 0
3 years ago
Physics. Need help. Brainlieast answer for most/ all of the answers answered
Mumz [18]

<u>ALL of the following work assumes NO AIR RESISTANCE:</u>

1). an object moving under the influence of only gravity, and not in orbit;  its horizontal velocity is constant, and its vertical motion is accelerated downward at 9.8 m/s²

2). a parabola

3). Horizontal: velocity is constant, acceleration is zero. . . . Vertical: acceleration is 9.8 m/s² downward, velocity depends on whether it was launched, thrown up, thrown down, dropped, etc.

4). a). the one that was thrown horizontally; b). both  hit the ground at the same time; c). both hit the ground with the same vertical velocity

5). a). zero; b). zero; c). gravity ... 9.8 m/s² down; d). 3.06 seconds; e). 4.38 m/s; f). 30 m/s g). no; gravity has no effect on horizontal motion

6). a). 1.8 seconds;   b). 13.1 meters;   c). 17.6 m/s down;   d). 7.3 m/s; gravity has no effect on horizontal motion

7). 45 m/s

8). without air resistance, the ball is traveling horizontally at 13 km/hr, and it lands back in your hand

9). a). 4.49 m/s;  b). 29.7 m/s

10). 7.24 meters

11).  700 meters

12).  A). 103.7 meters ( ! she's in big trouble ! );     B).  17.5 meters

3 0
3 years ago
The angular speed of the rotor in a centrifuge increases from 420 to 1420 rad/s in a time of 5.00 s. (a) Obtain the angle throug
9966 [12]

Answer:

a) θ = 2500 radians

b) α = 200 rad/s²

Explanation:

Using equations of motion,

θ = (w - w₀)t/2

θ = angle turned through = ?

w = final angular velocity = 1420 rad/s

w₀ = initial angular velocity = 420

t = time taken = 5s

θ = (1420 - 420) × 5/2 = 2500 rads

Again,

w = w₀ + αt

α = angular accelaration = ?

1420 = 420 + 5α

α = 1000/5 = 200 rad/s²

6 0
4 years ago
Read 2 more answers
Two charged particles are located on the x axis. The first is a charge 1Q at x 5 2a. The second is an unknown charge located at
sergejj [24]

Answer:

Q_2 = +/- 295.75*Q

Explanation:

Given:

- The charge of the first particle Q_1 = +Q

- The second charge = Q_2

- The position of first charge x_1 = 2a

- The position of the second charge x_2 = 13a

- The net Electric Field produced at origin is E_net = 2kQ / a^2

Find:

Explain how many values are possible for the unknown charge and find the possible values.

Solution:

- The Electric Field due to a charge is given by:

                               E = k*Q / r^2

Where, k: Coulomb's Constant

            Q: The charge of particle

            r: The distance from source

- The Electric Field due to charge 1:

                               E_1 = k*Q_1 / r^2

                               E_1 = k*Q / (2*a)^2

                               E_1 = k*Q / 4*a^2

- The Electric Field due to charge 2:

                               E_2 = k*Q_2 / r^2

                               E_2 = k*Q_2 / (13*a)^2

                               E_2 = +/- k*Q_2 / 169*a^2

- The two possible values of charge Q_2 can either be + or -. The Net Electric Field can be given as:

                               E_net = E_1 + E_2

                               2kQ / a^2 = k*Q_1 / 4*a^2 +/- k*Q_2 / 169*a^2

- The two equations are as follows:

        1:                   2kQ / a^2 = k*Q / 4*a^2 + k*Q_2 / 169*a^2

                               2Q = Q / 4 + Q_2 / 169

                               Q_2 = 295.75*Q

        2:                    2kQ / a^2 = k*Q / 4*a^2 - k*Q_2 / 169*a^2

                               2Q = Q / 4 - Q_2 / 169

                               Q_2 = -295.75*Q

- The two possible values corresponds to positive and negative charge Q_2.

7 0
3 years ago
Which of the following explains why international travelers need to use special adapters when plugging a device designed to oper
Darina [25.2K]

Answer:

The answer is: The increased voltage causes an increase in power usage, and the device will over-heat.

Explanation:

First, we must consider the variables of the electrical system that will allow us to respond. In this case, power, current and voltage, which are related by

P=VI

Where P=Power, V=Voltage, I=Current.

In the equation it can be observed that power is directly proportional to the system voltage. Thus, if the voltage increases as in this case, the power will also increase, which overheats the device and can cause damage to it.

8 0
3 years ago
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