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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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Which is not a family of the periodic table?
Vsevolod [243]

Answer:

anions

Explanation:

The other are families of the periodic table.

An anion is a negatively charged ion.

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3 years ago
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A kangaroo can jump over an object 2.46 m high. (a) Calculate its vertical speed (in m/s) when it leaves the ground.
Elenna [48]

Answer:

a) 6.95 m/s

b) 1.42 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

v^2-u^2=2as\\\Rightarrow -u^2=2as-v^2\\\Rightarrow -u^2=2\times -9.81\times 2.46-0^2\\\Rightarrow u=\sqrt{2\times 9.81\times 2.46}\\\Rightarrow u=6.95\ m/s

a) The vertical speed when it leaves the ground. is 6.95 m/s

v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-6.95}{-9.81}\\\Rightarrow t=0.71\ s

Time taken to reach the maximum height is 0.71 seconds

s=ut+\frac{1}{2}at^2\\\Rightarrow 2.46=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{2.46\times 2}{9.81}}\\\Rightarrow t=0.71\ s

Time taken to reach the ground from the maximum height is 0.71 seconds

b) Time it stayed in the air is 0.71+0.71 = 1.42 seconds

3 0
3 years ago
A teacher asks her students to jump off of the ground. Once the students complete the task, she says, "All of you just made Eart
Crazy boy [7]

Answer:

Explained below

Explanation:

A) Newton's first law of motion states that an object will remain at rest or continue in its current state of motion except it is acted upon by another force.

Now using this law, when you jump off the ground, the earth will move a tiny bit and accelerate due to the force applied by the jumping.

B) Newton's 2nd law states that the acceleration of a system is directly proportional to the net external force acting on that system, is in the same direction with it and also inversely proportional to the mass.

In this case, when one jumps, an external force is exerted on the earth and we are told it is directly proportional to the acceleration of the system which in this case it's the earth, then it means that there is some motion by the earth even though you didn't see it move.

C) Newton's third law of motion states that to every action, there is an equal and opposite reaction.

In this case the motion of the jumper will lead to an equal and opposite reaction of the earth.

8 0
2 years ago
An electron and a proton are held on an x axis, with the electron at x = + 1.000 m
mixas84 [53]

Answer:

  r2 = 1 m

therefore the electron that comes with velocity does not reach the origin, it stops when it reaches the position of the electron at x = 1m

Explanation:

For this exercise we must use conservation of energy

the electric potential energy is

          U = k \frac{q_1q_2}{r_{12}}

for the proton at x = -1 m

          U₁ =- k \frac{e^2 }{r+1}

for the electron at x = 1 m

          U₂ = k \frac{e^2 }{r-1}

starting point.

        Em₀ = K + U₁ + U₂

        Em₀ = \frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1}

final point

         Em_f = k e^2 ( -\frac{1}{r_2 +1} + \frac{1}{r_2 -1})

   

energy is conserved

        Em₀ = Em_f

        \frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1} = k e^2 (- \frac{1}{r_2 +1} + \frac{1}{r_2 -1})              

       

        \frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1} = k e²(  \frac{2}{(r_2+1)(r_2-1)} )

we substitute the values

½ 9.1 10⁻³¹ 450 + 9 10⁹ (1.6 10⁻¹⁹)² [ - \frac{1}{20+1} + \frac{1}{20-1} ) = 9 109 (1.6 10-19) ²( \frac{2}{r_2^2 -1} )

          2.0475 10⁻²⁸ + 2.304 10⁻³⁷ (5.0125 10⁻³) = 4.608 10⁻³⁷ ( \frac{1}{r_2^2 -1} )

          2.0475 10⁻²⁸ + 1.1549 10⁻³⁹ = 4.608 10⁻³⁷     \frac{1}{r_2^2 -1}

          \frac{2.0475 \ 10^{-28} }{1.1549 \ 10^{-37} } = \frac{1}{r_2^2 -1}

          r₂² -1 = (4.443 10⁸)⁻¹

           

          r2 = \sqrt{1 + 2.25 10^{-9}}

          r2 = 1 m

therefore the electron that comes with velocity does not reach the origin, it stops when it reaches the position of the electron at x = 1m

4 0
3 years ago
You are riding a bicycle up a gentle hill. It is fairly easy to increase your potential
Dahasolnce [82]
True or false: while riding a bicycle up a gentle hill, it fairly easy to increase your potential energy, but to increase your kinetic energy would ...
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