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Genrish500 [490]

3 years ago

12

Electrons in an x-ray tube are accelerated through 144 kV and directed toward a target to produce x-rays. Calculate the power of

the electron beam in this tube if it has a current of 16.0 mA

1 answer:

Katarina [22]3 years ago

7 0

Answer:

Power of the electron beam in the x-ray tube is 2304 W

Explanation:

Given;

voltage of the electron beam in the x-ray tube, V = 144 kV = 144 x 10³ V

current of the electron beam in the x-ray tube, I = 16.0 mA = 16 x 10⁻³ A

Power is given as the product of voltage and current in a circuit.

Power of the electron beam in the x-ray tube, P = V x I

Power of the electron beam in the x-ray tube, P = 144 x 10³ V x 16 x 10⁻³ A

P = 2304 W

Therefore, power of the electron beam in the x-ray tube is 2304 W

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An undiscovered planet, many lightyears from Earth, has one moon in a periodic orbit. This moon takes time TT on average to comp

gulaghasi [49]

Answer:

$a_{cp}=\frac{2\pi^2 d}{T^2}$

Explanation:

The equation for centripetal acceleration is

$a_{cp}=\frac{v^2}{r}$

Where v is the tangential velocity and r the radius of the orbit.

To calculate v we notice that the moon travels the circumference of the orbit $C=2\pi r$ in a time T, so we have

$v=\frac{C}{T} =\frac{2\pi r}{T}$

Putting all together:

$a_{cp}=\frac{v^2}{r}=\frac{(\frac{2\pi r}{T})^2}{r}=\frac{4\pi^2 r}{T^2}$

And since the radius r is d/2, where d is the diameter of the moon's orbit, we can write the formula asked:

$a_{cp}=\frac{4\pi^2 r}{T^2}=\frac{4\pi^2 d}{2T^2}=\frac{2\pi^2 d}{T^2}$

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3 years ago

Two vectors pointing right starting at the same position. Vector A is on top and longer. Vector B is shorter and below.

Oksana_A [137]

Answer:

The displacements are not equal. While their direction is the same, their length (i.e., distance or magnitude) is not. A is longer than B.

Explanation:

correct on e2020- sample respnse

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3 years ago

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What is a disadvantage of using nuclear power to produce electricity?

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3 years ago

Two charges, - Q0 and - 4Q0 are a distance d apart. These two charges are free to move but do not because there is a third charg

TEA [102]

To solve this problem we will use the equilibrium conditions in the Electrostatic Forces. In turn, we will use the concept formulated from Coulomb's laws to determine the intensity of the Forces and make the respective considerations.

Our values for the two charges are:

$q_1 = -Q_0$

$q_2 = -4Q_0$

As a general consideration we will start by determining that they are at a unit distance (1) separated from each other. And considering that both are negative charges, they will be subjected to repulsive force. Said equilibrium compensation will be achieved only by placing a third force between the two.

Let the third charge be $q_3 = +Q$ is placed at a distance x from $q_1$

$F_{1,3} = \frac{k(-Q_0)(+Q)}{x^2}$

The force on $q_3$ due to $q_2$ is

$F_{2,3} = \frac{k(-4Q_0)(+Q)}{1-x^2}$

The condition of equilibrium is

$F_{1,3} = F_{2,3}$

$\frac{k(-Q_0)(+Q)}{x^2}= \frac{k(-4Q_0)(+Q)}{1-x^2}$

$\frac{1}{x^2} = \frac{4}{(1-x)^2}$

$x = 0.331$ from $q_1$

To find the magnitude of $q_3$ we use $F_{1,2} = F_{1,3}$

$\frac{k(-Q_0)(4Q_0)}{1^2}= \frac{k(-Q_0)(Q)}{0.331^2}$

$Q = 0.43Q_0$

The magnitude of the third charge must be 0.43 the first charge $Q_0$

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3 years ago

I’m not sure how to do this

aleksley [76]

Do what?? i know how to do a lot of things to answer but there’s nothing to answer.‍♀️

6 0

4 years ago